4.3 Quantitative chemistry — 8462 spec coverage

4.3.1.1

Conservation of mass and balanced chemical equations

The law of conservation of mass states that atoms are neither created nor destroyed in a chemical reaction, so the total mass is unchanged in a closed system.
Balance a symbol equation by placing whole-number coefficients before formulae until each element has the same number of atoms on both sides.
For example, $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$ shows that two magnesium atoms and two oxygen atoms are present on each side.
Never change a subscript to balance an equation: that changes the identity of the substance rather than the quantity reacting.

Tier 1 · Easy

1 mark

ORIGINAL

Insert the smallest whole-number coefficients to balance $\mathrm{Al}+\mathrm{O}_2\rightarrow\mathrm{Al}_2\mathrm{O}_3$ .

Tier 2 · Standard

2 marks

ORIGINAL

Methane reacts completely with oxygen in a sealed vessel. The reactants have masses $4.0\,\mathrm{g}$ and $16.0\,\mathrm{g}$ . One product is $11.0\,\mathrm{g}$ of carbon dioxide. Calculate the mass of water formed.

Tier 3 · Hard

3 marks

ORIGINAL

Propane burns according to $\mathrm{C}_3\mathrm{H}_8+5\mathrm{O}_2\rightarrow3\mathrm{CO}_2+4\mathrm{H}_2\mathrm{O}$ . A sealed reaction uses $11.0\,\mathrm{g}$ of propane and forms $33.0\,\mathrm{g}$ of carbon dioxide plus $18.0\,\mathrm{g}$ of water. Determine the mass of oxygen used and show that the masses obey conservation.

4.3.1.2

Relative formula mass

Relative formula mass, $M_r$ , is the sum of the relative atomic masses of every atom shown in a formula.
Multiply each $A_r$ value by the number of that atom, including multipliers outside brackets, before adding the contributions.
Percentage by mass of an element is $\dfrac{\text{total }A_r\text{ of that element}}{M_r\text{ of the compound}}\times100$ .
A coefficient in an equation multiplies an entire formula; ignoring it when comparing equation masses is a common error.

Tier 1 · Easy

2 marks

ORIGINAL

Calculate the relative formula mass of $\mathrm{Ca(OH)}_2$ . Use $A_r$ : $\mathrm{Ca}=40$ , $\mathrm{O}=16$ , $\mathrm{H}=1$ .

Tier 2 · Standard

3 marks

ORIGINAL

Ammonium nitrate is $\mathrm{NH}_4\mathrm{NO}_3$ . Calculate its percentage by mass of nitrogen. Use $A_r$ : $\mathrm{N}=14$ , $\mathrm{H}=1$ , $\mathrm{O}=16$ .

Tier 3 · Hard

4 marks

ORIGINAL

A $17.1\,\mathrm{g}$ sample contains only aluminium sulfate, $\mathrm{Al}_2(\mathrm{SO}_4)_3$ . Calculate the mass of oxygen in the sample. Use $A_r$ : $\mathrm{Al}=27$ , $\mathrm{S}=32$ , $\mathrm{O}=16$ .

4.3.1.3

Mass changes when a reactant or product is a gas

An apparent mass change can occur in an open system when a gaseous reactant enters or a gaseous product escapes.
A metal can gain mass while reacting because oxygen particles from the air become part of the solid metal oxide.
A metal carbonate can lose measured mass on heating because carbon dioxide leaves, while the solid metal oxide remains.
Conservation of mass still holds when the gas is included; claiming that atoms or mass have disappeared is the common error.

Tier 1 · Easy

2 marks

ORIGINAL

A strip of magnesium has a mass of $6.0\,\mathrm{g}$ before heating in air and the magnesium oxide has a mass of $10.0\,\mathrm{g}$ . Explain the increase and find the mass added.

