1.1 ATOMIC STRUCTURE
Ionisation Energy
Section A
For each of the questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.
1. Which one of the following has the largest first ionisation energy?
A carbon
B fluorine
C neon
D nitrogen
2. The second ionisation energy for calcium is represented by the equation
A Ca(s) Ca2+ (g) + 2e-
B Ca(g) Ca 2+ (g) + 2e-
C Ca+(s) Ca2+ (g) + e-
D Ca+(g) Ca2+ (g) + e-
3.
4. The first six ionisation energies of an element Y are as follows:
786, 1580, 3230, 4360, 16 000, 20 000 kJ mol-1
Which one of the following statements is consistent with these data.
A Y is in Group I of the Periodic Table.
B The outer electronic configuration of an atom of Y is ns2np2
C Y forms a chloride with the molecular formula YCl
D The chemistry of Y and its compounds is similar to that of aluminium and its compounds.
5. In which one of the following series of elements is there the smallest variation in the value of the first ionisation energy?
A Li to F
B F to I
C K to Br
D Sc to Zn
6. Which one of the following processes requires the largest amount of energy?
A He (g) He+ (g) + e-
B Ne (g) Ne+ (g) + e-
C Na (g) Na+ (g) + e-
D Ca (g) Ca+ (g) + e-
7. Which one of the following statements is correct?
A The first ionisation energy of oxygen is greater than that of nitrogen.
B The second ionisation energy is always greater than the first ionisation energy for a chosen element.
C The first ionisation energy of an element. M. is the energy associated with the process M(s) M+ (aq) + e-
D Ionisation energy increases with increasing atomic number.
8. Which one of the following sets of ionisation energies corresponds to an element in Group II?
1st 2nd 3rd 4th 5th
A 580 1800 2700 11600 14800
B 1520 2700 3900 5800 7200
C 790 1600 3200 4400 16100
D 740 1500 7700 10500 13600
9. What frequency of radiation in Hz is required to ionise helium?
He He+ + e- ΔH = +2370 kJ mol-1
A 1.69 x 10-16
B 9.48 x 10-10
C 1.05 x 109
D 5.93 x 10
10.The first ionisation energy for carbon is greater than it is for sodium. One of the factors responsible is that the
A nuclear charge on carbon is greater
B shielding provided by the inner quantum shells of sodium is greater
C number of electrons in the outer quantum shell of carbon is greater than in sodium
D outer quantum shell is further from the nucleus in carbon atoms than in sodium atoms
11. The ground state electronic configurations of five elements are shown below. For which element would you expect the value of the first ionisation energy to be the greatest?
1s 2s 2p
A
B
C
D
12.An element has successive ionisation energies of 99, 1800, 14800 and 21000 kJ mol-1. To which group of the Periodic Table does the element belong?
A I
B II
C III
D IV
13. The first four ionisation energies of an element Z are 738, 1451, 7733 and 10541 kJ mol-1. Which one of the following ions is most likely to be formed when Z reacts with fluorine?
A Z+
B Z2+
C Z3+
D Z4+
14. The graph represents the variation in the first ionisation energy with atomic number.
The elements indicated by letters P, Q, R and S are all
A alkali metals
B halogens
C noble gases
D transition metals
15. The graph below shows the first six successive ionisation energies, Ie, of an element. In which Group of the Periodic Table is the element found?
A Group 11
B Group IV
C Group V
D Group VI
SECTION B
Answer all questions in the spaces provided
1. Here is some information concerning element X, of atomic number 31. In its natural state, it consists of a mixture of two isotopes XA and XB.
Isotope
Isotopic mass
Percentage abundance
XA
69.0
60.2
XB
71.0
39.8
Its first four ionisation energies are 580, 2000, 3000 and 6200 kJ mol-1.
(a) Calculate the value for the relative atomic mass of X correct to three significant figures.
Ar (X) = (69.0 x 60.2 + 71.0 x 39.8)/100 = 69.8
[3]
(b) Write down the electronic configuration of element X in its ground state.
X = 1s22s22p63s23p63d104s24p1 [1]
(c) In which group of the Periodic Table is X placed?
Group III [1]
(d) Explain why:
(i) The difference between the first and second ionisation energies is greater than that between the second and third ionisation energies.
p electron in higher energy level and better shielded. Therefore easier to remove. [2]
(ii) The difference between the third and fourth ionisation energies is much larger than that between the other successive energies.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]
2. The first five ionisation energies, in kJ mol-1, of two elements Q and R are shown below:
First
Second
Third
Fourth
Fifth
Q
600
1800
3000
12000
15700
R
520
5500
7100
9500
13500
(a)
(i) In which groups of the Periodic Table would you expect to find elements Q and R? explain your reasoning for element Q.
[3]
(ii) Explain why the element directly below R in the Periodic Table would be expected to have a smaller ionisation energy than R.
[3]
(b) Using element R define the term second ionisation energy and write an appropriate equation.
[3]
3. (a) The first ionisation energies for a sequence of elements are shown in the graph:
(i) State whether the first ionisation energy of potassium would be more or less than that of sodium. Explain your choice.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]
(ii) Explain the shape of the graph.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]
(b) The successive ionisation energies of two elements A and B are listed below:
units are kJ mol-1.
1st 2nd 3rd 4th 5th 6th
A 420 3100 4301 5891 7999 9512
B 589 1098 4811 6498 8080 10482
(i) Which group in the periodic table does element A belong to?
Explain briefly your reasoning.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]
(ii) What is the likely charge on an ion of element B? Explain your reasoning.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]
4. The graph below shows the first ionisation energies of the elements
from hydrogen to sodium.
(a) What is meant by the term first ionisation energy?
[2]
(b) (i) Explain why there is an overall rise in first ionisation energy
in going from lithium to neon.
[2]
(ii) Explain why the first ionisation energy falls markedly from helium to lithium.
[2]
(iii) Explain why there is a fall in the first ionisation energy from nitrogen to oxygen.
[2]
(iv) Explain why the first ionisation energy of sodium is less than that of lithium.
[2]
Friday, March 6, 2009
Electronic Structure
1.1 ATOMIC STRUCTURE
Electronic Structure
Section A
For each of the questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.
1. Which one of the following represents the electronic arrangement of the outer shell of the chlorine atom (atomic number, 17)?
3s
3p
A
¯
¯
¯
B
¯
¯
¯
C
¯
D
¯
2. Which one of the following is the electronic configuration of the bromine atom?
A 1s22s22p63s23p63d104s14p6
B 1s22s22p63s23p63d104s24p7
C 1s22s22p63s23p63d104s24p5
D 1s22s22p63s23p63d104s24p6
3. To which one of the following energy levels (denoted by n) does the electron in a hydrogen atom return when the emission spectrum is in the ultraviolet region;
A n= 1
B n = 2
C n = 3
D n = 4
4. Which one of the following does not represent the electronic configuration of an atom in its ground state?
A 1s2 2s2 2p3
B 1s2 2s2 2p4
C 1s2 2s2 2p6 3s1
D 1s2 2s2 2p6 3d1
5. Which one of the following is the maximum number of atomic orbitals having a principal quantum number of two?
A 2
B 4
C 8
D 9
6. Which one of the following contains no unpaired electrons in the ground state?
A Be
B N
C Si
D F
7. Which one of the following is the electronic structure of a metal with a maximum oxidation state of +3?
A 1s2 2s2 2p6 3s1
B 1s2 2s2 2p6 3s2 3p1
C 1s2 2s2 2p6 3s2 3p6 3d10 4s2
D 1s2 2s2 2p6 3s2 3p4
8. Which one of the following statements regarding electronic orbitals is correct?
A Each p-orbital can hold a maximum of six electrons.
B The 3p-orbitals have a higher energy level than the 3s-orbital.
C The three 3p-orbitals have slightly different energy levels.
D The 1s-orbital has the same size and shape as the 2s-orbital.
9. The number of electrons in the 3d orbital of the atom of atomic number 23 is
A 2
B 3
C 4
D 5
10. Which one of the following correctly represents the electronic structure of the named atom in its ground state?
A B C D
2p
2s
1s
2p
2s
1s
2p
2p
2s
1s
2s
1s
Lithium
Boron Carbon Nitrogen
11. The vanadium atom (atomic number 23) in its ground state has the electronic configuration:
A 1s2 2s2 2p6 3s2 3p6 3d3 4s2
B 1s2 2s2 2p6 3s2 3p6 3d2 4s3
C 1s2 2s2 2p6 3s2 3p6 3d1 4s2 4p2
D 1s2 2s2 2p6 3s2 3p6 3d2 4s2 4p1
12. Which one of the following does not have the electronic structure
1s2 2s2 2p6 3s2 3p6 3d10 4s24p6?
A Kr
B S2-
C Se2-
D Sr2+
13. The graph below shows the first ionisation energies for a series of elements whose atomic numbers increase in sequence. Which one could represent nitrogen?
D
1st ionisation C
energy
B
A
atomic number
14. Which one of the following is the electronic configuration of a titanium atom?
A 1s22s22p63s23p64s24p2
@B 1s22s22p63s23p63d24s2
C 1s22s22p63s23p63d'4s'
D 1s22s22p63s23p'4s'4p'
15. Which one of the following compounds contains two ions with different electronic configurations?
A CaCl2
B MgF2
C MgO
D NaCl
16. Which graph represents a plot of energy, E, against principal quantum number (energy level), n, for an electron in a hydrogen atom?
17. Which one of the following atoms contains no unpaired electrons in its ground state?
A Beryllium
B Boron
C Carbon
D Lithium
18. Which one of the following sub-shells does not exist?
A 2p
B 2d
C 3p
D 4d
19. Which one of the following represents the electronic structure of the chloride ion in its ground state?
20. Which one of the following represents the shape of a p-orbital?B
Section B
1. An outline of the periodic table is shown below.
(a) Indicate on it the positions of the s block, p block and d block elements. [2]
s P d
(b) Write the electronic configuration for
(i) a potassium atom
1s22s22p63s23p64s1
(ii) a sulphur atom
1s22s22p63s23p4 [2]
(c) Write the empirical formula for potassium sulphide indicating the charges on the ions.
K+2S2- [1]
2.
(a) Write the electronic configuration of an iron atom in the ground state
K+2S2- [1]
(b) Explain the term ground state.
K+2S2-
K+2S2-
K+2S2- [1]
3. The diagram below shows the first four energy levels in a sodium atom. Label these levels and, using arrows to represent electrons, show the electronic structure of sodium in the ground state.
[3]
4. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. Beryllium is used for "windows" in X-ray apparatus because it has the lowest stopping power per unit mass thickness of all suitable construction materials.
Calculate the energy of X-rays of frequency 3 x 1017 Hz passing through a "beryllium window".
K+2S2-
K+2S2-
K+2S2- [2]
5. The Periodic Table is divided into s, p and d blocks. Using the symbols s, p and d complete the table below.
element
block
beryllium
manganese
phosphorus
[2]
Electronic Structure
Section A
For each of the questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.
1. Which one of the following represents the electronic arrangement of the outer shell of the chlorine atom (atomic number, 17)?
3s
3p
A
¯
¯
¯
B
¯
¯
¯
C
¯
D
¯
2. Which one of the following is the electronic configuration of the bromine atom?
A 1s22s22p63s23p63d104s14p6
B 1s22s22p63s23p63d104s24p7
C 1s22s22p63s23p63d104s24p5
D 1s22s22p63s23p63d104s24p6
3. To which one of the following energy levels (denoted by n) does the electron in a hydrogen atom return when the emission spectrum is in the ultraviolet region;
A n= 1
B n = 2
C n = 3
D n = 4
4. Which one of the following does not represent the electronic configuration of an atom in its ground state?
A 1s2 2s2 2p3
B 1s2 2s2 2p4
C 1s2 2s2 2p6 3s1
D 1s2 2s2 2p6 3d1
5. Which one of the following is the maximum number of atomic orbitals having a principal quantum number of two?
A 2
B 4
C 8
D 9
6. Which one of the following contains no unpaired electrons in the ground state?
A Be
B N
C Si
D F
7. Which one of the following is the electronic structure of a metal with a maximum oxidation state of +3?
A 1s2 2s2 2p6 3s1
B 1s2 2s2 2p6 3s2 3p1
C 1s2 2s2 2p6 3s2 3p6 3d10 4s2
D 1s2 2s2 2p6 3s2 3p4
8. Which one of the following statements regarding electronic orbitals is correct?
A Each p-orbital can hold a maximum of six electrons.
B The 3p-orbitals have a higher energy level than the 3s-orbital.
C The three 3p-orbitals have slightly different energy levels.
D The 1s-orbital has the same size and shape as the 2s-orbital.
9. The number of electrons in the 3d orbital of the atom of atomic number 23 is
A 2
B 3
C 4
D 5
10. Which one of the following correctly represents the electronic structure of the named atom in its ground state?
A B C D
2p
2s
1s
2p
2s
1s
2p
2p
2s
1s
2s
1s
Lithium
Boron Carbon Nitrogen
11. The vanadium atom (atomic number 23) in its ground state has the electronic configuration:
A 1s2 2s2 2p6 3s2 3p6 3d3 4s2
B 1s2 2s2 2p6 3s2 3p6 3d2 4s3
C 1s2 2s2 2p6 3s2 3p6 3d1 4s2 4p2
D 1s2 2s2 2p6 3s2 3p6 3d2 4s2 4p1
12. Which one of the following does not have the electronic structure
1s2 2s2 2p6 3s2 3p6 3d10 4s24p6?
A Kr
B S2-
C Se2-
D Sr2+
13. The graph below shows the first ionisation energies for a series of elements whose atomic numbers increase in sequence. Which one could represent nitrogen?
D
1st ionisation C
energy
B
A
atomic number
14. Which one of the following is the electronic configuration of a titanium atom?
A 1s22s22p63s23p64s24p2
@B 1s22s22p63s23p63d24s2
C 1s22s22p63s23p63d'4s'
D 1s22s22p63s23p'4s'4p'
15. Which one of the following compounds contains two ions with different electronic configurations?
A CaCl2
B MgF2
C MgO
D NaCl
16. Which graph represents a plot of energy, E, against principal quantum number (energy level), n, for an electron in a hydrogen atom?
