The mole and the Avogadro constant, molar mass, empirical and molecular formulae, the ideal gas equation, concentrations of solutions, and reacting mass, gas volume and titration calculations including percentage yield and atom economy.

What this dot point is asking

CCEA wants you to use the mole and the Avogadro constant, calculate molar masses, find empirical and molecular formulae, apply the ideal gas equation, work with solution concentrations, and carry out reacting mass, gas volume and titration calculations, including percentage yield and atom economy.

The mole and molar mass

The key relationships are:

where $n$ is moles, $m$ is mass in g, $c$ is concentration in $\text{mol dm}^{-3}$, $V$ is volume (in $\text{dm}^3$ for solutions, $\text{m}^3$ for the gas equation), $p$ is pressure in Pa, $T$ is temperature in K, and $R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$.

Three points keep these formulae reliable. First, the volume $V$ means different things in different equations: in $n = c \times V$ it is the solution volume in $\text{dm}^3$, while in $pV = nRT$ it is the gas volume in $\text{m}^3$. Second, $24\ \text{dm}^3$ is the molar volume of any gas at room temperature and pressure, so for gases at RTP you can shortcut to $n = V / 24$ with $V$ in $\text{dm}^3$. Third, all amounts in a reaction are linked by the balanced equation: once you know the moles of one species, the stoichiometric ratios give every other.

Empirical and molecular formulae

The standard procedure

If you are given percentage composition, treat the percentages as masses in a $100\ \text{g}$ sample. Divide each by the $A_r$ to get moles, divide all results by the smallest, and round to whole numbers. If a ratio comes out close to $1.5$, multiply every figure by 2 rather than rounding, because the true ratio is a fraction. The empirical formula mass is then compared with the relative molecular mass (found from a mass spectrum or the ideal gas equation) to fix the multiplier $n$.

For combustion data, work back from the masses of $\text{CO}_2$ and $\text{H}_2\text{O}$: every mole of $\text{CO}_2$ holds one mole of C, and every mole of $\text{H}_2\text{O}$ holds two moles of H. Any remaining mass is usually oxygen.

Solutions, gases and titrations

For a titration, find the moles of the standard solution from $n = c \times V$, use the balanced equation to find the moles of the unknown, then calculate its concentration or mass. Always convert volumes from $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000. The titre is read to the nearest $0.05\ \text{cm}^3$, and only concordant titres (within $0.10\ \text{cm}^3$ of each other) are averaged before the calculation.

Efficiency: yield and atom economy

Atom economy measures how much of the reactant mass ends up in the useful product, which matters for green and sustainable chemistry. A reaction can have a high percentage yield but a low atom economy (if the equation produces a lot of waste product), or a perfect atom economy (an addition reaction, where there is only one product) but a poor yield in practice. The two figures answer different questions and are both reported in process design.

Examples in context

Example 1. Standardising a sodium hydroxide solution. Before a school laboratory uses sodium hydroxide for titrations, its true concentration is fixed by titrating it against a primary standard such as potassium hydrogen phthalate, which can be weighed accurately. A student weighs $2.04\ \text{g}$ of the acid salt ($M = 204\ \text{g mol}^{-1}$, so $0.0100\ \text{mol}$), dissolves it in water, and titrates against the alkali. The 1:1 reaction means the moles of alkali delivered at the endpoint equals $0.0100\ \text{mol}$, and dividing by the titre volume in $\text{dm}^3$ gives the standardised concentration. This is the practical reason that amount-of-substance calculations underpin every quantitative experiment.

Example 2. Atom economy in the industrial production of ethanol. The hydration of ethene, $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$, is an addition reaction with a single product, so its atom economy is $100\%$: every atom of reactant ends up in the ethanol. By contrast, fermentation, $\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$, has an atom economy of only $\frac{2 \times 46}{180} \times 100 = 51\%$ because half the carbon is lost as carbon dioxide. Comparing these two routes is a standard CCEA application of the atom economy idea to a real green-chemistry decision.

Try this

Q1. Calculate the amount in moles of $4.0\ \text{g}$ of sodium hydroxide ($M = 40\ \text{g mol}^{-1}$). [1 mark]

• Cue. $n = \frac{4.0}{40} = 0.10\ \text{mol}$.

Q2. Define atom economy. [1 mark]

• Cue. The mass of the desired product as a percentage of the total mass of all products.