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^- + 2SO₃^2- + 2H⁺ I₂ + 2SO₄^2- + H₂O

B 2IO₃^- + SO₃^2- + 2H⁺ I₂ + SO₄^2- + H₂O

C 2IO₃^- + 5SO₃^2- + 2H⁺ I₂ + 5SO₄^2- + H₂O

D 2IO₃^- + SO₃^2- + 10H⁺ I₂ + SO₄^2- + 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 1/12^th 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