17. Which one of the following atoms contains no unpaired electrons in its ground state?
A Beryllium
B Boron
C Carbon
D Lithium
18. Which one of the following sub-shells does not exist?
A 2p
B 2d
C 3p
D 4d
19. Which one of the following represents the electronic structure of the chloride ion in its ground state?
20. Which one of the following represents the shape of a p-orbital?B
Section B
1. An outline of the periodic table is shown below.
(a) Indicate on it the positions of the s block, p block and d block elements. [2]
s P d
(b) Write the electronic configuration for
(i) a potassium atom
1s22s22p63s23p64s1
(ii) a sulphur atom
1s22s22p63s23p4 [2]
(c) Write the empirical formula for potassium sulphide indicating the charges on the ions.
K+2S2- [1]
2.
(a) Write the electronic configuration of an iron atom in the ground state
K+2S2- [1]
(b) Explain the term ground state.
K+2S2-
K+2S2-
K+2S2- [1]
3. The diagram below shows the first four energy levels in a sodium atom. Label these levels and, using arrows to represent electrons, show the electronic structure of sodium in the ground state.
[3]
4. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. Beryllium is used for "windows" in X-ray apparatus because it has the lowest stopping power per unit mass thickness of all suitable construction materials.
Calculate the energy of X-rays of frequency 3 x 1017 Hz passing through a "beryllium window".
K+2S2-
K+2S2-
K+2S2- [2]
5. The Periodic Table is divided into s, p and d blocks. Using the symbols s, p and d complete the table below.
element
block
beryllium
manganese
phosphorus
[2]
ATOMIC STRUCTURE
1.1 ATOMIC STRUCTURE
Spectroscopy
Section A
For each of the questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.
1. The flame colour associated with potassium salts is
A Green
B Lilac
C Yellow
D Red
2. Which one of the following statements about the emission spectrum of hydrogen is correct?
A The number and/or position of lines in the spectrum cannot be influenced by external factors.
B The complete spectrum occurs in the visible region.
C The value for the ionisation energy of hydrogen can be determined from the spectrum.
D All the lines in the spectrum correspond to electron transitions from higher energy levels to the ground state.
3. Which one of the following diagrams represents the visible region of the atomic hydrogen spectrum?
A
Energy
B
Energy
C
Energy
D
Energy
4. The atomic spectrum of hydrogen shows
A a continuum in which individual lines are not apparent
B a series of lines which get closer together as the frequency increases
C equally spaced lines not confined to the visible spectrum
D equally spaced lines in groups separated by a continuum in the visible spectrum
5. In the line spectrum of atomic hydrogen in which region of the electromagnetic spectrum will lines not appear?
A infra-red
B ultraviolet
C visible
D X-ray
6. In which one of the following does the flame coloration correspond to the ion?
A Mg2+ yellow
B Sr2+ green
C Ba2+ white
D Ca2+ red
Section B
1. The presence of sodium ions in a sample of sea salt can be determined by flame colours.
(a) What is the flame colour associated with sodium compounds?
[1]
(b) The most prominent line in the visible spectrum of the atomic sodium spectrum has a frequency of 5.085 x 1014 Hz. Calculate the energy, E, of radiation with this frequency.
[2]
2. The Sun largely consists of a mixture of hydrogen and helium, the presence of each being detected by spectroscopy. The line emission spectrum of atomic hydrogen in the ultraviolet region of the electromagnetic spectrum is shown below.
frequency
(a) Explain why this spectrum consists of lines which are converging.
[5]
(b) Explain how the ionisation energy of atomic hydrogen can be calculated from the spectrum.
[2]
3. State the flame coloration for the following chlorides.
Metal chloride
colour
barium chloride
lithium chloride
potassium chloride
[3]
Spectroscopy
Section A
For each of the questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.
1. The flame colour associated with potassium salts is
A Green
B Lilac
C Yellow
D Red
2. Which one of the following statements about the emission spectrum of hydrogen is correct?
A The number and/or position of lines in the spectrum cannot be influenced by external factors.
B The complete spectrum occurs in the visible region.
C The value for the ionisation energy of hydrogen can be determined from the spectrum.
D All the lines in the spectrum correspond to electron transitions from higher energy levels to the ground state.
3. Which one of the following diagrams represents the visible region of the atomic hydrogen spectrum?
A
Energy
B
Energy
C
Energy
D
Energy
4. The atomic spectrum of hydrogen shows
A a continuum in which individual lines are not apparent
B a series of lines which get closer together as the frequency increases
C equally spaced lines not confined to the visible spectrum
D equally spaced lines in groups separated by a continuum in the visible spectrum
5. In the line spectrum of atomic hydrogen in which region of the electromagnetic spectrum will lines not appear?
A infra-red
B ultraviolet
C visible
D X-ray
6. In which one of the following does the flame coloration correspond to the ion?
A Mg2+ yellow
B Sr2+ green
C Ba2+ white
D Ca2+ red
Section B
1. The presence of sodium ions in a sample of sea salt can be determined by flame colours.
(a) What is the flame colour associated with sodium compounds?
[1]
(b) The most prominent line in the visible spectrum of the atomic sodium spectrum has a frequency of 5.085 x 1014 Hz. Calculate the energy, E, of radiation with this frequency.
[2]
2. The Sun largely consists of a mixture of hydrogen and helium, the presence of each being detected by spectroscopy. The line emission spectrum of atomic hydrogen in the ultraviolet region of the electromagnetic spectrum is shown below.
frequency
(a) Explain why this spectrum consists of lines which are converging.
[5]
(b) Explain how the ionisation energy of atomic hydrogen can be calculated from the spectrum.
[2]
3. State the flame coloration for the following chlorides.
Metal chloride
colour
barium chloride
lithium chloride
potassium chloride
[3]
ATOMIC STRUCTURE- MASS SPCTROMETER
1.1 ATOMIC STRUCTURE
Mass Spectrometry
Section A
For each of the questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.
1. Using a mass spectrometer, it is possible to determine the number of
A protons in an atom
B energy levels in an atom
C isotopes of an element
D neutrons in an atom
2. The accurate relative isotopic masses of five isotopes are
11H = 1.0078 21H = 2.0141 126C = 12.000 147N = 14.0031 168O = 15.9949
Using a high resolution mass spectrometer, a certain gas was found to have a relative molecular mass of 28.0172. The gas could be
A 147N2
B 126C2 11H4
C 21H 126C 147N
D 126C2 21H2
3. The mass spectrometer trace for naturally occurring magnesium is shown below. Assuming that all three peaks relate to ions with one positive charge, what is the relative atomic mass for magnesium?
A 24.2
B 24.3
C 24.4
D 24.7
4. The isotopic composition of a certain element X is 80% 24X, 10% 25X and 10% 26X. The relative-atomic mass of X is
A 24.25
B 24.30
C 24.33
D 24.67
5. The diagram below shows the mass spectrometer trace of the substance X in the region of relative mass 10 to 20 units.
Which of the following substances is X most likely to be?
A O2
B NH3
C H2O
D CH4
6. Which one of the following could not be obtained using a mass spectrometer?
A The number of electrons in an isotope of manganese
B The number of isotopes present in a sample of chlorine
C The RMM of a sample of cocaine
D The mass of an isotope of uranium
9 The mass spectrum of an element is as shown below.
Assuming each fragment has unit charge, what is the relative atomic mass of the element?
A 206.25
B 207.00
C 207.77
D 208.00
SECTION B
1. The mass spectrum of naturally occurring rubidium, Rb, is shown below.
From this spectrum calculate the relative atomic mass of rubidium.
100
55
0 m/z 85 87
[3]
Ar = 100 x 85 + 55 x 87 = 8500 + 4785 = 13285 = 85.71
155 155 155
2.
(a) An organic compound has the composition by mass.
Element
Percentage
Hydrogen
4.1
Carbon
24.2
Chlorine
71.7
(i) Calculate the empirical formula of the compound.
(ii) The mass spectrum of the compound showed the molecular ion peak at a mass/charge ratio of 98. Suggest a structural formula for the compound.
(iii) Suggest why smaller peaks may be found at 100 and 102.
[2]
3. A mixture of 21H2 and 8135Br2 was analysed in a mass spectrometer. The following pattern of lines due to singly-charged ions was obtained.
2 4 81 83 162
mass number on instrument scale
Suggest which ions give rise to each of these lines.
Mass number
Ion responsible
2
21H+
4
21H2+
81
8135Br+
83
21H8135Br+
162
8135Br2+
[3]
4. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. Beryllium is found in nature as 9Be.
(i) How could you show that naturally occurring beryllium consisted only of 9Be?
[2]
(ii) Draw the structure of 9Be in terms of the constituent particles of the atom.
[2]
4. (a) Complete the table below to show the relative masses and charges of a proton, a neutron and an electron. [3]
Relative mass
Relative charge
Proton
Electron
Neutron
(b) Describe the process by which particles are ionised in a mass spectrometer. [2]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
(c) Give two reasons why particles must be ionised before being analysed in a mass spectrometer. [2]
Reason 1 ……………………………………………………………………………………………………………………
Reason 2 ………………………………………………………………………………………………………………….
(d) A sample of boron contains 20% by mass of 10B and 80% by mass of 11B. Calculate the relative atomic mass of boron in this sample. [2]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
(e) Compound X contains only boron and hydrogen. The percentage by mass of boron in X is 81.2%. in the mass spectrum of X the peak at the largest value of m/z occurs at 54.
(i) Use the percentage by mass data to calculate the empirical formula of X. [3]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
(ii) Deduce the molecular formula of X. [1]
………………………………………………………………………………………………………………………………………
Mass Spectrometry
Section A
For each of the questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.
1. Using a mass spectrometer, it is possible to determine the number of
A protons in an atom
B energy levels in an atom
C isotopes of an element
D neutrons in an atom
2. The accurate relative isotopic masses of five isotopes are
11H = 1.0078 21H = 2.0141 126C = 12.000 147N = 14.0031 168O = 15.9949
Using a high resolution mass spectrometer, a certain gas was found to have a relative molecular mass of 28.0172. The gas could be
A 147N2
B 126C2 11H4
C 21H 126C 147N
D 126C2 21H2
3. The mass spectrometer trace for naturally occurring magnesium is shown below. Assuming that all three peaks relate to ions with one positive charge, what is the relative atomic mass for magnesium?
A 24.2
B 24.3
C 24.4
D 24.7
4. The isotopic composition of a certain element X is 80% 24X, 10% 25X and 10% 26X. The relative-atomic mass of X is
A 24.25
B 24.30
C 24.33
D 24.67
5. The diagram below shows the mass spectrometer trace of the substance X in the region of relative mass 10 to 20 units.
Which of the following substances is X most likely to be?
A O2
B NH3
C H2O
D CH4
6. Which one of the following could not be obtained using a mass spectrometer?
A The number of electrons in an isotope of manganese
B The number of isotopes present in a sample of chlorine
C The RMM of a sample of cocaine
D The mass of an isotope of uranium
9 The mass spectrum of an element is as shown below.
Assuming each fragment has unit charge, what is the relative atomic mass of the element?
A 206.25
B 207.00
C 207.77
D 208.00
SECTION B
1. The mass spectrum of naturally occurring rubidium, Rb, is shown below.
From this spectrum calculate the relative atomic mass of rubidium.
100
55
0 m/z 85 87
[3]
Ar = 100 x 85 + 55 x 87 = 8500 + 4785 = 13285 = 85.71
155 155 155
2.
(a) An organic compound has the composition by mass.
Element
Percentage
Hydrogen
4.1
Carbon
24.2
Chlorine
71.7
(i) Calculate the empirical formula of the compound.
(ii) The mass spectrum of the compound showed the molecular ion peak at a mass/charge ratio of 98. Suggest a structural formula for the compound.
(iii) Suggest why smaller peaks may be found at 100 and 102.
[2]
3. A mixture of 21H2 and 8135Br2 was analysed in a mass spectrometer. The following pattern of lines due to singly-charged ions was obtained.
2 4 81 83 162
mass number on instrument scale
Suggest which ions give rise to each of these lines.
Mass number
Ion responsible
2
21H+
4
21H2+
81
8135Br+
83
21H8135Br+
162
8135Br2+
[3]
4. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. Beryllium is found in nature as 9Be.
(i) How could you show that naturally occurring beryllium consisted only of 9Be?
[2]
(ii) Draw the structure of 9Be in terms of the constituent particles of the atom.
[2]
4. (a) Complete the table below to show the relative masses and charges of a proton, a neutron and an electron. [3]
Relative mass
Relative charge
Proton
Electron
Neutron
(b) Describe the process by which particles are ionised in a mass spectrometer. [2]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
(c) Give two reasons why particles must be ionised before being analysed in a mass spectrometer. [2]
Reason 1 ……………………………………………………………………………………………………………………
Reason 2 ………………………………………………………………………………………………………………….
(d) A sample of boron contains 20% by mass of 10B and 80% by mass of 11B. Calculate the relative atomic mass of boron in this sample. [2]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
(e) Compound X contains only boron and hydrogen. The percentage by mass of boron in X is 81.2%. in the mass spectrum of X the peak at the largest value of m/z occurs at 54.
(i) Use the percentage by mass data to calculate the empirical formula of X. [3]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
(ii) Deduce the molecular formula of X. [1]
………………………………………………………………………………………………………………………………………
AS level chemistry NOTES ATOMIC STRUCTURE
1. Atomic Structure
Electrons, protons and neutrons as the constituent particles of the atom. Their location in the atom, their relative masses and charges. Atomic number, mass number and isotopes.
When you have finished this section you should be able to:
· Describe the properties of protons, neutrons and electrons in terms of their relative charge and relative mass ;
· Understand the importance of these particles in the structure of the atom ;
· Define the terms atomic number, Z and mass number, A ;
· Use values for atomic number and mass number to calculate the number of protons and neutrons in the nucleus ;
· Explain the existence of isotopes
· Use isotopic symbols to describe the composition of the nucleus.
All atoms are electrically neutral. The number of electrons in the shells is the same as the number of protons in the nucleus. The mass is made up almost entirely from the masses of the protons and neutrons. The masses of the proton and neutron are virtually identical.
Properties of sub-atomic particles
Proton (p)
1
+1
Neutron (n)
1
0
Electron (e)
0 (1/1837)
-1
Evidence for particles
Electrons
J.J. Thompson (1897) experiments with cathode ray tubes. He discovers a beam of rays emitted from the cathode when an electric current passes through a gas at low pressure. The rays are deflected by electric and magnetic fields and consist of a stream of electrons.
