Answers · Relative masses

Where does the mass go? — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. RQ9 · Practice

Model answer. Reacting masses go in simple proportion: they keep the same ratio, so multiply or divide both by the same number. (a) 1.00 g is 4.00 g divided by 4, so the white solid is 2.56 ÷ 4 = 0.64 g. (b) 6.00 g is 1.5 times 4.00 g, so it is 1.5 × 2.56 = 3.84 g. Check: 0.64 g of every 1.00 g is the same ratio as 160 of every 250.

Checked automatically

(a) White solid from 1.00 g: 0.64 g (within ±0.01 g)
- Common wrong answer 0.36 : That is the water lost. The white solid is what stays: 2.56 g of every 4.00 g.
(b) White solid from 6.00 g: 3.84 g (within ±0.04 g)
- Common wrong answer 4.56 : Scale the white solid; do not subtract the same 1.44 g of water. 6.00 g of crystals lose 1.5 × 1.44 g.
- Common wrong answer 2.16 : That is the water lost, 1.5 × 1.44 g. The question asks for the white solid left.

A full answer includes

(a) I saw that 1.00 g is 4.00 g ÷ 4.
(a) I wrote 0.64 g.
(b) I saw that 6.00 g is 1.5 × 4.00 g.
(b) I wrote 3.84 g.
I did not subtract the same 1.44 g of water each time.

Also accept

(a) 0.64 g (b) 3.84 g

Watch for

“A bigger sample loses the same mass of water.” — Every formula unit loses its water, so 1.5 times the crystals lose 1.5 times the water: the masses keep the same ratio.
“Mass is conserved, so the basin keeps all its mass.” — Mass is conserved in the reaction, but the water leaves the basin as steam, so the basin gets lighter.

2. RQ10 · Practice

Model answer. The relative formula mass is the sum of the relative atomic masses, so 148 = A_r of M + 2 × (14 + 3 × 16). One NO₃ group is 14 + 48 = 62, and the 2 after the bracket doubles it: 124. So the A_r of M is 148 − 124 = 24, which is magnesium: the compound is magnesium nitrate, Mg(NO₃)₂. Check: 24 + 124 = 148.

Checked automatically

A_r of M: 24 (exact value)
- Common wrong answer 86 : The 2 after the bracket doubles the whole NO₃ group: 2 × 62 = 124, not 62.
- Common wrong answer 38 : The 2 doubles the nitrogen too: 2 × (14 + 3 × 16) = 124.
- Common wrong answer 12 : The 2 counts NO₃ groups, not M atoms. There is one M, so it has all of the 24 that is left.

A full answer includes

I worked out the two nitrate groups: 2 × (14 + 3 × 16) = 124.
I took 124 away from 148.
I wrote 24.
I identified M as magnesium.
I did not double only the oxygen atoms in the bracket.

Also accept

24; magnesium

Watch for

“The number after a bracket multiplies only the last symbol.” — The number after a bracket multiplies everything inside the bracket, the nitrogen as well as the oxygen.
“The 2 in M(NO₃)₂ means there are two atoms of M.” — The 2 comes after the bracket, so it counts NO₃ groups. The formula has one M.

3. RQ11 · Practice

Model answer. Fe₂O₃ is 2 × 56 + 3 × 16 = 160, and the equation makes 2Fe, 2 × 56 = 112. So 160 kg of iron(III) oxide makes 112 kg of iron. 320 kg is twice 160 kg, so it makes 2 × 112 = 224 kg of iron.

Checked automatically

Iron: 224 kg (within ±2%)
- Common wrong answer 112 kg: The equation makes 2Fe from one Fe₂O₃, and you have twice 160 kg.

A full answer includes

I found the M_r of Fe₂O₃: 2 × 56 + 3 × 16 = 160.
I used 2Fe from the equation: 2 × 56 = 112.
I saw that 320 kg is 2 × 160 kg.
I wrote 224 kg.
I did not use 56 for the iron: the equation makes 2Fe.

Also accept

224 kg

Watch for

“Use each M_r as it is and ignore the numbers in front of the formulae.” — Multiply each M_r by its big number in the balanced equation before you compare the masses.
“Units must be grams.” — The ratio works in any unit, as long as both masses are in the same one.

4. RQ12 · Practice

Model answer. M_r: Al 27, O₂ 32, Al₂O₃ 2 × 27 + 3 × 16 = 102. With the big numbers, 4 × 27 = 108 g of aluminium reacts with 3 × 32 = 96 g of oxygen to make 2 × 102 = 204 g of aluminium oxide (108 + 96 = 204). 306 g is 204 g multiplied by 1.5, so multiply the reactants by 1.5 too: 1.5 × 108 = 162 g of aluminium and 1.5 × 96 = 144 g of oxygen. Check: 162 + 144 = 306.

Checked automatically

(a) Aluminium: 162 g (within ±0.5 g)
- Common wrong answer 324 : The equation makes 2Al₂O₃: 108 g of aluminium makes 2 × 102 = 204 g of aluminium oxide, not 102 g.
- Common wrong answer 81 : Multiply by the big numbers: 4Al is 4 × 27 = 108 g, and 2Al₂O₃ is 2 × 102 = 204 g.
(b) Oxygen: 144 g (within ±0.5 g)
- Common wrong answer 288 : The equation makes 2Al₂O₃: 96 g of oxygen makes 2 × 102 = 204 g of aluminium oxide, not 102 g.
- Common wrong answer 72 : Oxygen is O₂: 3O₂ is 3 × 32 = 96 g, not 3 × 16.
- Common wrong answer 96 : 96 g of oxygen makes 204 g of aluminium oxide. You need 306 g, 1.5 times as much.