Tier 2 · Standard

3 marks

ORIGINAL

A student heats $12.5\,\mathrm{g}$ of calcium carbonate in an open tube. The equation is $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ . The solid left has a mass of $7.0\,\mathrm{g}$ . Calculate the mass change and explain it using particles.

Tier 3 · Hard

4 marks

ORIGINAL

A crucible and its contents gain $3.2\,\mathrm{g}$ while a metal is converted fully into its oxide. Predict the change in the combined mass of the crucible, contents and surrounding sealed chamber, and account for both observations.

4.3.1.4

Chemical measurements

Every measured result has uncertainty because instruments have limited resolution and repeated readings vary.
For repeats, calculate the mean after checking whether any result is anomalous and should be investigated.
A useful estimate is half the range, written as $\pm\dfrac{\text{maximum}-\text{minimum}}{2}$ about the mean.
Do not quote an uncertainty without a unit or keep unjustified extra decimal places in the reported result.

Tier 1 · Easy

3 marks

ORIGINAL

Three titre readings are $12.4$ , $12.6$ and $12.5\,\mathrm{cm}^3$ . Calculate their mean and estimate the uncertainty as half the range.

Tier 2 · Standard

4 marks

ORIGINAL

Five mass-loss results are $8.21$ , $8.25$ , $8.23$ , $8.20$ and $8.26\,\mathrm{g}$ . Represent their distribution by calculating the mean, range and half-range uncertainty.

Tier 3 · Hard

4 marks

ORIGINAL

A reaction-time experiment gives $31.4$ , $31.6$ , $31.5$ , $36.9$ and $31.3\,\mathrm{s}$ . Identify the anomalous reading, then report the mean of the consistent readings with a half-range uncertainty.

4.3.2.1

Moles (HT only)

The mole, symbol $\mathrm{mol}$ , measures amount of substance; one mole contains $6.02\times10^{23}$ stated particles.
The mass of one mole in grams is numerically equal to its $M_r$ , so $n=\dfrac{m}{M_r}$ and $m=nM_r$ .
For example, $9.0\,\mathrm{g}$ of water with $M_r=18$ is $9.0/18=0.50\,\mathrm{mol}$ .
State the particle type carefully: ionic substances have formula units and ions, not molecules.

Tier 1 · Easy

2 marks

ORIGINAL

Calculate the amount in moles in $9.0\,\mathrm{g}$ of water, $\mathrm{H}_2\mathrm{O}$ . Use $M_r=18$ .

Tier 2 · Standard

3 marks

ORIGINAL

Find the mass of $0.250\,\mathrm{mol}$ of sodium carbonate, $\mathrm{Na}_2\mathrm{CO}_3$ . Use $A_r$ : $\mathrm{Na}=23$ , $\mathrm{C}=12$ , $\mathrm{O}=16$ .

Tier 3 · Hard

4 marks

ORIGINAL

A sample contains $0.0150\,\mathrm{mol}$ of magnesium chloride, $\mathrm{MgCl}_2$ . Calculate the number of formula units and the number of chloride ions. Use the Avogadro constant $6.02\times10^{23}\,\mathrm{mol}^{-1}$ .

4.3.2.2

Amounts of substances in equations (HT only)

Balanced-equation coefficients give mole ratios, which can be used to connect masses of different substances.
Convert the known mass to moles, apply the coefficient ratio, then convert the required moles back to mass.
In $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ , one mole of calcium carbonate produces one mole of carbon dioxide.
Do not use a coefficient ratio directly on masses unless the molar masses happen to be equal.

Tier 1 · Easy

3 marks

ORIGINAL

For $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$ , calculate the mass of magnesium oxide made from $4.8\,\mathrm{g}$ of magnesium when oxygen is in excess. Use $A_r$ : $\mathrm{Mg}=24$ , $\mathrm{O}=16$ .