Protons
Electric discharges through gases at low pressure produces a stream of particles from the anode. Work with electric and magnetic fields shows them to be positively charged. Hydrogen gives the lightest particles, which are assumed to be protons.
Neutrons
Chadwick (1932) bombards beryllium with a-particles producing fast moving particles, which are not affected by electric or magnetic fields. The particles are neutrons.
Relative sizes of atoms and nuclei
Atoms consist of a very small, dense nucleus, around which the electrons circulate in a comparatively large volume. The volume of the nucleus is about 10-44 m3 and it is composed of two different types of particles, protons and neutrons. These are known collectively as nucleons.
An atom is small, but its nucleus is smaller still. The radius of an atom is of the order of 10-10 m, but the radius of a nucleus is of the order of 10-15 m.
Do the next exercises will help you to appreciate the difference in size.
Exercise 1
Suppose a football, diameter 22 cm, is scaled up so that it becomes as big as the earth, diameter 13000 km.
Calculate whether an atom of diameter 0.32 nm (3.2 x 10-10 m) will become as big as:
A a pin head, diameter 1mm
√B a 1p coin, diameter 1.9 cm
C a football, diameter 22cm
D a weather balloon, diameter 1.8 m
Exercise 2
If the nucleus of an atom were scaled up to the size of a pin head (say 1 mm diameter), how big would the atom be?
100 m (about the length of a football field
Since the mass of an atom is concentrated in its nucleus, the nucleus must be extremely dense. Estimate how dense it is by doing the next exercise.
Exercise 3
For atoms of elements at the beginning of the Periodic Table the volume of the nucleus, VN, is given by:
VN = 1.73 x (relative atomic mass) x 10-45 m3
Use this expression to calculate the density of the sodium nucleus:
(a) in kg m-3 [9.60 x 1017 kg m-3]
(b) in tonnes cm-3 [9.60 x 108 tonnes cm-3]
(Remember that 1 mol of sodium atoms weighs 23.0 g and contains 6.02 x 1023 atoms)
Exercise 4
Calculate:
(a) the volume occupied by a sodium atom (radius 1.86 x 10-10 m)
(b) the fraction of the volume occupied by the nucleus.
(Hint: assume that both the atom and its nucleus are spheres with volume given by 4πr3/3 )
Most of your body is empty space too. If all the spaces between the nuclei were squeezed out, you would be only half as big as a flea, although you would weigh the same.
With all this empty space why does any object appear solid? The electrons in an atom move very rapidly around the nucleus. The electrons effectively form a shield around the nucleus, marking the limits of the atom’s volume and making it seem solid.
Atomic number, Mass number and Isotopes
Atomic number and mass number give us important information about an atom and are particularly useful in distinguishing one isotope of an element from another.
Atomic number
The atomic number (Z) of an atom is the number of protons in the nucleus.
Mass number
The mass number (A) is the total number of particles in the nucleus.
Isotopes
Isotopes are atoms of an element that have different numbers of neutrons in the nucleus i.e. they have the same atomic number but different mass numbers.
e.g. 3517Cl and 3717 Cl.
Exercise 5
The table shows the mass number and number of neutrons in the nucleus, for four atoms, W, X, Y and Z.
W X Y Z
Mass number 36 39 40 40
Neutrons in nucleus 18 20 21 22
a) Write down the atomic numbers of the four atoms.W=18; X=19; Y=19;Z=18
b) Which of the four atoms are isotopes of the same element? W and Z;X and Y
c) Use your Periodic Table to write isotopic symbols (e.g. 2713Al) for the four atoms.
3618W; 3919X; 4019Y; 4018Z
Relative atomic mass, relative isotopic mass and relative molecular mass. The carbon-12 standard. The use of the mass spectrometer to obtain accurate atomic masses. (Details of the workings of the mass spectrometer are not required). Deduction of Relative Molecular Mass from a molecular ion peak. (Limited to ions with single charges).
When you have finished this section you should be able to:
· Calculate the masses of coins relative to a chosen standard ;
· Express masses in a variety of units ;
· Define the terms relative atomic mass (Ar), relative isotopic mass and relative molecular mass (Mr) in terms of carbon-12 ;
Relative atomic mass
Atoms are so small that their masses, expressed in grams, are difficult to work with. Some examples are listed in Table 1 below
Table 1
Element
Average mass of atom g
H
1.67355 x 10-24
He
6.64605 x 10-24
Li
1.15217 x 10-23
C
1.99436 x 10-23
O
2.65659 x 10-23
Na
3.81730 x 10-23
Ar
6.63310 x 10-23
U
3.95233 x 10-22
The mass of an atom expressed as relative atomic mass (Ar) is much more manageable.
Exercise 6
Collect as many British coins of each kind (1p, 2p, 5p, 10p) as you can. Weigh a group of each kind to the nearest 0.01 g and calculate the average mass of each of the denominations to the nearest 0.001 g.
Enter your results in Table 2.
Table 2
Coin
Number of coins
Total mass g
Average mass g
1p
12
43.12
3.593
2p
9
63.49
7.054
5p
8
45.16
5.645
10p
10
110.68
11.068
You can calculate the relative mass of each of the coins using
Relative mass of coin = average mass of coin
mass of standard
Fill in column 1 of Table 3
Exercise 7
Table 3
Coin
1 Average mass g
Relative mass
2 Mass OCU
3 Mass CCU
4 Mass SCU
1p
3.593
1.000
2p
7.054
5p
5.645
10p
11.068
(a) Define the unit of mass, the OCU (for one-penny coin unit).
Let one OCU equal the average mass of a one penny coin.
1.00 OCU = g
(b) Calculate the relative mass of each type of coin on the OCU scale using
Relative mass of coin = average mass of coin
mass of OCU
Fill in column 2
Exercise 8
(a) Define a second unit of mass, the SCU (for silver coin unit) .
Using the data in Table 1 calculate
Average mass of 5p coin = g
1.000 SCU = average mass of 5p coin
1.000 SCU = g
(c) Calculate the relative mass of each type of coin using the expression:
Relative mass of coin = average mass of coin
Mass of SCU
Record the values in column 4 of Table 1.
Exercise 9
According to the Royal Mint, the mass of a newly minted 2p coin is 7.128 g. We define a second unit, the CCU (copper coin unit), as one half the mass of a newly minted 2p coin.
1.000 CCU = ½ x 7.128 g
1.000 CCU = 3.564 g
Using the defined value for the CCU, calculate the relative masses for column 3.
The relative atomic mass scale
You have now completed a series of exercises using coins to illustrate how relative mass changes as the choice of standard changes. Now you will do a similar exercise using masses of atoms instead of coins, where you calculate the relative atomic masses on the hydrogen, oxygen and carbon-12 scales
Exercise 10
Use the values in Table 1 to calculate atomic masses relative to
(a) hydrogen,
(b) oxygen,
(c) carbon-12
in a similar way to that in which you calculated relative masses of coins.
Complete Table 4.
Some values are included as a check.
(The mass of an atom of carbon-12 = 1.99252 x 10-23 g).
Table 4
Element
Relative atomic mass (Ar)
H scale
O scale
12C scale
H
1.00000
1.00794
1.00790
He
3.97123
4.00276
4.00260
Li
6.88459
6.93924
6.93897
C
11.9169
12.0115
12.01110
O
15.8740
16.0000
15.9994
Na
22.8096
22.9907
22.9898
Ar
39.6349
39.9496
39.9480
U
236.164
238.039
238.030
In most of you’re A level work, you use relative atomic masses expressed to three significant figures (e.g. He = 4.00, O = 16.0, U = 238). To this degree of precision, the oxygen scale and the carbon-12 scale can be regarded as the same, but you should not use the hydrogen scale as it differs so much from the others.
RELATIVE ATOMIC MASS (Ar)
Atoms are so light that their actual masses are not used. Instead, each atom is compared to a standard atom. The atom chosen as standard is the most common isotope of carbon, carbon-12.
The carbon-12 atom has a mass of exactly 12.0000 units and all other atoms are given a mass relative to the carbon-12 standard.
For example, a magnesium atom is twice as heavy as a carbon-12 atom.
The relative atomic mass of an element
relative atomic mass of an element = mass of one atom of the element
(1/12) x mass of one atom of carbon-12
RELATIVE MOLECULAR MASS (Mr)
The Relative Molecular Mass of a compound is the sum of the relative atomic masses of all the atoms in a molecule of a compound.
For example: find the Relative Molecular Mass of sulphuric acid.
Formula H 2SO4
2 atoms of H = 2.00
1 atom of S = 32.0
4 atoms of O = 64.0
TOTAL = 98.0
R.M.M. sulphuric acid is 98.
RELATIVE FORMULA MASS
The Relative Formula Mass of an ionic compound equals the sum of the Relative Atomic Masses of all the atoms in a formula unit of the compound.
For example: find the Relative Formula Mass of magnesium chloride.
Formula MgCl2
1 atom of Mg = 24.0
2 atoms of Cl = 71.0
TOTAL = 95.0
R.F.M. magnesium chloride is 95.0.
Find the Relative Formula Mass of hydrated copper sulphate crystals, CuSO4.5H2O.
1 atom of copper Cu = 64.0
1 atom of sulphur S = 32.0
10 atoms of hydrogen H = 10.0
9 atoms of oxygen O = 144
TOTAL = 250
R.F.M. hydrated copper sulphate is 250.The Mass Spectrometer
When you have finished this section you should be able to:
· Understand the principles of a simple mass spectrometer, limited to ionisation, acceleration, deflection and detection ;
· Identify peaks on a simple mass spectrum and use them to calculate the relative abundance and masses of ions ;
· Calculate the relative atomic mass of an element from (a) a mass spectrum, (b) percentage abundance data ;
· Sketch a mass spectrum, given relevant data.
The mass spectrometer is used to obtain accurate atomic masses by measuring the mass and relative abundance of the isotopes of an atom.
1. The sample is vaporised – atoms must be in a gaseous state.
2. Positive ions are formed. Atoms are bombarded by electrons and positive ions are formed .
X(g) X+(g) + e-
3. The positive ions are accelerated by an electric field. The slits restrict the ions to a narrow beam.
4. A magnetic field deflects the ions depending on their mass/charge ratio. Ions with high m/z ratio are deflected less than those with low m/e ratio.
5. Ions with the correct m/z ratio pass through the slit and arrive at the detector.
6. The charge received at the detector is amplified and turned into a sizeable electric current.
7. The electric current operates a pen recorder, which traces a peak on the recording.
8. If the magnetic field is kept constant while the accelerating electric field varies continually, ions of different m/z ratio are deflected one after the other into the detector and a trace is obtained.
Interpreting Mass Spectra
E g. the mass spectrum of rubidium Rb.
Height of peak
m/z 85 87
Note:
1. The height of each peak is proportional to the amount of each isotope present (i.e. it’s relative abundance).
2. The m/z ratio for each peak is found from the accelerating voltage for each peak. Many ions have a +1 charge so that the m/z ratio is numerically equal to mass m of the ion.
Exercise 11
Refer to the diagram of the mass spectrum of rubidium previously to answer this question.
(a) Describe the two isotopes of rubidium using isotopic symbols.
8737Rb; 8537Rb
(b) What information can you get from the heights of the peaks on the mass spectrum?
The height of each peak is proportional to the relative abundance of the isotope it represents. In this case 8537Rb is more than twice as abundant as 8737Rb
Calculating the relative atomic mass of an element
1. Measure the height of each peak.
85 Rb = 5.82 cm
87 Rb = 2.25 cm
Therefore the ratio 85 Rb : 87 Rb is
5.82 : 2.25
2. Calculate the percentage relative abundance
% abundance = amount of isotope x 100
total amount of all isotopes
% 85 Rb = 5.82 x 100 = 72.1 %
(5.82 + 2.25)
% 87 Rb = = 2.25 x 100 = 27.9 %
(5.82 + 2.25)
3. Calculate the Ar
Ar (Rb) = (72.1 x 85) + (27.9 x 87) = 85.6
100
Exercise 12
Use the mass spectrum shown below to calculate:
(a) the percentage of each isotope present in a sample of naturally occurring lithium;
(b) the relative atomic mass of lithium.
3 4 5 6 7 8
mass/charge ratio
Exercise 13
The mass spectrum of neon consists of three lines corresponding to mass/charge ratios of 20, 21 and 22 with relative intensities of 0.910; 0.0026; 0.088 respectively.
Calculate the relative atomic mass of neon. [20.2]
Exercise 14
The percentage abundance of the stable isotopes of chromium are:
5024Cr – 4.31%; 5224Cr – 83.76%; 5324Cr – 9.55%; 5424Cr – 2.38%.
(a) Sketch the mass spectrum that would be obtained from naturally occurring chromium.
(b) Calculate the relative atomic mass of chromium, correct to three significant figures.
(c) Label each peak on the mass spectrum using isotopic symbols.
The next exercise involves a molecular element.
Exercise 15The element chlorine has isotopes of mass number 35 and 37 in the approximate proportion 3:1.
30 40 50 60 70 80
mass/charge ratio
Interpret the mass spectrum of gaseous chlorine shown above indicating the formula (including mass number) and charges of the ions responsible for each peak.
Exercise 16
Calculate the relative atomic mass of potassium, which consists of 93.0% 39K and 7.0% 41K.
Additional Exercises
(a) Chlorine consists of isotopes of relative masses 34.97 and 36.96 with natural abundances of 75.77% and 24.23% respectively.
Calculate the mean relative atomic mass of naturally-occurring chlorine.
35.45
(b) Calculate the relative atomic mass of natural lithium which consists of 7.4% of 6Li (relative atomic mass 6.02) and 92.6% of 7Li (relative atomic mass 7.02). 6.95
(c) Copper (atomic number 29) has two isotopes, the first of relative atomic mass 62.9 and abundance 65%, the second of relative atomic mass 64.9 and abundance 35%.
Calculate the mean relative atomic mass of naturally-occurring copper. 63.6
(d) Using mass spectrometry, the element gallium has been found to consist of 60.4 per cent of an isotope of atomic mass 68.93 and 39.6 per cent of an isotope of atomic mass 70.92.
Calculate, to three significant figures, the relative atomic mass of gallium.
69.7
Mass Spectra of Molecules
Molecules produce more complex mass spectra than atomic spectra. The simplest ion produced is the parent molecule with one electron removed.
M (g) M+ (g) + e-
M+ is referred to as the molecular ion.
