Relative Masses of Atoms and Molecules Worked ExamplesIGCSE Chemistry 0620

(a) The standard is an atom of carbon-12, $^{12}\text{C}$. Relative masses compare with $\tfrac{1}{12}$ of the mass of one carbon-12 atom. On this scale a magnesium-24 atom has relative mass $2 \times 12 = 24$, which is why it balances two carbon-12 atoms.

(b) $A_r$ is the average mass of the isotopes of an element compared to $\tfrac{1}{12}$ of the mass of an atom of carbon-12.

(c) $A_r$ is an average that takes account of how common each isotope is. A mixture of atoms of mass 35 and 37 has an average between the two — here closer to 35, because chlorine-35 is more abundant.

Count the atoms. The subscript 3 outside the bracket multiplies everything inside it: Al 2, S $1 \times 3 = 3$, O $4 \times 3 = 12$.

Use $A_r$: Al = 27, S = 32, O = 16.

The answer is D. A ignores the bracket completely (one S, four O). B multiplies only the S by 3. C multiplies only the O by 3.

(a) Count the atoms in the diagram: C 2, H 4, O 2 (molecular formula $\text{C}_2\text{H}_4\text{O}_2$).

(b)(i) Mg 1, N 2, O 6.

(ii) N 2, H 8, S 1, O 4.

(c) Magnesium nitrate is an ionic compound. It is made of ions in a lattice, not molecules, so it has no molecular mass — the relative formula mass is the sum of the $A_r$ values in its formula.

Multiply each isotope's mass by its percentage and add. (The percentages add up to 100, so divide by 100.)

Divide by the total abundance.

Check: the answer lies between 85 and 87 and closer to 85, the more abundant isotope. ✓

Treat the ratio as 5 atoms in total: 1 atom of mass 10 and 4 atoms of mass 11. Find the total mass.

Divide by the total number of atoms in the ratio, $1 + 4 = 5$ — not by 100.

(a) The nucleon number of the atoms is $9 + 10 = 19$, exactly equal to $A_r$. An average equal to one mass means there is only one mass to average: fluorine has only one (naturally occurring) isotope.

(b) Weighted average of the two isotopes:

(a) Subtract the known atoms, then divide by the number of X atoms.

$A_r = 56$ is iron, Fe (consistent with iron(III) oxide).

(b) $A_r(\text{Y}) = 74.5 - 35.5 = 39$: potassium, K — which is in Group I. ✓

(c) $A_r(\text{Z}) = 34 - 3(1) = 31$: phosphorus, P.

(a) Use $A_r$: N = 14, H = 1, Br = 80.

(b) At the same temperature, a gas with a lower relative molecular mass diffuses faster. Ammonia (17) is much lighter than hydrogen bromide (81), so ammonia travels further along the tube in the same time — the gases meet nearer the hydrobromic acid end.

(c) The two gases combine to form a solid.

(a) The point at $1.28$ g of copper ($1.52$ g of oxide) lies below the line; the line predicts $1.60$ g. Less oxide formed than expected: not all of the copper reacted — for example it was not heated for long enough (or not heated to constant mass).

(b) Go up from $1.12$ g on the horizontal axis to the line, then across: $1.40$ g of copper(II) oxide.

(c) From the equation, $2 \times 64 = 128$ g of copper makes $2 \times 80 = 160$ g of copper(II) oxide — the same ratio as $64 : 80$. Scale in simple proportion:

Let the percentage of silver-107 be $x$. Then the percentage of silver-109 is $100 - x$.

Multiply out and solve.

Quick check without algebra: $107.9$ lies $0.9$ above 107 and $1.1$ below 109 — the gap is 2. The share of silver-107 is $1.1 \div 2 = 0.55 = 55\%$. Check: $(107 \times 55 + 109 \times 45) \div 100 = 107.9$. ✓

(a) Mass of iron(III) oxide in the ore:

$M_r(\text{Fe}_2\text{O}_3) = 2(56) + 3(16) = 160$. In every 160 tonnes, $2 \times 56 = 112$ tonnes is iron.

Scale in simple proportion:

(b) From the equation, 160 tonnes of $\text{Fe}_2\text{O}_3$ reacts with $3 \times 28 = 84$ tonnes of CO.

A $\text{Cl}_2$ molecule contains two atoms, each of which can be either isotope. List the combinations:

chlorine-35 + chlorine-35: $70$; chlorine-35 + chlorine-37: $72$; chlorine-37 + chlorine-37: $74$. Three different values: C.

D counts 35–37 and 37–35 as different, but they are the same molecule. A thinks of the average, 71 — that is the $M_r$ used in calculations, but no single molecule has a mass of 71. B misses the mixed molecule.