Reacting masses and volumes (of solutions and gases)

1. Overview

A balanced equation gives the ratio in which particles react: a ratio of moles, not of grams. This section turns that mole ratio into real quantities: masses, gas volumes, and volumes and concentrations of solutions. It also covers percentage yield, limiting and excess reagents, working backwards from measurements to an equation, and significant figures. Nearly every later calculation in the course is built on these steps.

Key Definitions

• Molar mass (M): the mass of one mole of a substance, in g mol⁻¹; numerically equal to its relative atomic, molecular or formula mass.
• Stoichiometry: the mole ratio in which substances react and are formed in a reaction, given by the balancing numbers in the balanced equation.
• Concentration of a solution: the amount of solute, in moles, dissolved in 1 dm³ of solution (mol dm⁻³); a mass concentration is the mass of solute in 1 dm³ of solution (g dm⁻³).
• Molar volume of a gas (Vm): the volume occupied by one mole of any gas, assumed ideal, at a stated temperature and pressure: 24.0 dm³ mol⁻¹ at room conditions and 22.4 dm³ mol⁻¹ at s.t.p. (273 K and 101 kPa).
• Avogadro's law: equal volumes of gases at the same temperature and pressure contain equal numbers of particles, so gas volumes are in the same ratio as amounts in moles.
• Standard solution: a solution whose concentration is known accurately, usually made by dissolving a weighed mass of solid and making the solution up to the mark in a volumetric flask.
• Limiting reagent: the reactant that is completely used up first in a reaction, so it fixes the maximum amount of product that can form.
• Excess reagent: a reactant present in more than the amount needed to react with all of the limiting reagent, so some is left over at the end.
• Theoretical yield: the maximum mass or amount of product that the limiting reagent could give, calculated from the balanced equation, assuming complete reaction and no losses.
• Percentage yield: the actual yield of product divided by the theoretical yield, multiplied by 100.

Content

Moles: the bridge between quantities

Convert every measured quantity into an amount in moles, $n$. The balanced equation then gives the mole ratio, which takes you from one substance to another.

You are given	Find moles with	Units
a mass, $m$	$n = \frac{m}{M}$	$m$ in g; $M$ in g mol⁻¹
a gas volume at room conditions or s.t.p.	$n = \frac{V}{V_{\text{m}}}$	$V$ in dm³; $V_{\text{m}}$ = 24.0 or 22.4 dm³ mol⁻¹
a gas at a stated temperature and pressure	$n = \frac{pV}{RT}$ (topic 4.1)	$p$ in Pa; $V$ in m³; $T$ in K
a volume of solution	$n = c \times \frac{V}{1000}$	$c$ in mol dm⁻³; $V$ in cm³
a number of particles, $N$	$n = \frac{N}{L}$	$L$ = 6.022 × 10²³ mol⁻¹

The molar mass, $M$, is the sum of the $A_{\text{r}}$ values of all the atoms in the formula, with the unit g mol⁻¹. For water, $M = 2(1.0) + 16.0 = 18.0$ g mol⁻¹.

Reacting masses from formulas

A formula fixes the ratio of atoms, so it also fixes the ratio of masses. The percentage by mass of an element in a compound is:

$$\text{percentage by mass} = \frac{\text{number of atoms of the element} \times A_{\text{r}}}{M_{\text{r}} \text{ of the compound}} \times 100$$

Molybdenum(VI) oxide, MoO₃, has $M_{\text{r}} = 95.9 + 3(16.0) = 143.9$. It is $\frac{95.9}{143.9} \times 100 = 66.6\%$ molybdenum by mass, so 1.00 kg of the oxide can give at most 0.666 kg of the metal, whatever method is used.

Reacting masses from equations

Masses do not react in the ratio of the balancing numbers; moles do. To get from the mass of one substance to the mass of another, use three steps:

1. Moles of the substance you know: $n = m/M$.
2. Mole ratio from the equation: multiply by (balancing number of the wanted substance) ÷ (balancing number of the known substance).
3. Back to mass: $m = n \times M$ for the wanted substance.

Percentage yield

The theoretical yield is the most product that the limiting reagent could give. The actual yield is what you collect.

$$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$

Both yields must be in the same unit: both masses, or both amounts in moles. Yields are below 100% because:

• the reaction does not go to completion, for example a reversible reaction that reaches equilibrium;
• side reactions make other products;
• some product is lost when it is filtered, transferred or purified.

A yield above 100% usually means the product is impure, for example still damp with water or solvent: no reaction can make more than the theoretical yield.