It is possible for the molecular ion to break apart to give fragment ions.
M+ (g) Y+ (g) + Z+ (g) etc.
The relative molecular mass of a molecule can be determined from the molecular ion peak.
e.g. The mass spectrum of copper (II) nitrate
molecular ions
peak
height
Electrons, protons and neutrons as the constituent particles of the atom. Their location in the atom, their relative masses and charges. Atomic number, mass number and isotopes.
When you have finished this section you should be able to:
· Describe the properties of protons, neutrons and electrons in terms of their relative charge and relative mass ;
· Understand the importance of these particles in the structure of the atom ;
· Define the terms atomic number, Z and mass number, A ;
· Use values for atomic number and mass number to calculate the number of protons and neutrons in the nucleus ;
· Explain the existence of isotopes
· Use isotopic symbols to describe the composition of the nucleus.
All atoms are electrically neutral. The number of electrons in the shells is the same as the number of protons in the nucleus. The mass is made up almost entirely from the masses of the protons and neutrons. The masses of the proton and neutron are virtually identical.
Properties of sub-atomic particles
Proton (p)
1
+1
Neutron (n)
1
0
Electron (e)
0 (1/1837)
-1
Evidence for particles
Electrons
J.J. Thompson (1897) experiments with cathode ray tubes. He discovers a beam of rays emitted from the cathode when an electric current passes through a gas at low pressure. The rays are deflected by electric and magnetic fields and consist of a stream of electrons.
Protons
Electric discharges through gases at low pressure produces a stream of particles from the anode. Work with electric and magnetic fields shows them to be positively charged. Hydrogen gives the lightest particles, which are assumed to be protons.
Neutrons
Chadwick (1932) bombards beryllium with a-particles producing fast moving particles, which are not affected by electric or magnetic fields. The particles are neutrons.
Relative sizes of atoms and nuclei
Atoms consist of a very small, dense nucleus, around which the electrons circulate in a comparatively large volume. The volume of the nucleus is about 10-44 m3 and it is composed of two different types of particles, protons and neutrons. These are known collectively as nucleons.
An atom is small, but its nucleus is smaller still. The radius of an atom is of the order of 10-10 m, but the radius of a nucleus is of the order of 10-15 m.
Do the next exercises will help you to appreciate the difference in size.
Exercise 1
Suppose a football, diameter 22 cm, is scaled up so that it becomes as big as the earth, diameter 13000 km.
Calculate whether an atom of diameter 0.32 nm (3.2 x 10-10 m) will become as big as:
A a pin head, diameter 1mm
√B a 1p coin, diameter 1.9 cm
C a football, diameter 22cm
D a weather balloon, diameter 1.8 m
Exercise 2
If the nucleus of an atom were scaled up to the size of a pin head (say 1 mm diameter), how big would the atom be?
100 m (about the length of a football field
Since the mass of an atom is concentrated in its nucleus, the nucleus must be extremely dense. Estimate how dense it is by doing the next exercise.
Exercise 3
For atoms of elements at the beginning of the Periodic Table the volume of the nucleus, VN, is given by:
VN = 1.73 x (relative atomic mass) x 10-45 m3
Use this expression to calculate the density of the sodium nucleus:
(a) in kg m-3 [9.60 x 1017 kg m-3]
(b) in tonnes cm-3 [9.60 x 108 tonnes cm-3]
(Remember that 1 mol of sodium atoms weighs 23.0 g and contains 6.02 x 1023 atoms)
Exercise 4
Calculate:
(a) the volume occupied by a sodium atom (radius 1.86 x 10-10 m)
(b) the fraction of the volume occupied by the nucleus.
(Hint: assume that both the atom and its nucleus are spheres with volume given by 4πr3/3 )
Most of your body is empty space too. If all the spaces between the nuclei were squeezed out, you would be only half as big as a flea, although you would weigh the same.
With all this empty space why does any object appear solid? The electrons in an atom move very rapidly around the nucleus. The electrons effectively form a shield around the nucleus, marking the limits of the atom’s volume and making it seem solid.
Atomic number, Mass number and Isotopes
Atomic number and mass number give us important information about an atom and are particularly useful in distinguishing one isotope of an element from another.
Atomic number
The atomic number (Z) of an atom is the number of protons in the nucleus.
Mass number
The mass number (A) is the total number of particles in the nucleus.
Isotopes
Isotopes are atoms of an element that have different numbers of neutrons in the nucleus i.e. they have the same atomic number but different mass numbers.
e.g. 3517Cl and 3717 Cl.
Exercise 5
The table shows the mass number and number of neutrons in the nucleus, for four atoms, W, X, Y and Z.
W X Y Z
Mass number 36 39 40 40
Neutrons in nucleus 18 20 21 22
a) Write down the atomic numbers of the four atoms.W=18; X=19; Y=19;Z=18
b) Which of the four atoms are isotopes of the same element? W and Z;X and Y
c) Use your Periodic Table to write isotopic symbols (e.g. 2713Al) for the four atoms.
3618W; 3919X; 4019Y; 4018Z
Relative atomic mass, relative isotopic mass and relative molecular mass. The carbon-12 standard. The use of the mass spectrometer to obtain accurate atomic masses. (Details of the workings of the mass spectrometer are not required). Deduction of Relative Molecular Mass from a molecular ion peak. (Limited to ions with single charges).
When you have finished this section you should be able to:
· Calculate the masses of coins relative to a chosen standard ;
· Express masses in a variety of units ;
· Define the terms relative atomic mass (Ar), relative isotopic mass and relative molecular mass (Mr) in terms of carbon-12 ;
Relative atomic mass
Atoms are so small that their masses, expressed in grams, are difficult to work with. Some examples are listed in Table 1 below
Table 1
Element
Average mass of atom g
H
1.67355 x 10-24
He
6.64605 x 10-24
Li
1.15217 x 10-23
C
1.99436 x 10-23
O
2.65659 x 10-23
Na
3.81730 x 10-23
Ar
6.63310 x 10-23
U
3.95233 x 10-22
The mass of an atom expressed as relative atomic mass (Ar) is much more manageable.
Exercise 6
Collect as many British coins of each kind (1p, 2p, 5p, 10p) as you can. Weigh a group of each kind to the nearest 0.01 g and calculate the average mass of each of the denominations to the nearest 0.001 g.
Enter your results in Table 2.
Table 2
Coin
Number of coins
Total mass g
Average mass g
1p
12
43.12
3.593
2p
9
63.49
7.054
5p
8
45.16
5.645
10p
10
110.68
11.068
You can calculate the relative mass of each of the coins using
Relative mass of coin = average mass of coin
mass of standard
Fill in column 1 of Table 3
Exercise 7
Table 3
Coin
1 Average mass g
Relative mass
2 Mass OCU
3 Mass CCU
4 Mass SCU
1p
3.593
1.000
2p
7.054
5p
5.645
10p
11.068
(a) Define the unit of mass, the OCU (for one-penny coin unit).
Let one OCU equal the average mass of a one penny coin.
1.00 OCU = g
(b) Calculate the relative mass of each type of coin on the OCU scale using
Relative mass of coin = average mass of coin
mass of OCU
Fill in column 2
Exercise 8
(a) Define a second unit of mass, the SCU (for silver coin unit) .
Using the data in Table 1 calculate
Average mass of 5p coin = g
1.000 SCU = average mass of 5p coin
1.000 SCU = g
(c) Calculate the relative mass of each type of coin using the expression:
Relative mass of coin = average mass of coin
Mass of SCU
Record the values in column 4 of Table 1.
Exercise 9
According to the Royal Mint, the mass of a newly minted 2p coin is 7.128 g. We define a second unit, the CCU (copper coin unit), as one half the mass of a newly minted 2p coin.
1.000 CCU = ½ x 7.128 g
1.000 CCU = 3.564 g
Using the defined value for the CCU, calculate the relative masses for column 3.
The relative atomic mass scale
You have now completed a series of exercises using coins to illustrate how relative mass changes as the choice of standard changes. Now you will do a similar exercise using masses of atoms instead of coins, where you calculate the relative atomic masses on the hydrogen, oxygen and carbon-12 scales
Exercise 10
Use the values in Table 1 to calculate atomic masses relative to
(a) hydrogen,
(b) oxygen,
(c) carbon-12
in a similar way to that in which you calculated relative masses of coins.
Complete Table 4.
Some values are included as a check.
(The mass of an atom of carbon-12 = 1.99252 x 10-23 g).
Table 4
Element
Relative atomic mass (Ar)
H scale
O scale
12C scale
H
1.00000
1.00794
1.00790
He
3.97123
4.00276
4.00260
Li
6.88459
6.93924
6.93897
C
11.9169
12.0115
12.01110
O
15.8740
16.0000
15.9994
Na
22.8096
22.9907
22.9898
Ar
39.6349
39.9496
39.9480
U
236.164
238.039
238.030
In most of you’re A level work, you use relative atomic masses expressed to three significant figures (e.g. He = 4.00, O = 16.0, U = 238). To this degree of precision, the oxygen scale and the carbon-12 scale can be regarded as the same, but you should not use the hydrogen scale as it differs so much from the others.
RELATIVE ATOMIC MASS (Ar)
Atoms are so light that their actual masses are not used. Instead, each atom is compared to a standard atom. The atom chosen as standard is the most common isotope of carbon, carbon-12.
The carbon-12 atom has a mass of exactly 12.0000 units and all other atoms are given a mass relative to the carbon-12 standard.
For example, a magnesium atom is twice as heavy as a carbon-12 atom.
The relative atomic mass of an element
relative atomic mass of an element = mass of one atom of the element
(1/12) x mass of one atom of carbon-12
RELATIVE MOLECULAR MASS (Mr)
The Relative Molecular Mass of a compound is the sum of the relative atomic masses of all the atoms in a molecule of a compound.
For example: find the Relative Molecular Mass of sulphuric acid.
Formula H 2SO4
2 atoms of H = 2.00
1 atom of S = 32.0
4 atoms of O = 64.0
TOTAL = 98.0
R.M.M. sulphuric acid is 98.
RELATIVE FORMULA MASS
The Relative Formula Mass of an ionic compound equals the sum of the Relative Atomic Masses of all the atoms in a formula unit of the compound.
For example: find the Relative Formula Mass of magnesium chloride.
Formula MgCl2
1 atom of Mg = 24.0
2 atoms of Cl = 71.0
TOTAL = 95.0
R.F.M. magnesium chloride is 95.0.
Find the Relative Formula Mass of hydrated copper sulphate crystals, CuSO4.5H2O.
1 atom of copper Cu = 64.0
1 atom of sulphur S = 32.0
10 atoms of hydrogen H = 10.0
9 atoms of oxygen O = 144
TOTAL = 250
R.F.M. hydrated copper sulphate is 250.The Mass Spectrometer
When you have finished this section you should be able to:
· Understand the principles of a simple mass spectrometer, limited to ionisation, acceleration, deflection and detection ;
· Identify peaks on a simple mass spectrum and use them to calculate the relative abundance and masses of ions ;
· Calculate the relative atomic mass of an element from (a) a mass spectrum, (b) percentage abundance data ;
· Sketch a mass spectrum, given relevant data.
The mass spectrometer is used to obtain accurate atomic masses by measuring the mass and relative abundance of the isotopes of an atom.
1. The sample is vaporised – atoms must be in a gaseous state.
2. Positive ions are formed. Atoms are bombarded by electrons and positive ions are formed .
X(g) X+(g) + e-
3. The positive ions are accelerated by an electric field. The slits restrict the ions to a narrow beam.
4. A magnetic field deflects the ions depending on their mass/charge ratio. Ions with high m/z ratio are deflected less than those with low m/e ratio.
5. Ions with the correct m/z ratio pass through the slit and arrive at the detector.
6. The charge received at the detector is amplified and turned into a sizeable electric current.
7. The electric current operates a pen recorder, which traces a peak on the recording.
8. If the magnetic field is kept constant while the accelerating electric field varies continually, ions of different m/z ratio are deflected one after the other into the detector and a trace is obtained.
Interpreting Mass Spectra
E g. the mass spectrum of rubidium Rb.
Height of peak
m/z 85 87
Note:
1. The height of each peak is proportional to the amount of each isotope present (i.e. it’s relative abundance).
2. The m/z ratio for each peak is found from the accelerating voltage for each peak. Many ions have a +1 charge so that the m/z ratio is numerically equal to mass m of the ion.
Exercise 11
Refer to the diagram of the mass spectrum of rubidium previously to answer this question.
(a) Describe the two isotopes of rubidium using isotopic symbols.
8737Rb; 8537Rb
(b) What information can you get from the heights of the peaks on the mass spectrum?
The height of each peak is proportional to the relative abundance of the isotope it represents. In this case 8537Rb is more than twice as abundant as 8737Rb
Calculating the relative atomic mass of an element
1. Measure the height of each peak.
85 Rb = 5.82 cm
87 Rb = 2.25 cm
Therefore the ratio 85 Rb : 87 Rb is
5.82 : 2.25
2. Calculate the percentage relative abundance
% abundance = amount of isotope x 100
total amount of all isotopes
% 85 Rb = 5.82 x 100 = 72.1 %
(5.82 + 2.25)
% 87 Rb = = 2.25 x 100 = 27.9 %
(5.82 + 2.25)
3. Calculate the Ar
Ar (Rb) = (72.1 x 85) + (27.9 x 87) = 85.6
100
Exercise 12
Use the mass spectrum shown below to calculate:
(a) the percentage of each isotope present in a sample of naturally occurring lithium;
(b) the relative atomic mass of lithium.
3 4 5 6 7 8
mass/charge ratio
Exercise 13
The mass spectrum of neon consists of three lines corresponding to mass/charge ratios of 20, 21 and 22 with relative intensities of 0.910; 0.0026; 0.088 respectively.
Calculate the relative atomic mass of neon. [20.2]
Exercise 14
The percentage abundance of the stable isotopes of chromium are:
5024Cr – 4.31%; 5224Cr – 83.76%; 5324Cr – 9.55%; 5424Cr – 2.38%.
(a) Sketch the mass spectrum that would be obtained from naturally occurring chromium.
(b) Calculate the relative atomic mass of chromium, correct to three significant figures.
(c) Label each peak on the mass spectrum using isotopic symbols.
The next exercise involves a molecular element.
Exercise 15The element chlorine has isotopes of mass number 35 and 37 in the approximate proportion 3:1.
