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