Worked example 1: molybdenum from its oxide

Question: Molybdenum is made by heating molybdenum(VI) oxide in a stream of excess hydrogen. The overall equation is:

MoO₃(s) + 3H₂(g) → Mo(s) + 3H₂O(g)

A batch of 55.0 kg of MoO₃ gives 33.2 kg of molybdenum. ($A_{\text{r}}$: Mo 95.9, O 16.0)

(a) Calculate the percentage yield. (b) Calculate the minimum volume of hydrogen, measured at room conditions, needed to reduce all 55.0 kg of MoO₃.

Step 1: moles of MoO₃. Convert to grams first: 55.0 kg = 55.0 × 10³ g, and $M$(MoO₃) = 143.9 g mol⁻¹. $$n(\text{MoO}_3) = \frac{55.0 \times 10^3}{143.9} = 382.21\ \text{mol}$$

Step 2: theoretical yield of Mo. The ratio MoO₃ : Mo is 1 : 1, so $n(\text{Mo}) = 382.21$ mol. $$m(\text{Mo}) = 382.21 \times 95.9 = 36\,654\ \text{g} = 36.65\ \text{kg}$$ Check with the formula: $55.0 \times \frac{95.9}{143.9} = 36.65$ kg, the same answer.

Step 3: percentage yield. $$\text{percentage yield} = \frac{33.2}{36.65} \times 100 = 90.6\%$$

Step 4: hydrogen. The ratio MoO₃ : H₂ is 1 : 3. $$n(\text{H}_2) = 3 \times 382.21 = 1146.6\ \text{mol} \qquad V = 1146.6 \times 24.0 = 27\,519\ \text{dm}^3$$

Answer: (a) percentage yield = 90.6% (3 s.f.); (b) 27.5 m³ ($2.75 \times 10^4$ dm³) of hydrogen.

Volumes of gases

Avogadro's law: equal volumes of gases at the same temperature and pressure contain equal numbers of particles. So:

• At a given temperature and pressure, one mole of any gas fills the same volume, the molar volume, $V_{\text{m}}$: 24.0 dm³ mol⁻¹ (24 000 cm³ mol⁻¹) at room conditions and 22.4 dm³ mol⁻¹ at s.t.p. (273 K and 101 kPa).
• For gases at the same temperature and pressure, volumes react in the same ratio as moles, so you can apply the balancing numbers straight to volumes, in any unit.

Strictly, both apply to ideal gases, but real gases at room conditions come close enough (topic 4.1 explains why). Use 24.0 dm³ mol⁻¹ only when a question says room conditions, and 22.4 dm³ mol⁻¹ only when it says s.t.p. If it gives a temperature and pressure instead, use $pV = nRT$ (topic 4.1).

Ethene burns as C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), with water a liquid at room temperature.

• Volume to volume: 20.0 cm³ of ethene needs 3 × 20.0 = 60.0 cm³ of oxygen and gives 2 × 20.0 = 40.0 cm³ of carbon dioxide, all at the same temperature and pressure.
• Mass to volume: 3.50 g of ethene ($M$ = 28.0 g mol⁻¹) is 0.125 mol. It gives 2 × 0.125 = 0.250 mol of CO₂, which fills 0.250 × 24.0 = 6.00 dm³ at room conditions.

Gas volumes when hydrocarbons burn

For the complete combustion of any hydrocarbon, CₓHᵧ:

$$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\,\text{CO}_2 + \frac{y}{2}\,\text{H}_2\text{O}$$

Each carbon atom needs one O₂ molecule to become CO₂, and every four hydrogen atoms need one O₂ molecule to become two H₂O.

A typical experiment burns a gaseous hydrocarbon in a known excess of oxygen, then measures the gas volume twice:

1. After cooling to room temperature, the water has condensed. Treat its volume as zero: one mole of liquid water fills about 18 cm³, against 24 000 cm³ for a mole of gas (the small amount left as vapour is also ignored). The gas left is CO₂ plus unreacted O₂.
2. After passing through aqueous sodium hydroxide, the acidic CO₂ has been absorbed: CO₂(g) + 2NaOH(aq) → Na₂CO₃(aq) + H₂O(l). Only the unreacted O₂ is left.

The fall in volume at stage 2 is the volume of CO₂: divide it by the volume of hydrocarbon to get $x$. Divide the oxygen used, (oxygen at the start) − (oxygen left), by the volume of hydrocarbon to get $x + \frac{y}{4}$, and so $y$. If volumes are measured above 100 °C instead, the water stays as steam and counts as a gas.