30 40 50 60 70 80
mass/charge ratio
Interpret the mass spectrum of gaseous chlorine shown above indicating the formula (including mass number) and charges of the ions responsible for each peak.
Exercise 16
Calculate the relative atomic mass of potassium, which consists of 93.0% 39K and 7.0% 41K.
Additional Exercises
(a) Chlorine consists of isotopes of relative masses 34.97 and 36.96 with natural abundances of 75.77% and 24.23% respectively.
Calculate the mean relative atomic mass of naturally-occurring chlorine.
35.45
(b) Calculate the relative atomic mass of natural lithium which consists of 7.4% of 6Li (relative atomic mass 6.02) and 92.6% of 7Li (relative atomic mass 7.02). 6.95
(c) Copper (atomic number 29) has two isotopes, the first of relative atomic mass 62.9 and abundance 65%, the second of relative atomic mass 64.9 and abundance 35%.
Calculate the mean relative atomic mass of naturally-occurring copper. 63.6
(d) Using mass spectrometry, the element gallium has been found to consist of 60.4 per cent of an isotope of atomic mass 68.93 and 39.6 per cent of an isotope of atomic mass 70.92.
Calculate, to three significant figures, the relative atomic mass of gallium.
69.7
Mass Spectra of Molecules
Molecules produce more complex mass spectra than atomic spectra. The simplest ion produced is the parent molecule with one electron removed.
M (g) M+ (g) + e-
M+ is referred to as the molecular ion.
It is possible for the molecular ion to break apart to give fragment ions.
M+ (g) Y+ (g) + Z+ (g) etc.
The relative molecular mass of a molecule can be determined from the molecular ion peak.
e.g. The mass spectrum of copper (II) nitrate
molecular ions
peak
height
moles -test/ As level
TEST 2
Calculations and Equations
1. Balance the equation
FeCl3 + H2S S + FeCl2 + HCl
[5]
2. Write a balanced symbol equation for the reaction
sodium carbonate + hydrochloric acid sodium chloride + carbon dioxide + water
[5]
3. A mixture containing 2.80 g of iron and 2.00 g of sulphur is heated together.
What mass of iron(II) sulphide , FeS, is produced?
Fe (s) + S (s) FeS (s)
Calculations and Equations
1. Balance the equation
FeCl3 + H2S S + FeCl2 + HCl
[5]
2. Write a balanced symbol equation for the reaction
sodium carbonate + hydrochloric acid sodium chloride + carbon dioxide + water
[5]
3. A mixture containing 2.80 g of iron and 2.00 g of sulphur is heated together.
What mass of iron(II) sulphide , FeS, is produced?
Fe (s) + S (s) FeS (s)
moles -test/ As level
CCEA MODULE 1
TEST
Calculations and Equations
1. What mass of material is there in each of the following?
(a) 2.00 mol of SO3 molecules
(b) 0.0300 mol of Cl atoms
(c) 9.00 mo1 of SO42- ions
(d) 0.150 mol of MgSO4.7H2O molecules [4]
2. What amount of each substance is contained in the following?
(a) 31.0 g of P4 molecules
(b) 1.00 x 1022 atoms of Cu
(c) 70.0 g of Fe2+ ions
(d) 9.00 x 1024 molecules of C2H5OH [4]
3. The mass of one molecule of a compound is 2.19 x 10-22 g.
What is the molar mass of the compound? [2]
4. What mass of aluminium, Al, is required to produce 1000 g of iron, Fe, according to the equation:
3Fe3O4(s) + 8Al(s) 4A12O3(s) + 9Fe(s) [3]
5. A solution is made containing 2.38 g of magnesium chloride, MgC12, in 500 cm3 of solution.
(a) What is the concentration of magnesium chloride, MgC12, in this solution?
(b) What is the concentration of chloride ions in this solution? [3]
6. A solution is made by dissolving 8.50 g of sodium nitrate, NaNO3, and
16.40 g of calcium nitrate, Ca(NO3)2, in enough water to give 2000 cm3 of solution.
(a) What is the concentration of sodium ions in the solution?
(b) What is the concentration of nitrate ions in the solution? [4]
7. A hydrated aluminium sulphate A12(SO4)3.xH2O, contains 8.10% of
aluminium by mass.
Find the value of x [5]
[TOTAL: 25 marks]
TEST
Calculations and Equations
1. What mass of material is there in each of the following?
(a) 2.00 mol of SO3 molecules
(b) 0.0300 mol of Cl atoms
(c) 9.00 mo1 of SO42- ions
(d) 0.150 mol of MgSO4.7H2O molecules [4]
2. What amount of each substance is contained in the following?
(a) 31.0 g of P4 molecules
(b) 1.00 x 1022 atoms of Cu
(c) 70.0 g of Fe2+ ions
(d) 9.00 x 1024 molecules of C2H5OH [4]
3. The mass of one molecule of a compound is 2.19 x 10-22 g.
What is the molar mass of the compound? [2]
4. What mass of aluminium, Al, is required to produce 1000 g of iron, Fe, according to the equation:
3Fe3O4(s) + 8Al(s) 4A12O3(s) + 9Fe(s) [3]
5. A solution is made containing 2.38 g of magnesium chloride, MgC12, in 500 cm3 of solution.
(a) What is the concentration of magnesium chloride, MgC12, in this solution?
(b) What is the concentration of chloride ions in this solution? [3]
6. A solution is made by dissolving 8.50 g of sodium nitrate, NaNO3, and
16.40 g of calcium nitrate, Ca(NO3)2, in enough water to give 2000 cm3 of solution.
(a) What is the concentration of sodium ions in the solution?
(b) What is the concentration of nitrate ions in the solution? [4]
7. A hydrated aluminium sulphate A12(SO4)3.xH2O, contains 8.10% of
aluminium by mass.
Find the value of x [5]
[TOTAL: 25 marks]
moles - As/ gce chemistry notes
Equations and Calculations
SECTION A
1. Iodate ions in the presence of acid oxidise sulphite ions to sulphate ions. Which one of the following equations correctly represents this reaction?
A IO3- + 2SO32- + 2H+ I2 + 2SO42- + H2O
B 2IO3- + SO32- + 2H+ I2 + SO42- + H2O
C 2IO3- + 5SO32- + 2H+ I2 + 5SO42- + H2O
D 2IO3- + SO32- + 10H+ I2 + SO42- + 5H2O
2. If 8.7 g of potassium sulphate is dissolved in water and made up to 200 cm3, what is the concentration of the solution in mol dm-3?
A 0.05
B 0.20
C 0.25
D 0.50
3. On heating with soda lime. sodium ethanoate decomposes in accordance with the following equation.
CH3COONa (s) + NaOH (s) CH4 (g) + Na2 CO3 (s)
If 8.2 g of sodium ethanoate produced 560 cm3 of methane (at s.t.p.). which one of the following is the percentage yield of the reaction;
A 2.5
B 4.0
C 12.0
D 25.0
4. What mass of iron(II) ammonium sulphate-6-water,Fe(NH4)2(SO4)3.6H2O, (relative molecular mass = 392) must be dissolved in water to produce 1 litre of a solution which is 0.05M with respect to Fe2+(aq)?
A 2.80
B 14.2
C 19.6
D 28.0
5.D 0.0640 g of an unknown sulphate yielded 0.1165 g of barium sulphate when appropriately treated with barium chloride. Which one of the following is the percentage of sulphate in the unknown?
A 75
B 45
C 30
D 25
6. Equal masses of sulphur dioxide and sulphur trioxide are mixed. The mole fraction of sulphur trioxide will be
(Relative atomic masses: O=16, S=32)
A 0.22
B 0.44
C 0.50
D 0.56
E 0.80
7. Calculate the number of protons present in the nuclei of the atoms contained in 2.24 dm3 of oxygen gas, O2, (atomic number of oxygen is 8) at standard temperature and pressure, given that one mole of gas under these conditions occupies 22.4 dm3 and the value of the Avogadro constant L = 6.0 x 1023 mol-1.
It equals
A 9.6 x 1023
B 6.0 x 1022
C 4.8 x 1023
D 9.6 x 1024
8. If L is the Avogadro constant the number of ions in 17.43 g of solid K2SO4 (relative molecular mass 174.3) will be
A L/30
B L/10
C 3L/10
D 30L
9. Chemical analysis showed that a sample of a substance contained only carbon, hydrogen and nitrogen in the following amounts: carbon, 0.060 g; hydrogen, 0.025 g and nitrogen, 0.070 g. What is the empirical formula for this substance?
A CH2N2
B CH3N2
C CH4N2
D CH5N
Section B
1. (a) State what is meant by relative atomic mass.
[2] (b) (i) Calculate the relative formula mass of washing soda Na 2CO3.10H20
[1]
(ii) On heating, the washing soda loses all the water present. Calculate the mass of the residue when 57.2 g of washing soda is completely dehydrated.
[2]
2. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. The mineral beryl, Be3Al2Si6O16, is the main source of beryllium.
Calculate the percentage, by mass, of beryllium in beryl.
% [3]
3. Calcium oxalate may be converted to oxalic acid and calcium sulphate by reaction with dilute sulphuric acid:
CaC2O4 + H2SO4 H2C2O4 + CaSO4
The calcium sulphate precipitated can be hydrated to make plaster of Paris, CaSO4.xH2O. Calculate the value of x in the formula if 1.45 g of plaster of Paris is heated to dryness and 1.36 g of CaSO4 remains.
x = [3]
4. The sulphur content of cystine is 26.7 %. Given that cystine contains two sulphur atoms, what is the molar mass of cystine? Show your working clearly.
[3]
SECTION B
1. (a) Explain why 235 92U and 238 92U are regarded as isotopes of uranium.
Same atomic number but different mass numbers/Same number of protons but different numbers of neutrons
[2]
(b) The natural abundance of the two isotopes of uranium is:
235 92U 0.72% 238 92U 99.28%
Use the values to calculate the relative atomic mass of uranium to two decimal places.
Ar(U) = [(235 x 0.72) + (238 x 99.28)]/100 = 237.98
[2]
(c) Sketch the mass spectrum (not to scale) of naturally occurring uranium. Label the axes. [3]
Relative abundance
235 238 m/z
2. (a) State what is meant by relative atomic mass.
The mass of one atom of an element compared to 1/12th the mass of one atom of carbon-12
[2]
(b) Calculate the relative formula mass of washing soda Na2CO3.10H2O
Na2CO3.10H2O = (23 x 2) + 12 + (16 x 3) + 10 (2 + 16) = 286
[2]
(c) On heating, the washing soda crystals lose all the water present. Calculate the mass of the residue when 57.2 g of washing soda is completely dehydrated.
Mass of Na2CO3 in 1mol crystals = 106 g
% Na2CO3 = 106 x 100/286 = 37.1%
Mass of residue = 57.2 g x 37.1 % =21.2 g
[2]
3. An eccentric chemist weighed a piece of paper, wrote his name on it, and re-weighed it. The results were
Weight of paper = 0.8041 g
Weight of paper + name = 0.8053 g
If he wrote his name with pure graphite (a form of carbon), how many carbon atoms were used to write his name?
Mass of graphite used = 0.8053 – 0.8041 = 0.0012 g (1.2 x 10-3 g)
No of C atoms = 1.2 x 10-3 g x 6.02 x 1023 mol-1
12
= 6.02 x 1019
SECTION A
1. Iodate ions in the presence of acid oxidise sulphite ions to sulphate ions. Which one of the following equations correctly represents this reaction?
A IO3- + 2SO32- + 2H+ I2 + 2SO42- + H2O
B 2IO3- + SO32- + 2H+ I2 + SO42- + H2O
C 2IO3- + 5SO32- + 2H+ I2 + 5SO42- + H2O
D 2IO3- + SO32- + 10H+ I2 + SO42- + 5H2O
2. If 8.7 g of potassium sulphate is dissolved in water and made up to 200 cm3, what is the concentration of the solution in mol dm-3?
A 0.05
B 0.20
C 0.25
D 0.50
3. On heating with soda lime. sodium ethanoate decomposes in accordance with the following equation.
CH3COONa (s) + NaOH (s) CH4 (g) + Na2 CO3 (s)
If 8.2 g of sodium ethanoate produced 560 cm3 of methane (at s.t.p.). which one of the following is the percentage yield of the reaction;
A 2.5
B 4.0
C 12.0
D 25.0
4. What mass of iron(II) ammonium sulphate-6-water,Fe(NH4)2(SO4)3.6H2O, (relative molecular mass = 392) must be dissolved in water to produce 1 litre of a solution which is 0.05M with respect to Fe2+(aq)?
A 2.80
B 14.2
C 19.6
D 28.0
5.D 0.0640 g of an unknown sulphate yielded 0.1165 g of barium sulphate when appropriately treated with barium chloride. Which one of the following is the percentage of sulphate in the unknown?
A 75
B 45
C 30
D 25
6. Equal masses of sulphur dioxide and sulphur trioxide are mixed. The mole fraction of sulphur trioxide will be
(Relative atomic masses: O=16, S=32)
A 0.22
B 0.44
C 0.50
D 0.56
E 0.80
7. Calculate the number of protons present in the nuclei of the atoms contained in 2.24 dm3 of oxygen gas, O2, (atomic number of oxygen is 8) at standard temperature and pressure, given that one mole of gas under these conditions occupies 22.4 dm3 and the value of the Avogadro constant L = 6.0 x 1023 mol-1.
It equals
A 9.6 x 1023
B 6.0 x 1022
C 4.8 x 1023
D 9.6 x 1024
8. If L is the Avogadro constant the number of ions in 17.43 g of solid K2SO4 (relative molecular mass 174.3) will be
A L/30
B L/10
C 3L/10
D 30L
9. Chemical analysis showed that a sample of a substance contained only carbon, hydrogen and nitrogen in the following amounts: carbon, 0.060 g; hydrogen, 0.025 g and nitrogen, 0.070 g. What is the empirical formula for this substance?
A CH2N2
B CH3N2
C CH4N2
D CH5N
Section B
1. (a) State what is meant by relative atomic mass.
[2] (b) (i) Calculate the relative formula mass of washing soda Na 2CO3.10H20
[1]
(ii) On heating, the washing soda loses all the water present. Calculate the mass of the residue when 57.2 g of washing soda is completely dehydrated.
[2]
2. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. The mineral beryl, Be3Al2Si6O16, is the main source of beryllium.