A full answer includes

I found the equation masses: 4 × 27 = 108 g of Al and 3 × 32 = 96 g of O₂ make 2 × 102 = 204 g of Al₂O₃.
I saw that 306 g is 1.5 × 204 g.
(a) I wrote 162 g of aluminium.
(b) I wrote 144 g of oxygen.
I did not use 102 for the aluminium oxide: the equation makes 2Al₂O₃.

Also accept

(a) 162 g (b) 144 g

Watch for

“Use each M_r as it is and ignore the numbers in front of the formulae.” — Multiply each M_r by its big number in the balanced equation before you compare the masses.
“You can only scale from a reactant to a product.” — The ratio works both ways: if you know the product's mass, scale each reactant's mass by the same number.

5. RQ13 · Practice

Options as numbered on the pupil sheet: 1 — Multiply the product's mass by that same number; 2 — Write the balanced equation; 3 — Work out the Mr of the reactant and of the product; 4 — Multiply each Mr by its big number in the equation; 5 — Find the number that turns the equation's mass of reactant into the mass you have.

Model answer. First write the balanced equation. Then work out the M_r of the reactant you know and the product you want, and multiply each by its big number in the equation; this gives the masses that react. Then find the number that turns the equation's mass of reactant into the mass you have, and multiply the product's mass by the same number.

A full answer includes

I put "Write the balanced equation" first.
I worked out the M_r values before multiplying by the big numbers.
I found the scaling number before multiplying the product's mass.
I put "Multiply the product's mass by that same number" last.
I did not put the scaling step before the equation masses: there is nothing to scale until they are known.

Also accept

a, b, c, d, e

Watch for

“Use each M_r as it is and ignore the numbers in front of the formulae.” — Multiply each M_r by its big number in the balanced equation before you compare the masses.

6. RQ14 · Practice

Model answer. The mass fell after heating 1 and fell again after heating 2, so water was still leaving after heating 1. Heatings 2 and 3 both give 51.31 g: the mass is constant, so all the water has gone. Record 51.31 g. The water lost is 52.10 − 51.31 = 0.79 g.

A full answer includes

I chose 51.31 g.
I saw that heatings 2 and 3 gave the same reading.
I did not use the mean of the heated readings: 51.46 g was read while water was still leaving.

Also accept

51.31 g, because two readings in a row agree.

Watch for

“One heating drives off all the water.” — The mass fell again after heating 2, so water was still leaving after heating 1.
“The final mass is the mean of the readings after heating.” — The first heated reading still had water in it. Use the reading once two readings in a row agree.
“Mass is conserved, so the basin keeps its starting mass.” — Mass is conserved in the change, but the water leaves the basin as steam, so the balance reads less.

7. RT2 · Apply

Model answer. Na₂CO₃ is 46 + 12 + 48 = 106. The water in one formula unit is 286 − 106 = 180, and each water molecule is 2 × 1 + 16 = 18, so x = 180 ÷ 18 = 10. So 286 g of crystals leave 106 g of sodium carbonate and lose 180 g of water. 10.0 g of crystals keeps the same ratio: the sodium carbonate is 106 × 10.0 ÷ 286 = 3.706… = 3.71 g, and the water is 180 × 10.0 ÷ 286 = 6.293… = 6.29 g, to 3 significant figures. Check: 3.71 + 6.29 = 10.00.

Checked automatically

(a) M_r of Na₂CO₃: 106 (exact value)
- Common wrong answer 83 : Na₂ is two sodium atoms: 2 × 23.
(b) x: 10 (exact value)
- Common wrong answer 180 : That is the mass of the water. Divide by 18 for the number of water molecules.
- Common wrong answer 16 : 286 ÷ 18 counts the sodium carbonate as water too. Take away 106 first.
(c) Sodium carbonate left: 3.71 g (3 significant figures)
- Common wrong answer 6.29 : That is the water lost. The sodium carbonate is 106 of every 286.
- Common wrong answer 5.89 : Compare the crystals, 286, with the sodium carbonate, 106, not with the water, 180.
(d) Water lost: 6.29 g (3 significant figures)
- Common wrong answer 3.71 : That is the sodium carbonate left. The water is 180 of every 286.
- Common wrong answer 16.98 : Compare the crystals, 286, with the water, 180, not the sodium carbonate with the water.

A full answer includes

(a) I added 2 × 23 + 12 + 3 × 16.
(a) I wrote 106.
(b) I took 106 from 286 to find the mass of water: 180.
(b) I divided 180 by 18, the M_r of water.
(b) I wrote 10.
(b) I did not give 180: that is the mass of the water, not the number of molecules.
(c) I worked out 106 × 10.0 ÷ 286.
(c) I wrote 3.71 g.
(d) I worked out 180 × 10.0 ÷ 286.
(d) I wrote 6.29 g.
I checked that 3.71 + 6.29 = 10.0.

Also accept

(a) 106 (b) 10 (c) 3.71 g (d) 6.29 g

Watch for

“The dot in a hydrate formula means multiply.” — The dot means "together with": add the mass of the water molecules to the rest.
“x is the mass of water in the formula.” — x counts water molecules. Divide the mass of water by 18, the M_r of one water molecule.
“Mass is conserved, so the basin keeps all its mass.” — Mass is conserved in the reaction, but the water leaves the basin as steam, so the basin gets lighter.