Worked example 2: identifying a hydrocarbon from gas volumes

Question: 12.0 cm³ of a gaseous hydrocarbon, CₓHᵧ, is burned completely in 100 cm³ of oxygen, an excess. After cooling to room temperature, the gas volume is 76.0 cm³. After passing through aqueous sodium hydroxide, 28.0 cm³ of gas remains. All volumes are measured at room conditions. Deduce the molecular formula of the hydrocarbon.

Step 1: identify each gas. The 28.0 cm³ left at the end is unreacted oxygen. The gas absorbed by the alkali is CO₂. $$V(\text{CO}_2) = 76.0 - 28.0 = 48.0\ \text{cm}^3$$

Step 2: find x. Volumes are in the mole ratio, so 12.0 cm³ of CₓHᵧ gives $12.0x$ cm³ of CO₂. $$x = \frac{48.0}{12.0} = 4$$

Step 3: find y from the oxygen used. $$V(\text{O}_2\ \text{used}) = 100 - 28.0 = 72.0\ \text{cm}^3 \qquad x + \frac{y}{4} = \frac{72.0}{12.0} = 6$$ $$\frac{y}{4} = 6 - 4 = 2 \qquad y = 8$$

Step 4: check. For C₄H₈ + 6O₂ → 4CO₂ + 4H₂O, 12.0 cm³ of C₄H₈ uses 72.0 cm³ of O₂ and makes 48.0 cm³ of CO₂, leaving 28.0 + 48.0 = 76.0 cm³ after cooling, as measured.

Answer: the molecular formula is C₄H₈.

Volumes and concentrations of solutions

Concentration is the amount of solute per unit volume of solution:

$$c = \frac{n}{V} \qquad n = c \times \frac{V}{1000}$$

Here $c$ is in mol dm⁻³ and $n$ in mol. In the first form $V$ is in dm³; in the second it is in cm³ (1 dm³ = 1000 cm³). $V$ is the volume of the whole solution, not of the water added.

• Mass concentration: multiply by $M$. A 0.250 mol dm⁻³ solution of potassium hydroxide ($M$ = 56.1 g mol⁻¹) holds 0.250 × 56.1 = 14.0 g dm⁻³.
• Dilution: adding water changes the volume but not the moles of solute, so $c_1V_1 = c_2V_2$. Making 40.0 cm³ of 0.850 mol dm⁻³ hydrochloric acid up to 250.0 cm³ gives $c_2 = \frac{0.850 \times 40.0}{250.0} = 0.136$ mol dm⁻³.

Titration calculations

In a titration, a standard solution reacts with a measured volume of another solution, and the equation links the two.

1. Find the mean titre from concordant results (titres within 0.10 cm³ of each other).
2. Moles of the reagent in the burette that reacted: $n = c \times V/1000$.
3. Mole ratio from the equation: this gives the moles of the other reactant in the pipetted portion, often 25.0 cm³.
4. Scale up if that portion came from a larger flask. 25.0 cm³ taken from a 250.0 cm³ volumetric flask holds one tenth of the solute, so multiply by 10.
5. Convert to what the question asks for: a concentration, a mass, an $M_{\text{r}}$ or a percentage purity, $\frac{\text{mass of pure substance}}{\text{mass of sample}} \times 100$.

Worked example 3: the reacting ratio of an acid

Question: 1.53 g of hexanedioic acid, HOOC(CH₂)₄COOH ($M$ = 146.0 g mol⁻¹), is dissolved in water and made up to 250.0 cm³ in a volumetric flask. 25.0 cm³ portions are titrated with 0.105 mol dm⁻³ sodium hydroxide, and the mean of the concordant titres is 19.95 cm³. Deduce how many moles of NaOH react with one mole of the acid, and write the equation.

Step 1: moles of NaOH. $$n(\text{NaOH}) = 0.105 \times \frac{19.95}{1000} = 2.095 \times 10^{-3}\ \text{mol}$$

Step 2: moles of acid in one 25.0 cm³ portion. $$n(\text{acid in flask}) = \frac{1.53}{146.0} = 0.010\,48\ \text{mol} \qquad n(\text{acid in portion}) = \frac{0.010\,48}{10} = 1.048 \times 10^{-3}\ \text{mol}$$

Step 3: the ratio. $$\frac{n(\text{NaOH})}{n(\text{acid})} = \frac{2.095 \times 10^{-3}}{1.048 \times 10^{-3}} = 2.00$$

The unrounded ratio, 1.999, is 0.05% from 2. That is smaller than the uncertainty in the titre alone: two burette readings, each ±0.05 cm³, give ±0.10 cm³, about 0.5% of 19.95 cm³. So the ratio is 2 : 1.