Calculate the percentage, by mass, of beryllium in beryl.
% [3]
3. Calcium oxalate may be converted to oxalic acid and calcium sulphate by reaction with dilute sulphuric acid:
CaC2O4 + H2SO4 H2C2O4 + CaSO4
The calcium sulphate precipitated can be hydrated to make plaster of Paris, CaSO4.xH2O. Calculate the value of x in the formula if 1.45 g of plaster of Paris is heated to dryness and 1.36 g of CaSO4 remains.
x = [3]
4. The sulphur content of cystine is 26.7 %. Given that cystine contains two sulphur atoms, what is the molar mass of cystine? Show your working clearly.
[3]
SECTION B
1. (a) Explain why 235 92U and 238 92U are regarded as isotopes of uranium.
Same atomic number but different mass numbers/Same number of protons but different numbers of neutrons
[2]
(b) The natural abundance of the two isotopes of uranium is:
235 92U 0.72% 238 92U 99.28%
Use the values to calculate the relative atomic mass of uranium to two decimal places.
Ar(U) = [(235 x 0.72) + (238 x 99.28)]/100 = 237.98
[2]
(c) Sketch the mass spectrum (not to scale) of naturally occurring uranium. Label the axes. [3]
Relative abundance
235 238 m/z
2. (a) State what is meant by relative atomic mass.
The mass of one atom of an element compared to 1/12th the mass of one atom of carbon-12
[2]
(b) Calculate the relative formula mass of washing soda Na2CO3.10H2O
Na2CO3.10H2O = (23 x 2) + 12 + (16 x 3) + 10 (2 + 16) = 286
[2]
(c) On heating, the washing soda crystals lose all the water present. Calculate the mass of the residue when 57.2 g of washing soda is completely dehydrated.
Mass of Na2CO3 in 1mol crystals = 106 g
% Na2CO3 = 106 x 100/286 = 37.1%
Mass of residue = 57.2 g x 37.1 % =21.2 g
[2]
3. An eccentric chemist weighed a piece of paper, wrote his name on it, and re-weighed it. The results were
Weight of paper = 0.8041 g
Weight of paper + name = 0.8053 g
If he wrote his name with pure graphite (a form of carbon), how many carbon atoms were used to write his name?
Mass of graphite used = 0.8053 – 0.8041 = 0.0012 g (1.2 x 10-3 g)
No of C atoms = 1.2 x 10-3 g x 6.02 x 1023 mol-1
12
= 6.02 x 1019
moles - As/ gce chemistry notes
Molar Volume of Gases
Section A
1. 10 cm3 of a gaseous hydrocarbon needed 30 cm3 of oxygen for complete combustion and 20 cm3 of carbon dioxide were produced. What is the molecular formula of the hydrocarbon?
A CH4
B C2H2
C C2H4
D C2H6
2. What mass of iodine will be produced if 200 cm3 of chlorine gas is passed into 100 cm3 of 0.2 mol dm-3 potassium iodide solution, at 20 oC and a pressure of one atmosphere?
A 1.06 g
B 2.12 g
C 2.54 g
D 5.08 g
3. 10 cm3 of a gas G were mixed with 100 cm3 (an excess) of oxygen in a sealed container at 20 oC and one atmosphere pressure. The gas was then ignited electrically and burned completely in the oxygen. After the reaction the volume of the gas mixture, measured at the same temperature and pressure, was found to be 105 cm3. Which one of the following was G?
A Methane, CH4
B Hydrogen, H2
C Carbon monoxide, CO
D Ethene, C2H4
Section B
Carbon dioxide is a molecular substance. It has properties characteristic of a molecule with low molecular mass. Water dissolves about 1.75 times its own volume of carbon dioxide at 20 oC and a pressure of one atmosphere. Solutions of carbon dioxide are weakly acidic and this can explain why carbon dioxide is used in the preparation of effervescent drinks such as lemonade.
(a) (i) Calculate the relative molecular mass of carbon dioxide. [1]
(ii) The word “relative” means that the mass is compared with something else. What is the mass of carbon dioxide compared with and why? [2]
(iii) Calculate the mass of carbon dioxide dissolved in 150 cm3 of water at 20 oC and atmospheric pressure.
(One mole of any gas occupies a volume of 24 dm3 at 20 oC and a pressure of one atmosphere.) [2]
(iv) Use this value to calculate the molarity of the solution. [2]
Section A
1. 10 cm3 of a gaseous hydrocarbon needed 30 cm3 of oxygen for complete combustion and 20 cm3 of carbon dioxide were produced. What is the molecular formula of the hydrocarbon?
A CH4
B C2H2
C C2H4
D C2H6
2. What mass of iodine will be produced if 200 cm3 of chlorine gas is passed into 100 cm3 of 0.2 mol dm-3 potassium iodide solution, at 20 oC and a pressure of one atmosphere?
A 1.06 g
B 2.12 g
C 2.54 g
D 5.08 g
3. 10 cm3 of a gas G were mixed with 100 cm3 (an excess) of oxygen in a sealed container at 20 oC and one atmosphere pressure. The gas was then ignited electrically and burned completely in the oxygen. After the reaction the volume of the gas mixture, measured at the same temperature and pressure, was found to be 105 cm3. Which one of the following was G?
A Methane, CH4
B Hydrogen, H2
C Carbon monoxide, CO
D Ethene, C2H4
Section B
Carbon dioxide is a molecular substance. It has properties characteristic of a molecule with low molecular mass. Water dissolves about 1.75 times its own volume of carbon dioxide at 20 oC and a pressure of one atmosphere. Solutions of carbon dioxide are weakly acidic and this can explain why carbon dioxide is used in the preparation of effervescent drinks such as lemonade.
(a) (i) Calculate the relative molecular mass of carbon dioxide. [1]
(ii) The word “relative” means that the mass is compared with something else. What is the mass of carbon dioxide compared with and why? [2]
(iii) Calculate the mass of carbon dioxide dissolved in 150 cm3 of water at 20 oC and atmospheric pressure.
(One mole of any gas occupies a volume of 24 dm3 at 20 oC and a pressure of one atmosphere.) [2]
(iv) Use this value to calculate the molarity of the solution. [2]
moles - As/ gce chemistry notes
CALCULATIONS FROM EQUATIONS
Amounts of substances, the mole , molar mass, molar volume of gases (24 litre/dm3) at room temperature and pressure, the Avogadro constant and their use in calculations. (Determination of the Avogadro constant is not required.)
A major problem confronting a chemist when carrying out reactions is to try and understand what they see in the reaction vessels in terms of what is happening between individual atoms and ions, considering their small size;
• The diameter of an atom is about 10 -10 m
OR
• Two million hydrogen atoms would cover an average full stop.
• If a single drop of water could be magnified to the size of the Earth, then on the same scale its atoms would be the size of golf balls.
The masses of atoms
Atoms are very small. A single hydrogen atom for instance, weighs about 2 x 10-24 g (0.000 000 000 000 000 000 000 002g). Numbers this small are awkward to use, so instead of using the actual masses of atoms, a simpler way of expressing the mass of an atom has been found.
Relative atomic mass Ar
When you have finished this section you should be able to:
· Define relative atomic mass in terms of carbon-12
· Deduce relative atomic masses for the elements from the Periodic Table
Since hydrogen is the smallest and lightest atom it was originally used as the standard atom against which all other atoms would be compared.
Relative atomic mass = mass of one atom of an element
mass of one atom of hydrogen
Thus Ar(H) = 1
Ar(O) = 16 each oxygen atom is 16 times heavier than a hydrogen atom
Ar(S) = 32
Ar(C) = 12
Relative atomic masses are now determined by mass spectrometry (see later), and since volatile carbon compounds are used a lot in mass spectrometry, the mass of an atom of 126C is now taken as the reference standard. The modern definition of relative atomic mass is:
Relative atomic mass = mass of one atom of an element
1/12 the mass of one atom of carbon-12
On this scale, carbon-12 has a relative atomic mass of 12.000000, carbon has a relative atomic mass of 12.011011 and hydrogen has a relative atomic mass of 1.00797.
Relative atomic masses are not always whole numbers due to the existence of isotopes. e.g.. Ar(Cl) = 35.5
Note that since relative atomic masses are ratios they have no units.
Approximate relative atomic masses can be found from the mass numbers in your Periodic Table.
Relative molecular mass Mr
When you have finished this section you should be able to:
· Calculate the relative molecular mass of a substance given its formula and a table of relative atomic masses.
A molecule consists of atoms joined together. The mass of a molecule can be found by adding up the masses of the atoms it contains.
The relative molecular mass of a compound is the sum of the relative atomic masses of all the atoms in a molecule of the compound.
carbon dioxide CO2
1 atom of C, Ar(C) = 12 12
2 atoms of O, Ar(O) = 16 32
Mr(CO2) = 44
i.e. 1 molecule of CO2 weighs 44 times as much as 1 atom of hydrogen.
What is the relative molecular mass of sulphuric acid H2SO4?
Mr(H2SO4) = (2 x1) + 32 + (4 x 16) = 98
What is the relative molecular mass of magnesium nitrate Mg(NO3)2
Mr[Mg(NO3)2] = 24 + {2 x (14 + 3 x 16)}
= 24 + {2 x (14 + 48)}
= 148
What is the relative molecular mass of copper sulphate crystals CuSO4.5H2O.
Mr[CuSO4.5H2O] = 64 + 32 + (4 x 16) + 5 x {(2 x 1) + 16}
= 64 + 32 + 64 + (5 x 18)
= 250
Many compounds consist of ions, not molecules. For ionic compounds the formula represents a formula unit , rather than a molecule of the compound. A formula unit for sodium sulphate is Na2SO4. The term relative formula mass can be used for ionic compounds.
Exercise 1
Work out the relative molecular masses of these compounds:
(a) NaOH [40]
(b) KCl [ ]
(c) MgO
(d) Ca(OH)2
(e) HNO3
(f) CuCO3
(g) NH4NO3
(h) CaSO4
(i) Na2CO3.10H2O
(j) Mg(HCO3)2
THE MOLE
The mole is a convenient way of describing a large number of objects (in this case atoms, molecules or ions)
c.f. a dozen eggs, a gross of apples, a ream of paper or a bag of sugar.
It is similar in principle to the way a bank cashier 'counts' the number of coins in a bag by weighing the whole bag.
Suppose
A 2p coin is twice as heavy as a 1p coin.
A 5p coin is five times as heavy as a 1p coin.
A 10p coin is ten times as heavy as a 1p coin.
If a shopkeeper gives the cashier four bags of coins with the following weights
1p bag (100g), 2p bag (200g), 5p bag (500g) and 10p bag (1000g)
What do you know about the number of coins in each bag?
Consider
1p 2p 5p 10p
1 coin 1 2 5 10
2 coins 2 4 10 20
3 coins 3 6 15 30
4 coins 4 8 20 40
Whatever the actual weight of the coins the ratio of weights is always constant 1 : 2 : 5 : 10
As long as equal numbers of coins are taken their weights will always be in the same ratio.
Moles of Atoms
Now consider the following atoms
H C S Mg Ca
Ar 1 12 32 24 40
1 atom
2 atoms
10 atoms Ratio of their weights will always be the same
100 atoms as the ratio of their Ar.
1000 atoms
1 million atoms
??? atoms 1g 12g 32g 24g 40g
Atoms, like coins, can be counted by weighing. The only difference is that millions and millions of atoms are needed before we get a weight that registers on a normal balance.
The number of atoms in 1g of hydrogen atoms is 602 000 000 000 000 000 000 000 (6 x 1023 ) and this amount is called one mole.
A mole of any substance always contains this number (called the Avogadro number) of particles.
Therefore:
Mass of one mole of carbon atoms is 12g.
Mass of one mole of sulphur atoms is 32g.
Mass of one mole of magnesium atoms is 24g.
Mass of one mole of calcium atoms is 40g.
What is the mass of 5 moles of fluorine atoms?
Ar(F) = 19
1 mole of fluorine atoms weighs 19g
5 moles of fluorine atoms weigh 5 x 19g = 95g
How many moles of atoms are there in 1.6g of copper.
Ar(Cu) = 64
1 mole of copper weighs 64g
1.6g copper must be less than 1 mole
Amount of copper = 1.6g
64g
= 0.025 mol
Mass of 1 mole of an element = relative atomic mass in grams
Mass of 1 mole of a compound = relative molecular mass in grams
Number of moles = mass of substance
mass of 1 mole of substance
Exercise 1
1. What is the mass of 1 mole of (a) sodium atoms
(b) cobalt atoms
(c) lead atoms
2. What is the mass of 0.1 moles of (a) barium atoms
(b) copper atoms
(c) tin atoms
Moles of compounds
What is the mass of 2 mol of sulphuric acid?
Mr(H2SO4) = 98
1 mol of sulphuric acid weighs 98g
2 mol sulphuric acid weighs 2 x98g = 196g
What is the mass of 0.1 mol of water?
Mr(H2O) = 18
1 mol water weighs 18g
0.1 mol water weighs 0.1 x 18g = 1.8g
How many moles of sodium hydroxide is 8.0g?
Mr(NaOH) = 40
40g of NaOH is 1 mol
8.0g is less than 1 mol
Amount of NaOH = 8.0g
40g
= 0.2 mol
Exercise 2
1. Calculate the mass of 2 moles of
(a) sodium carbonate Na2CO3
(b) potassium hydroxide KOH
2. Calculate the mass of
(a) 1 mol of sodium chloride NaCl
(b) 0.5 mol of magnesium hydroxide Mg(OH)2
(c) 4 mol of iron (II) chloride
(d) 2.5 mol of sodium carbonate
(e) 0.1 mol of zinc (II) chloride
Exercise 3
Calculate the amount of each of the following:
(a) 30.0 g of oxygen molecules, O2.
(b) 31.0 g of phosphorus molecules, P4.
(c) 50.0 g of calcium carbonate, CaCO3.
Exercise 4
Calculate the mass of each of the following:
(a) 2.50 mol of hydrogen, H2.
(b) 0.500 mol of sodium chloride, NaCl.
(c) 0.250 mol of carbon dioxide, CO2.
Exercise 5
A sample of ammonia, NH3, weighs 1.00 g.
(a) What amount of ammonia is contained in this sample?
(b) What mass of sulphur dioxide, SO2, contains the same number of molecules as are in 1.00 g of ammonia?
