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