Answer: 2 mol of NaOH react with 1 mol of the acid, so both -COOH groups react:

HOOC(CH₂)₄COOH(aq) + 2NaOH(aq) → NaOOC(CH₂)₄COONa(aq) + 2H₂O(l)

Limiting reagent and excess reagent

Unless reactants are mixed in the exact ratio of the equation, one runs out first. That limiting reagent decides how much product forms; some of the excess reagent is left over.

To find the limiting reagent:

1. Work out the moles of each reactant.
2. Divide each by its balancing number in the equation.
3. The reactant with the smallest result is the limiting reagent.

Calculate every product from the limiting reagent only. For the excess left over, work out how much of the excess reagent reacts and subtract it from the amount at the start. Do not simply pick the reactant with fewer moles: the ratio matters, as the next example shows.

Worked example 4: aluminium and iodine

Question: 0.810 g of aluminium powder is mixed with 10.0 g of iodine, and the reaction goes to completion.

2Al(s) + 3I₂(s) → 2AlI₃(s)

(a) Identify the limiting reagent. (b) Calculate the mass of aluminium iodide formed. (c) Calculate the mass of the excess reagent left over. ($A_{\text{r}}$: Al 27.0, I 126.9)

Step 1: moles of each reactant. $M(\text{I}_2) = 2 \times 126.9 = 253.8$ g mol⁻¹. $$n(\text{Al}) = \frac{0.810}{27.0} = 0.0300\ \text{mol} \qquad n(\text{I}_2) = \frac{10.0}{253.8} = 0.039\,40\ \text{mol}$$

Step 2: divide by the balancing numbers. $$\text{Al: } \frac{0.0300}{2} = 0.0150 \qquad \text{I}_2\text{: } \frac{0.039\,40}{3} = 0.013\,13$$

Iodine gives the smaller value, so iodine is the limiting reagent, even though there are more moles of iodine than of aluminium. Check: 0.0300 mol of Al would need 1.5 × 0.0300 = 0.0450 mol of I₂, more than the 0.0394 mol present.

Step 3: product from the limiting reagent. $n(\text{AlI}_3) = \frac{2}{3} \times 0.039\,40 = 0.026\,27$ mol, and $M(\text{AlI}_3) = 27.0 + 3(126.9) = 407.7$ g mol⁻¹. $$m(\text{AlI}_3) = 0.026\,27 \times 407.7 = 10.71\ \text{g}$$

Step 4: excess aluminium. The ratio Al : I₂ is 2 : 3, so the aluminium used is also 0.026 27 mol, which is 0.026 27 × 27.0 = 0.709 g. $$m(\text{Al left}) = 0.810 - 0.709 = 0.101\ \text{g}$$

Answer: (a) iodine; (b) 10.7 g of AlI₃ (3 s.f.); (c) 0.101 g of aluminium is left over. As a check, 10.71 + 0.10 = 10.81 g, the total mass at the start.

Deducing stoichiometric relationships

You can also work backwards: measure the amounts that react, then deduce the equation.

1. Convert each measured quantity to moles, or use gas volumes directly.
2. Divide every amount by the smallest one.
3. Round to whole numbers only when the values are within experimental error of them. If one is close to a half, multiply them all by 2 (1 : 1.5 becomes 2 : 3).
4. Use the whole numbers as balancing numbers, then balance the rest.

For example, 1.30 g of zinc powder is added to excess silver nitrate solution, and 4.30 g of silver forms. $n(\text{Zn}) = 1.30/65.4 = 0.0199$ mol and $n(\text{Ag}) = 4.30/107.9 = 0.0399$ mol, a ratio of 1 : 2.00. So each zinc atom replaces two silver ions:

Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)

Significant figures and rounding

• Give the number of significant figures (s.f.) a question asks for. If it does not say, use the same number as the least precise data, or one more: data to 3 s.f. (0.810 g, 10.0 g) justify 10.7 g, not 10.709 g or 11 g.
• Exact numbers do not limit precision: the 2 in 2NaOH, the 1000 in cm³ to dm³, the 100 in a percentage.
• Carry unrounded values through the calculation and round only the final answer. In Worked example 4, rounding $n(\text{I}_2)$ to 0.04 mol at the start gives 10.9 g of AlI₃ instead of 10.7 g.
• Significant figures are not decimal places. A titre of 19.95 cm³ has 4 s.f. and 2 decimal places. 0.0300 mol has 3 s.f.: zeros before the 3 do not count, zeros after it do.