Calculating reacting masses
A chemical equation such as
N2 (g) + 3H2 (g) 2NH3 (g)
is a kind of chemical balance sheet; it tells us that one mole of nitrogen reacts with three moles moles of hydrogen to yield two moles of ammonia. (it does not tell us anything about the rate of the reaction or the conditions necessary to bring it about).
Such an equation is an essential starting point for many experiments and calculations; it tells us the proportions in which the substances react and the products are formed.
When you have finished this section you should be able to:
· Do simple reacting mass calculations based on a given chemical equation.
Worked Example
What mass of iodine will react completely with 10.0 g of aluminium?
The problem is a bit more complicated than those you have done previously because it involves several steps. Each step is simple but you may not immediately see where to start. One approach to solving multi-step problems is given below. You may find the approach useful to you in solving more difficult problems.
Ask yourself three questions:
1. What do I know?
In this case the answer would be:
(a) the equation for the reaction
(b) the mass of aluminium used
In some problems you will be given the equation; in this case the equation is not provided and you would be expected to write it down from your general chemical knowledge.
2. What can I get from what I know?
(a) from the equation , you can find the ratio of reacting amounts
(b) from the mass of aluminium you can calculate the amount, provided you can look up the molar mass.
3. Can I now see how to get the final answer?
Usually the answer will be ‘yes’, but you may have ask Question 2 again, now you know more than you did at the start.
(a) from the amount of aluminium and the ratio of reacting masses, you can calculate the amount of iodine
(b) from the amount of iodine, you can get the mass, using the molar mass.
Now we will go through each step, doing the necessary calculations.
1. the balanced equation for the reaction is
2Al (s) + 3I2 (s) 2AlI3 (s)
this equation tells us that 2 mol of Al reacts with 3 mol of I2; so we can write the ratio
amount of Al = 2
amount of I2 3
2. calculate the amount of aluminium using n=m/M.
n = 10.0 g = 0.370 mol
27.0 g mol-1
3. calculate the amount of iodine which reacts with this amount of aluminium by substituting in the expression based on the equation.
amount of Al = 2
amount of I2 3
amount of iodine = 3/2 x 0.370 mol = 0.555 mol
4. Calculate the mass of iodine from the amount using n = m/M in the form m = nM.
M = nM = 0.555 mol x 254 g mol-1 = 141 g
Example 2
What mass of magnesium oxide can be obtained from the complete combustion of 12g of magnesium?
2Mg (s) + O2 (g) 2MgO (s)
2 mol 1 mol 2 mol
Since Ar (Mg) = 24 Mr (MgO) = 40
2 x 24g Mg 2 x 40g MgO
48g magnesium forms 80g magnesium oxide
1g magnesium forms 80g magnesium oxide
48
12g magnesium forms 12 x 80g magnesium oxide
48
= 20 g magnesium oxide
c.f. 3 packets of crisps cost 39p. What do 7 packets of crisps cost?
Exercise 6
(a) What mass of magnesium would react completely with 16.0 g of sulphur?
Mg (s) + S (s) MgS (s)
(b) What mass of oxygen would be produced by completely decomposing 4.25 g of sodium nitrate, NaNO3?
2NaNO3 (s) 2NaNO2 (s) + O2 (g)
(c) What mass of phosphorus(V) oxide, P2O5, would be formed by complete oxidation of 4.00 g of phosphorus?
4P (s) + 5O2 (g) 2P2O5 (s)
(d) When 0.27 g of aluminium is added to excess copper(II) sulphate solution, 0.96 g of copper is precipitated. Deduce the equation for the reaction.
Molar Volumes of Gases
Calculation of empirical and molecular formulae from analytical data and molar masses
The empirical formula of a compound is the simplest form of the ration of the atoms of different elements in it. The molecular formula tells the actual number of each kind of atom in one molecule of the substance.
For example, the molecular formula of phosphorus(V) oxide is P4O10, whereas its empirical formula is P2O5.
When you have finished this section you should be able to:
· Calculate the empirical formula of a compound given either
(a) the masses of constituents in a sample, or
(b) the composition in terms of the mass percentages of the constituents.
Calculating empirical formula from the masses of constituents
To determine the empirical formula of a compound, we must first calculate the amount of each substance present in a sample and then calculate the simplest whole number ratio of the amounts.
It is convenient to set out the results in the form of a table. In the following example we will go through the procedure step by step, establishing the procedure as we go.
Worked Example
An 18.3 g sample of a hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its empirical formula.
Solution
1. List the mass of each component and its molar mass. Although water is a molecule, in the calculation treat it in the same way as we do atoms.
Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
2. From this information calculate the amount of each substance present using the expression n = m/M.
Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
Amount /mol
4.0/40.0
= 0.10
7.1/35.5
= 0.20
7.2/18.0
= 0.40
This result means that in the given sample there is 0.10 mol of calcium, 0.20 mol of chlorine and 0.40 mol of water.
3. Calculate the relative amount of each substance by dividing each amount by the smallest amount.
Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
Amount /mol
4.0/40.0
= 0.10
7.1/35.5
= 0.20
7.2/18.0
= 0.40
Amount/smallest amount
= relative amount
0.10/0.10
= 1.0
0.20/0.10
= 2.0
0.40/0.10
=4.0
The relative amounts are in the simple ratio 1:2:4.
From this result you can see that the empirical formula is CaCl2.4H2O
Exercise 1
A sample of hydrated compound was analysed and found to contain 2.10 g of cobalt, 1.14 g of sulphur, 2.28 g of oxygen and 4.50 g of water.
Calculate its empirical formula.
CoSO4.7H2O
A modification of this type of problem is to determine the ratio of the amount of water to the amount of anhydrous compound.
Exercise 2
10.00 g of hydrated barium chloride is heated until all the water is driven off. The mass of anhydrous compound is 8.53 g.
Determine the value of x in BaCl2.xH2O.
BaCl2.2H2O x=2
You should be prepared for variations to this type of problem.
Exercise 3
When 127 g of copper combine with oxygen. 143 g of an oxide are formed. What is the empirical formula of the oxide?
[notice here that the mass of oxygen is not given to you – you must obtain it by subtraction]
Calculating empirical formula from percentage composition by mass
The result of the analysis of a compound may also be given in terms of the percentage composition by mass.
Worked Example
An organic compound was analysed and was found to have the following percentage composition by mass: 48.8% carbon, 13.5% hydrogen and 37.7% nitrogen.
Calculate the empirical formula of the compound.
Solution
If we assume the mass of the sample is 100.0 g, we can write immediately the mass of each substance: 48.8 g carbon, 13.5 g hydrogen and 37.7 g nitrogen. Then we set up a table as before.
C
H
N
Mass /g
48.8
13.5
37.7
Molar mass /g mol-1
12.0
1.00
37.7
Amount /mol
48.8/12.0
= 4.07
13.5/1.00
= 13.5
37.7/14
= 2.69
Amount/smallest amount
= relative amount
4.07/2.69
= 1.51
13.5/2.69
= 5.02
2.69/2.69
= 1.00
Simplest ratio of relative amounts
3
10
2
Empirical formula = C3H10N2
Values close to whole numbers are ‘rounded off’ in order to get a simple ratio. This is justified because small differences from whole numbers are probably due to experimental error. In the above example however we cannot justify rounding off 1.51 to 1 or 2, but we can obtain a simple ratio by multiplying the relative amounts by two.
Try the following exercises where you must decide whether to round off or multiply by a factor.
Exercise 4
A compound of carbon, hydrogen and oxygen contains 40.0% carbon, 6.6% hydrogen and 53.4% oxygen.
Calculate its empirical formula.
CH2O
Exercise 5
Determine the formula of a mineral with the following mass composition:
Na = 12.1%, Al = 14.2%, Si = 22.1%, O = 42.1%, H2O = 9.48%.
Na2Al2Si3O10.2H2O
Exercise 6
A 10.0 g sample of a compound contained 3.91 g of carbon, 0.87 g of hydrogen and the remainder is oxygen.
Calculate the empirical formula of the compound.
C3H8O3
Calculation of reacting masses and volumes of substances, including examples in which some reactants are in excess.
In some chemical reactions it may be that one or more of the reactants is in excess and is not completely used up in the reaction. The amount of product is determined by the amount of the reactant which is not in excess and is therefore used up completely in the reaction. This is called the limiting reactant.
Example 1
5.00 g of iron and 5.00 g of sulphur are heated together to form iron(II) sulphide. What mass of product is formed.
Fe (s) + S (s) FeS (s)
1 mol 1 mol 1 mol
56 g 32 g 88 g
Amount of Fe = 5/56 mol = 0.0893 mol
Amount of S = 5/32 mol = 0.156 mol
There is not enough Fe to react with 0.156 mol of S so Fe is the limiting reactant.
0.0893 mol Fe forms 0.0893 mol of FeS
Mass of FeS = 0.0893 x 88 g = 7.86 g
Exercise 1
(a) In the blast furnace, the overall reaction is
2Fe2O3 (s) + 3C (s) 3CO2 (g) + 4Fe (s)
what is the maximum mass of iron that can be obtained from700 tonnes of iron(III) oxide and 70 tonnes of coke? (1 tonne = 1000 kg)
436 tonnes
(b) In the manufacture of calcium carbide
CaO (s) + 3C (s) CaC2 (s) + CO (g)
What is the maximum mass of calcium carbide that can be obtained from 40 kg of quicklime and 40 kg of coke?
46 kg
(c) In the manufacture of the fertiliser ammonium sulphate
H2SO4 (aq) + 2NH3 (g) (NH4)2SO4 (aq)
What is the maximum mass of ammonium sulphatethat can be obtained from 2.0 kg of sulphuric acid and 1.0 kg of ammonia?
2.7 kg
(d) In the Solvay process, ammonia is recovered by the reaction
2NH4Cl (s) CaO (s) CaCl2 (s) + H2O (g) + 2NH3 (g)
What is the maximum amount of ammonia that can be recovered from 2.00 x 103 kg of ammonium chloride and 500 kg of quicklime?
304 kg
(e) In the Thermit reaction
2Al (s) + Cr2O3 (s) 2Cr (s) + Al2O3 (s)
Calculate the percentage yield when 180 g of chromium are obtained from a reaction between 100 g of aluminium and 400 g of chromium(III) oxide.
93.5%
Amounts of substances, the mole , molar mass, molar volume of gases (24 litre/dm3) at room temperature and pressure, the Avogadro constant and their use in calculations. (Determination of the Avogadro constant is not required.)
A major problem confronting a chemist when carrying out reactions is to try and understand what they see in the reaction vessels in terms of what is happening between individual atoms and ions, considering their small size;
• The diameter of an atom is about 10 -10 m
OR
• Two million hydrogen atoms would cover an average full stop.
• If a single drop of water could be magnified to the size of the Earth, then on the same scale its atoms would be the size of golf balls.
The masses of atoms
Atoms are very small. A single hydrogen atom for instance, weighs about 2 x 10-24 g (0.000 000 000 000 000 000 000 002g). Numbers this small are awkward to use, so instead of using the actual masses of atoms, a simpler way of expressing the mass of an atom has been found.
Relative atomic mass Ar
When you have finished this section you should be able to:
· Define relative atomic mass in terms of carbon-12
· Deduce relative atomic masses for the elements from the Periodic Table
Since hydrogen is the smallest and lightest atom it was originally used as the standard atom against which all other atoms would be compared.
Relative atomic mass = mass of one atom of an element
mass of one atom of hydrogen
Thus Ar(H) = 1
Ar(O) = 16 each oxygen atom is 16 times heavier than a hydrogen atom
Ar(S) = 32
Ar(C) = 12
Relative atomic masses are now determined by mass spectrometry (see later), and since volatile carbon compounds are used a lot in mass spectrometry, the mass of an atom of 126C is now taken as the reference standard. The modern definition of relative atomic mass is:
Relative atomic mass = mass of one atom of an element
1/12 the mass of one atom of carbon-12
On this scale, carbon-12 has a relative atomic mass of 12.000000, carbon has a relative atomic mass of 12.011011 and hydrogen has a relative atomic mass of 1.00797.
Relative atomic masses are not always whole numbers due to the existence of isotopes. e.g.. Ar(Cl) = 35.5
Note that since relative atomic masses are ratios they have no units.
Approximate relative atomic masses can be found from the mass numbers in your Periodic Table.
Relative molecular mass Mr
When you have finished this section you should be able to:
· Calculate the relative molecular mass of a substance given its formula and a table of relative atomic masses.
A molecule consists of atoms joined together. The mass of a molecule can be found by adding up the masses of the atoms it contains.
The relative molecular mass of a compound is the sum of the relative atomic masses of all the atoms in a molecule of the compound.
carbon dioxide CO2
1 atom of C, Ar(C) = 12 12
2 atoms of O, Ar(O) = 16 32
Mr(CO2) = 44
i.e. 1 molecule of CO2 weighs 44 times as much as 1 atom of hydrogen.
What is the relative molecular mass of sulphuric acid H2SO4?
Mr(H2SO4) = (2 x1) + 32 + (4 x 16) = 98
What is the relative molecular mass of magnesium nitrate Mg(NO3)2
Mr[Mg(NO3)2] = 24 + {2 x (14 + 3 x 16)}
= 24 + {2 x (14 + 48)}
= 148
What is the relative molecular mass of copper sulphate crystals CuSO4.5H2O.
Mr[CuSO4.5H2O] = 64 + 32 + (4 x 16) + 5 x {(2 x 1) + 16}
= 64 + 32 + 64 + (5 x 18)
= 250
Many compounds consist of ions, not molecules. For ionic compounds the formula represents a formula unit , rather than a molecule of the compound. A formula unit for sodium sulphate is Na2SO4. The term relative formula mass can be used for ionic compounds.
Exercise 1
Work out the relative molecular masses of these compounds:
(a) NaOH [40]
(b) KCl [ ]
(c) MgO
(d) Ca(OH)2
(e) HNO3
(f) CuCO3
(g) NH4NO3
(h) CaSO4
(i) Na2CO3.10H2O
(j) Mg(HCO3)2
THE MOLE
The mole is a convenient way of describing a large number of objects (in this case atoms, molecules or ions)
c.f. a dozen eggs, a gross of apples, a ream of paper or a bag of sugar.
It is similar in principle to the way a bank cashier 'counts' the number of coins in a bag by weighing the whole bag.
Suppose
A 2p coin is twice as heavy as a 1p coin.