Tier 2 · Standard

3 marks

ORIGINAL

Calcium carbonate decomposes as $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ . Calculate the mass of carbon dioxide made from $25.0\,\mathrm{g}$ of calcium carbonate. Use $M_r(\mathrm{CaCO}_3)=100$ and $M_r(\mathrm{CO}_2)=44$ .

Tier 3 · Hard

4 marks

ORIGINAL

Ammonia is oxidised by $4\mathrm{NH}_3+5\mathrm{O}_2\rightarrow4\mathrm{NO}+6\mathrm{H}_2\mathrm{O}$ . Calculate the mass of water formed from $10.2\,\mathrm{g}$ of ammonia when oxygen is in excess. Use $M_r(\mathrm{NH}_3)=17$ and $M_r(\mathrm{H}_2\mathrm{O})=18$ .

4.3.2.3

Using moles to balance equations (HT only)

Experimental masses can reveal balancing coefficients after each mass is converted into moles.
Divide all mole amounts by the smallest value to obtain a simple ratio, then multiply every value if fractions remain.
For mole amounts $0.20:0.60:0.40$ , division by $0.20$ gives the whole-number ratio $1:3:2$ .
Rounding a ratio too early can produce incorrect coefficients; keep enough significant figures until the ratio is clear.

Tier 1 · Easy

2 marks

ORIGINAL

Nitrogen, hydrogen and ammonia occur in amounts $0.20$ , $0.60$ and $0.40\,\mathrm{mol}$ respectively. Use these amounts to balance $\mathrm{N}_2+\mathrm{H}_2\rightarrow\mathrm{NH}_3$ .

Tier 2 · Standard

4 marks

ORIGINAL

Iron and oxygen form iron(III) oxide. The reacting masses are $11.2\,\mathrm{g}$ of $\mathrm{Fe}$ , $4.8\,\mathrm{g}$ of $\mathrm{O}_2$ and $16.0\,\mathrm{g}$ of $\mathrm{Fe}_2\mathrm{O}_3$ . Determine the balanced equation. Use $M_r$ : $\mathrm{Fe}=56$ , $\mathrm{O}_2=32$ , $\mathrm{Fe}_2\mathrm{O}_3=160$ .

Tier 3 · Hard

5 marks

ORIGINAL

Ethane burns in oxygen. A complete reaction uses $3.0\,\mathrm{g}$ of $\mathrm{C}_2\mathrm{H}_6$ and $11.2\,\mathrm{g}$ of $\mathrm{O}_2$ , producing $8.8\,\mathrm{g}$ of $\mathrm{CO}_2$ and $5.4\,\mathrm{g}$ of $\mathrm{H}_2\mathrm{O}$ . Determine the balanced equation. Use $M_r$ : $30$ , $32$ , $44$ and $18$ in the same order.

4.3.2.4

Limiting reactants (HT only)

The limiting reactant is used up completely and therefore fixes the maximum amount of product that can form.
Compare available moles with the balanced-equation ratio; the smaller mass is not necessarily the limiting amount.
Once the limiting reactant is known, use its moles and the coefficient ratio to calculate the product amount.
An excess reactant remains after the reaction, so using its full starting amount to calculate product overestimates the yield.

Tier 1 · Easy

2 marks

ORIGINAL

For $\mathrm{H}_2+\mathrm{Cl}_2\rightarrow2\mathrm{HCl}$ , a mixture contains $3.0\,\mathrm{mol}$ of hydrogen and $2.0\,\mathrm{mol}$ of chlorine. Identify the limiting reactant and calculate the amount of hydrogen chloride formed.

Tier 2 · Standard

4 marks

ORIGINAL

Magnesium reacts by $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$ . A vessel contains $7.2\,\mathrm{g}$ of magnesium and $6.4\,\mathrm{g}$ of oxygen. Determine the limiting reactant and the mass of magnesium oxide. Use $A_r$ : $\mathrm{Mg}=24$ , $\mathrm{O}=16$ .