Key Equations

Amount from mass ($m$ in g; $M$ in g mol⁻¹, numerically equal to $M_{\text{r}}$): $$n = \frac{m}{M}$$

Amount of gas ($V$ in dm³; $V_{\text{m}}$ = 24.0 dm³ mol⁻¹ at room conditions or 22.4 dm³ mol⁻¹ at s.t.p.; at other stated conditions use $pV = nRT$, topic 4.1): $$n = \frac{V}{V_{\text{m}}}$$

Amount in solution ($c$ in mol dm⁻³; $V$ in cm³): $$n = c \times \frac{V}{1000}$$

Mass concentration (in g dm⁻³; $c$ in mol dm⁻³; $M$ in g mol⁻¹): $$\text{mass concentration} = c \times M$$

Dilution ($c$ in mol dm⁻³; the same volume unit on both sides; moles of solute unchanged): $$c_1V_1 = c_2V_2$$

Percentage by mass of an element (from the formula): $$\text{percentage by mass} = \frac{\text{number of atoms} \times A_{\text{r}}}{M_{\text{r}}} \times 100$$

Percentage yield (both yields in g, or both in mol): $$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$

Complete combustion of a hydrocarbon (volumes of gases at the same temperature and pressure are in the same ratio): $$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\,\text{CO}_2 + \frac{y}{2}\,\text{H}_2\text{O}$$

Common Mistakes to Avoid

• Mistake: stopping at the moles of the substance you were given. Fix: multiply by the mole ratio. In Worked example 1, 382.21 mol of MoO₃ needs 3 × 382.21 = 1146.6 mol of H₂; leaving out the 3 gives 9.17 m³ of hydrogen instead of 27.5 m³.
• Mistake: using 24.0 dm³ mol⁻¹ whatever the conditions. Fix: 24.0 is for room conditions only and 22.4 is for s.t.p. only. 1.00 dm³ of gas at s.t.p. is 1.00 ÷ 22.4 = 0.0446 mol, not 0.0417 mol. At any other stated conditions, use $pV = nRT$.
• Mistake: ignoring the dilution when a portion was taken from a volumetric flask. Fix: in Worked example 3, the 25.0 cm³ portion held $1.048 \times 10^{-3}$ mol of acid, one tenth of the 0.010 48 mol in the flask. Using the whole flask gives a ratio of 0.200, which makes no chemical sense.
• Mistake: choosing the limiting reagent by comparing moles directly. Fix: divide by the balancing numbers first. In Worked example 4 there is more iodine (0.0394 mol) than aluminium (0.0300 mol), yet iodine runs out first; working from the aluminium predicts 12.2 g of AlI₃, not the 10.7 g that forms.
• Mistake: treating the gas left after the alkali as a product. Fix: it is the unreacted excess oxygen. In Worked example 2, the 28.0 cm³ left is O₂; the CO₂ is the 48.0 cm³ that the alkali removed.
• Mistake: writing the percentage yield upside down. Fix: actual ÷ theoretical. In Worked example 1, 33.2 ÷ 36.65 × 100 = 90.6%; the inverted 36.65 ÷ 33.2 × 100 = 110% is impossible for a pure product.
• Mistake: rounding at every step. Fix: keep the calculator value until the end. In Worked example 3, rounding the moles of acid to $1.0 \times 10^{-3}$ (2 s.f.) gives a ratio of 2.09, almost 5% from 2 and far outside the 0.5% uncertainty in the titre, so the 2 : 1 ratio is no longer clear.

Exam Tips

• Calculate: put each amount on its own labelled line, such as $n(\text{NaOH}) = 0.105 \times 19.95/1000 = 2.095 \times 10^{-3}$ mol, so one slip does not hide the rest of your working.
• Give the final answer with its unit (mol, g, dm³, cm³ or mol dm⁻³) and to the significant figures asked for (see Significant figures and rounding above).
• Deduce an equation or formula by showing the moles (or volumes), the division by the smallest, and the whole-number ratio before you write the equation.
• Show by calculation that a reagent is in excess: work out how much of it would react with all of the other reagent, and compare that with the amount present, quoting both numbers.
• Gas volumes: check whether the question says room conditions, s.t.p. or gives a temperature and pressure, and name the gases present at each stage of a combustion experiment.