A 5p coin is five times as heavy as a 1p coin.
A 10p coin is ten times as heavy as a 1p coin.
If a shopkeeper gives the cashier four bags of coins with the following weights
1p bag (100g), 2p bag (200g), 5p bag (500g) and 10p bag (1000g)
What do you know about the number of coins in each bag?
Consider
1p 2p 5p 10p
1 coin 1 2 5 10
2 coins 2 4 10 20
3 coins 3 6 15 30
4 coins 4 8 20 40
Whatever the actual weight of the coins the ratio of weights is always constant 1 : 2 : 5 : 10
As long as equal numbers of coins are taken their weights will always be in the same ratio.
Moles of Atoms
Now consider the following atoms
H C S Mg Ca
Ar 1 12 32 24 40
1 atom
2 atoms
10 atoms Ratio of their weights will always be the same
100 atoms as the ratio of their Ar.
1000 atoms
1 million atoms
??? atoms 1g 12g 32g 24g 40g
Atoms, like coins, can be counted by weighing. The only difference is that millions and millions of atoms are needed before we get a weight that registers on a normal balance.
The number of atoms in 1g of hydrogen atoms is 602 000 000 000 000 000 000 000 (6 x 1023 ) and this amount is called one mole.
A mole of any substance always contains this number (called the Avogadro number) of particles.
Therefore:
Mass of one mole of carbon atoms is 12g.
Mass of one mole of sulphur atoms is 32g.
Mass of one mole of magnesium atoms is 24g.
Mass of one mole of calcium atoms is 40g.
What is the mass of 5 moles of fluorine atoms?
Ar(F) = 19
1 mole of fluorine atoms weighs 19g
5 moles of fluorine atoms weigh 5 x 19g = 95g
How many moles of atoms are there in 1.6g of copper.
Ar(Cu) = 64
1 mole of copper weighs 64g
1.6g copper must be less than 1 mole
Amount of copper = 1.6g
64g
= 0.025 mol
Mass of 1 mole of an element = relative atomic mass in grams
Mass of 1 mole of a compound = relative molecular mass in grams
Number of moles = mass of substance
mass of 1 mole of substance
Exercise 1
1. What is the mass of 1 mole of (a) sodium atoms
(b) cobalt atoms
(c) lead atoms
2. What is the mass of 0.1 moles of (a) barium atoms
(b) copper atoms
(c) tin atoms
Moles of compounds
What is the mass of 2 mol of sulphuric acid?
Mr(H2SO4) = 98
1 mol of sulphuric acid weighs 98g
2 mol sulphuric acid weighs 2 x98g = 196g
What is the mass of 0.1 mol of water?
Mr(H2O) = 18
1 mol water weighs 18g
0.1 mol water weighs 0.1 x 18g = 1.8g
How many moles of sodium hydroxide is 8.0g?
Mr(NaOH) = 40
40g of NaOH is 1 mol
8.0g is less than 1 mol
Amount of NaOH = 8.0g
40g
= 0.2 mol
Exercise 2
1. Calculate the mass of 2 moles of
(a) sodium carbonate Na2CO3
(b) potassium hydroxide KOH
2. Calculate the mass of
(a) 1 mol of sodium chloride NaCl
(b) 0.5 mol of magnesium hydroxide Mg(OH)2
(c) 4 mol of iron (II) chloride
(d) 2.5 mol of sodium carbonate
(e) 0.1 mol of zinc (II) chloride
Exercise 3
Calculate the amount of each of the following:
(a) 30.0 g of oxygen molecules, O2.
(b) 31.0 g of phosphorus molecules, P4.
(c) 50.0 g of calcium carbonate, CaCO3.
Exercise 4
Calculate the mass of each of the following:
(a) 2.50 mol of hydrogen, H2.
(b) 0.500 mol of sodium chloride, NaCl.
(c) 0.250 mol of carbon dioxide, CO2.
Exercise 5
A sample of ammonia, NH3, weighs 1.00 g.
(a) What amount of ammonia is contained in this sample?
(b) What mass of sulphur dioxide, SO2, contains the same number of molecules as are in 1.00 g of ammonia?
Calculating reacting masses
A chemical equation such as
N2 (g) + 3H2 (g) 2NH3 (g)
is a kind of chemical balance sheet; it tells us that one mole of nitrogen reacts with three moles moles of hydrogen to yield two moles of ammonia. (it does not tell us anything about the rate of the reaction or the conditions necessary to bring it about).
Such an equation is an essential starting point for many experiments and calculations; it tells us the proportions in which the substances react and the products are formed.
When you have finished this section you should be able to:
· Do simple reacting mass calculations based on a given chemical equation.
Worked Example
What mass of iodine will react completely with 10.0 g of aluminium?
The problem is a bit more complicated than those you have done previously because it involves several steps. Each step is simple but you may not immediately see where to start. One approach to solving multi-step problems is given below. You may find the approach useful to you in solving more difficult problems.
Ask yourself three questions:
1. What do I know?
In this case the answer would be:
(a) the equation for the reaction
(b) the mass of aluminium used
In some problems you will be given the equation; in this case the equation is not provided and you would be expected to write it down from your general chemical knowledge.
2. What can I get from what I know?
(a) from the equation , you can find the ratio of reacting amounts
(b) from the mass of aluminium you can calculate the amount, provided you can look up the molar mass.
3. Can I now see how to get the final answer?
Usually the answer will be ‘yes’, but you may have ask Question 2 again, now you know more than you did at the start.
(a) from the amount of aluminium and the ratio of reacting masses, you can calculate the amount of iodine
(b) from the amount of iodine, you can get the mass, using the molar mass.
Now we will go through each step, doing the necessary calculations.
1. the balanced equation for the reaction is
2Al (s) + 3I2 (s) 2AlI3 (s)
this equation tells us that 2 mol of Al reacts with 3 mol of I2; so we can write the ratio
amount of Al = 2
amount of I2 3
2. calculate the amount of aluminium using n=m/M.
n = 10.0 g = 0.370 mol
27.0 g mol-1
3. calculate the amount of iodine which reacts with this amount of aluminium by substituting in the expression based on the equation.
amount of Al = 2
amount of I2 3
amount of iodine = 3/2 x 0.370 mol = 0.555 mol
4. Calculate the mass of iodine from the amount using n = m/M in the form m = nM.
M = nM = 0.555 mol x 254 g mol-1 = 141 g
Example 2
What mass of magnesium oxide can be obtained from the complete combustion of 12g of magnesium?
2Mg (s) + O2 (g) 2MgO (s)
2 mol 1 mol 2 mol
Since Ar (Mg) = 24 Mr (MgO) = 40
2 x 24g Mg 2 x 40g MgO
48g magnesium forms 80g magnesium oxide
1g magnesium forms 80g magnesium oxide
48
12g magnesium forms 12 x 80g magnesium oxide
48
= 20 g magnesium oxide
c.f. 3 packets of crisps cost 39p. What do 7 packets of crisps cost?
Exercise 6
(a) What mass of magnesium would react completely with 16.0 g of sulphur?
Mg (s) + S (s) MgS (s)
(b) What mass of oxygen would be produced by completely decomposing 4.25 g of sodium nitrate, NaNO3?
2NaNO3 (s) 2NaNO2 (s) + O2 (g)
(c) What mass of phosphorus(V) oxide, P2O5, would be formed by complete oxidation of 4.00 g of phosphorus?
4P (s) + 5O2 (g) 2P2O5 (s)
(d) When 0.27 g of aluminium is added to excess copper(II) sulphate solution, 0.96 g of copper is precipitated. Deduce the equation for the reaction.
Molar Volumes of Gases
Calculation of empirical and molecular formulae from analytical data and molar masses
The empirical formula of a compound is the simplest form of the ration of the atoms of different elements in it. The molecular formula tells the actual number of each kind of atom in one molecule of the substance.
For example, the molecular formula of phosphorus(V) oxide is P4O10, whereas its empirical formula is P2O5.
When you have finished this section you should be able to:
· Calculate the empirical formula of a compound given either
(a) the masses of constituents in a sample, or
(b) the composition in terms of the mass percentages of the constituents.
Calculating empirical formula from the masses of constituents
To determine the empirical formula of a compound, we must first calculate the amount of each substance present in a sample and then calculate the simplest whole number ratio of the amounts.
It is convenient to set out the results in the form of a table. In the following example we will go through the procedure step by step, establishing the procedure as we go.
Worked Example
An 18.3 g sample of a hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its empirical formula.
Solution
1. List the mass of each component and its molar mass. Although water is a molecule, in the calculation treat it in the same way as we do atoms.
Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
2. From this information calculate the amount of each substance present using the expression n = m/M.
Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
Amount /mol
4.0/40.0
= 0.10
7.1/35.5
= 0.20
7.2/18.0
= 0.40
This result means that in the given sample there is 0.10 mol of calcium, 0.20 mol of chlorine and 0.40 mol of water.
3. Calculate the relative amount of each substance by dividing each amount by the smallest amount.
Ca
Cl
H2O
Mass /g
4.0
7.1
7.2
Molar mass /g mol-1
40.0
35.5
18.0
Amount /mol
4.0/40.0
= 0.10
7.1/35.5
= 0.20
7.2/18.0
= 0.40
Amount/smallest amount
= relative amount
0.10/0.10
= 1.0
0.20/0.10
= 2.0
0.40/0.10
=4.0
The relative amounts are in the simple ratio 1:2:4.
From this result you can see that the empirical formula is CaCl2.4H2O
Exercise 1
A sample of hydrated compound was analysed and found to contain 2.10 g of cobalt, 1.14 g of sulphur, 2.28 g of oxygen and 4.50 g of water.
Calculate its empirical formula.
CoSO4.7H2O
A modification of this type of problem is to determine the ratio of the amount of water to the amount of anhydrous compound.
Exercise 2
10.00 g of hydrated barium chloride is heated until all the water is driven off. The mass of anhydrous compound is 8.53 g.
Determine the value of x in BaCl2.xH2O.
BaCl2.2H2O x=2
You should be prepared for variations to this type of problem.
Exercise 3
When 127 g of copper combine with oxygen. 143 g of an oxide are formed. What is the empirical formula of the oxide?
[notice here that the mass of oxygen is not given to you – you must obtain it by subtraction]
Calculating empirical formula from percentage composition by mass
The result of the analysis of a compound may also be given in terms of the percentage composition by mass.
Worked Example
An organic compound was analysed and was found to have the following percentage composition by mass: 48.8% carbon, 13.5% hydrogen and 37.7% nitrogen.
Calculate the empirical formula of the compound.
Solution
If we assume the mass of the sample is 100.0 g, we can write immediately the mass of each substance: 48.8 g carbon, 13.5 g hydrogen and 37.7 g nitrogen. Then we set up a table as before.
C
H
N
Mass /g
48.8
13.5
37.7
Molar mass /g mol-1
12.0
1.00
37.7
Amount /mol
48.8/12.0
= 4.07
13.5/1.00
= 13.5
37.7/14
= 2.69
Amount/smallest amount
= relative amount
4.07/2.69
= 1.51
13.5/2.69
= 5.02
2.69/2.69
= 1.00
Simplest ratio of relative amounts
3
10
2
Empirical formula = C3H10N2
Values close to whole numbers are ‘rounded off’ in order to get a simple ratio. This is justified because small differences from whole numbers are probably due to experimental error. In the above example however we cannot justify rounding off 1.51 to 1 or 2, but we can obtain a simple ratio by multiplying the relative amounts by two.
Try the following exercises where you must decide whether to round off or multiply by a factor.
Exercise 4
A compound of carbon, hydrogen and oxygen contains 40.0% carbon, 6.6% hydrogen and 53.4% oxygen.
Calculate its empirical formula.
CH2O
Exercise 5
Determine the formula of a mineral with the following mass composition:
Na = 12.1%, Al = 14.2%, Si = 22.1%, O = 42.1%, H2O = 9.48%.
Na2Al2Si3O10.2H2O
Exercise 6
A 10.0 g sample of a compound contained 3.91 g of carbon, 0.87 g of hydrogen and the remainder is oxygen.
Calculate the empirical formula of the compound.
C3H8O3
Calculation of reacting masses and volumes of substances, including examples in which some reactants are in excess.
In some chemical reactions it may be that one or more of the reactants is in excess and is not completely used up in the reaction. The amount of product is determined by the amount of the reactant which is not in excess and is therefore used up completely in the reaction. This is called the limiting reactant.
Example 1
5.00 g of iron and 5.00 g of sulphur are heated together to form iron(II) sulphide. What mass of product is formed.
Fe (s) + S (s) FeS (s)
1 mol 1 mol 1 mol
56 g 32 g 88 g
Amount of Fe = 5/56 mol = 0.0893 mol
Amount of S = 5/32 mol = 0.156 mol
There is not enough Fe to react with 0.156 mol of S so Fe is the limiting reactant.
0.0893 mol Fe forms 0.0893 mol of FeS
Mass of FeS = 0.0893 x 88 g = 7.86 g
Exercise 1
(a) In the blast furnace, the overall reaction is
2Fe2O3 (s) + 3C (s) 3CO2 (g) + 4Fe (s)
what is the maximum mass of iron that can be obtained from700 tonnes of iron(III) oxide and 70 tonnes of coke? (1 tonne = 1000 kg)
436 tonnes
(b) In the manufacture of calcium carbide
CaO (s) + 3C (s) CaC2 (s) + CO (g)
What is the maximum mass of calcium carbide that can be obtained from 40 kg of quicklime and 40 kg of coke?
46 kg
(c) In the manufacture of the fertiliser ammonium sulphate
H2SO4 (aq) + 2NH3 (g) (NH4)2SO4 (aq)
What is the maximum mass of ammonium sulphatethat can be obtained from 2.0 kg of sulphuric acid and 1.0 kg of ammonia?
2.7 kg
(d) In the Solvay process, ammonia is recovered by the reaction
2NH4Cl (s) CaO (s) CaCl2 (s) + H2O (g) + 2NH3 (g)
What is the maximum amount of ammonia that can be recovered from 2.00 x 103 kg of ammonium chloride and 500 kg of quicklime?
304 kg
(e) In the Thermit reaction
2Al (s) + Cr2O3 (s) 2Cr (s) + Al2O3 (s)
Calculate the percentage yield when 180 g of chromium are obtained from a reaction between 100 g of aluminium and 400 g of chromium(III) oxide.
93.5%
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