Tier 3 · Hard

5 marks

ORIGINAL

Ammonia forms by $\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3$ . A reactor receives $14.0\,\mathrm{g}$ of nitrogen and $2.40\,\mathrm{g}$ of hydrogen. Calculate the ammonia mass and the mass of excess reactant left. Use $M_r$ : $\mathrm{N}_2=28$ , $\mathrm{H}_2=2$ , $\mathrm{NH}_3=17$ .

4.3.2.5

Concentration of solutions

Mass concentration in $\mathrm{g\,dm}^{-3}$ is the mass of dissolved solute divided by the solution volume: $c=m/V$ .
Convert volumes before calculating: $1000\,\mathrm{cm}^3=1\,\mathrm{dm}^3$ .
For example, $6.0\,\mathrm{g}$ in $0.20\,\mathrm{dm}^3$ has concentration $6.0/0.20=30\,\mathrm{g\,dm}^{-3}$ .
Use the volume of the final solution, not the volume of solvent added or the mass of the whole solution.

Tier 1 · Easy

2 marks

ORIGINAL

A solution contains $12.0\,\mathrm{g}$ of solute in $0.400\,\mathrm{dm}^3$ . Calculate its concentration in $\mathrm{g\,dm}^{-3}$ .

Tier 2 · Standard

3 marks

ORIGINAL

A fertiliser solution has concentration $18.0\,\mathrm{g\,dm}^{-3}$ . Calculate the solute mass in $250\,\mathrm{cm}^3$ of solution.

Tier 3 · Hard

4 marks

ORIGINAL

A beaker initially contains $3.60\,\mathrm{g}$ of dissolved salt in $150\,\mathrm{cm}^3$ of solution. Water is added until the concentration is $12.0\,\mathrm{g\,dm}^{-3}$ . Calculate the volume of water added, assuming volumes are additive.

4.3.3.1

Percentage yield (chemistry only)

Percentage yield compares the actual product obtained with the maximum theoretical product: $\dfrac{\text{actual yield}}{\text{theoretical yield}}\times100$ .
Yield may be below $100\%$ because a reversible reaction is incomplete, side reactions occur, or product is lost during separation.
For example, an actual mass of $8.0\,\mathrm{g}$ from a theoretical $10.0\,\mathrm{g}$ gives an $80\%$ yield.
Use actual over theoretical, not the reverse, and compare quantities in the same unit.

Tier 1 · Easy

2 marks

ORIGINAL

A preparation has a theoretical product mass of $9.0\,\mathrm{g}$ and an actual product mass of $7.2\,\mathrm{g}$ . Calculate the percentage yield.

Tier 2 · Standard

3 marks

ORIGINAL

A process makes $20.4\,\mathrm{kg}$ of product at a percentage yield of $68.0\%$ . Calculate the theoretical product mass.

Tier 3 · Hard

4 marks

ORIGINAL

Two batches have theoretical product masses of $45.0\,\mathrm{g}$ and $30.0\,\mathrm{g}$ . Their actual masses are $36.0\,\mathrm{g}$ and $19.5\,\mathrm{g}$ . Determine the combined percentage yield.

4.3.3.2

Atom economy (chemistry only)

Atom economy measures the proportion of reactant atoms that become the desired product in the balanced equation.
Calculate it using $\dfrac{M_r\text{ of desired product from the equation}}{\text{sum of }M_r\text{ of all reactants from the equation}}\times100$ .
A high atom economy reduces unwanted by-products, conserves resources and can lower disposal costs.
Include equation coefficients when totaling formula masses; atom economy is not the same as percentage yield.

Tier 1 · Easy

2 marks

ORIGINAL

Calcium oxide is the desired product in $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ . Calculate the atom economy using $M_r(\mathrm{CaCO}_3)=100$ and $M_r(\mathrm{CaO})=56$ .

Tier 2 · Standard

3 marks

ORIGINAL

Chlorine is the desired product in $2\mathrm{NaCl}+2\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{Cl}_2+\mathrm{H}_2+2\mathrm{NaOH}$ . Calculate the atom economy. Use $M_r$ : $\mathrm{NaCl}=58.5$ , $\mathrm{H}_2\mathrm{O}=18.0$ , $\mathrm{Cl}_2=71.0$ .

Tier 3 · Hard

5 marks

ORIGINAL

Titanium is the desired product in $\mathrm{TiCl}_4+4\mathrm{Na}\rightarrow\mathrm{Ti}+4\mathrm{NaCl}$ . Calculate the atom economy and the theoretical titanium mass represented by $250\,\mathrm{kg}$ of reactants in this ratio. Use $A_r$ : $\mathrm{Ti}=48$ , $\mathrm{Cl}=35.5$ , $\mathrm{Na}=23$ .

4.3.4

Using concentrations of solutions in mol/dm3 (chemistry only) (HT only)

Molar concentration is amount of solute per solution volume: $c=n/V$ , with $c$ in $\mathrm{mol\,dm}^{-3}$ and $V$ in $\mathrm{dm}^3$ .
Use $n=cV$ to find moles, then use $m=nM_r$ when a solute mass is required.
Reacting-solution calculations use the balanced-equation mole ratio between the two dissolved substances.
Convert $\mathrm{cm}^3$ to $\mathrm{dm}^3$ by dividing by $1000$ before multiplying by concentration.

Tier 1 · Easy

2 marks

ORIGINAL

A solution contains $0.0750\,\mathrm{mol}$ of solute in $250\,\mathrm{cm}^3$ . Calculate its concentration in $\mathrm{mol\,dm}^{-3}$ .

Tier 2 · Standard

4 marks

ORIGINAL

Calculate the sodium hydroxide mass in $150\,\mathrm{cm}^3$ of a $0.400\,\mathrm{mol\,dm}^{-3}$ solution. Use $M_r(\mathrm{NaOH})=40.0$ .

Tier 3 · Hard

5 marks

ORIGINAL

$25.0\,\mathrm{cm}^3$ of sulfuric acid reacts exactly with $32.0\,\mathrm{cm}^3$ of $0.150\,\mathrm{mol\,dm}^{-3}$ sodium hydroxide. Use $\mathrm{H}_2\mathrm{SO}_4+2\mathrm{NaOH}\rightarrow\mathrm{Na}_2\mathrm{SO}_4+2\mathrm{H}_2\mathrm{O}$ to calculate the acid concentration.

4.3.5

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

Equal mole amounts of gases occupy equal volumes at the same temperature and pressure.
At room temperature and pressure, one mole of any gas occupies $24\,\mathrm{dm}^3$ , so $V=24n$ .
Balanced coefficients give gas-volume ratios directly when all gaseous substances are compared under the same conditions.
Keep units consistent: $1000\,\mathrm{cm}^3=1\,\mathrm{dm}^3$ , and do not use $24\,\mathrm{dm}^3$ before converting mass to moles.

Tier 1 · Easy

2 marks

ORIGINAL

Calculate the volume occupied by $0.350\,\mathrm{mol}$ of a gas at room temperature and pressure.

Tier 2 · Standard

3 marks

ORIGINAL

Calculate the volume of carbon dioxide at room temperature and pressure produced by $8.80\,\mathrm{g}$ of the gas. Use $M_r(\mathrm{CO}_2)=44.0$ .

Tier 3 · Hard

4 marks

ORIGINAL

Carbon monoxide reacts by $2\mathrm{CO}+\mathrm{O}_2\rightarrow2\mathrm{CO}_2$ . A mixture contains $36.0\,\mathrm{dm}^3$ of carbon monoxide and $24.0\,\mathrm{dm}^3$ of oxygen under the same conditions. Calculate the carbon dioxide volume and the volume of reactant gas left in excess after complete reaction.