Quantitative Chemistry and Titrations for the ESAT

Relative Molar Mass and the Mole

Chemical reactions involve vast numbers of particles that are measured by weighing. The fundamental unit for the amount of substance is the mole (mol). One mole of any substance contains the Avogadro constant, $N_A$, which is approximately $6.022 \times 10^{23}$ particles per mole. The mass of one mole of a substance is its relative atomic mass ($A_r$) or relative molar mass ($M_r$) expressed in grams.

The relative molar mass ($M_r$) is determined by summing the relative atomic masses of all atoms within a formula unit. For example, for barium nitrate, $Ba(NO_3)_2$, the calculation is $137 + 2 \times (14.0 + (3 \times 16.0)) = 261$. In hydrated compounds like $CuSO_4 \cdot 5H_2O$, the mass of the water of crystallisation must be included: $63.5 + 32.0 + (4 \times 16.0) + 5 \times (18.0) = 249.5$.

Converting Mass to Moles

The relationship between mass, molar mass, and the number of moles is defined by the formula:

$\text{number of moles} = \frac{\text{mass in g}}{\text{molar mass in g mol}^{-1}}$

When working with larger scales, such as kilograms or tonnes, units must be converted to grams. There are $1,000$ grams in a kilogram and $1,000,000$ grams in a tonne. For instance, to find the moles in $5.4$ tonnes of aluminium: $\frac{5.4 \times 10^6}{27.0} = 200,000$ mol.

Percentage Composition and Formulae

The percentage composition by mass of an element in a compound is calculated using the formula:

$\% \text{ element} = \frac{A_r \times \text{number of atoms of that element}}{M_r \text{ of the compound}} \times 100$

An empirical formula represents the simplest whole number ratio of atoms in a compound. It can be found from percentage masses by dividing each percentage by the element $A_r$ to find the molar ratio, then simplifying to the smallest integers. A molecular formula shows the actual number of atoms and is a multiple of the empirical formula, found by comparing the empirical formula mass to the actual $M_r$ of the compound.

Reacting Masses and Limiting Reactants

Balanced equations provide the molar ratio of reactants and products. This allows for the calculation of the maximum mass of a product formed from a given mass of reactant. A limiting reactant is the substance that is completely consumed first, thereby determining the maximum theoretical yield of the product. To identify it, calculate the moles of all reactants and compare them to the stoichiometric ratios in the balanced equation.

Molar Gas Volume

For an ideal gas, the volume occupied depends on temperature and pressure rather than the identity of the gas. At room temperature and pressure (rtp), one mole of any gas occupies $24$ dm$^3$. The number of moles is calculated as:

$\text{moles of gas} = \frac{\text{volume}}{\text{molar volume}}$

Solutions and Titrations

Concentration measures the amount of solute in a specific volume of solution, typically expressed in mol dm$^{-3}$ or g dm$^{-3}$. Note that $1$ dm$^3$ is equal to $1,000$ cm$^3$.

$\text{amount in mol} = \text{concentration in mol dm}^{-3} \times \text{volume in dm}^3$

Titration calculations use the known concentration and volume of one solution to find the unknown concentration of another. The process involves:

1. Calculating the moles of the reactant with a known concentration.
2. Using the balanced equation to find the molar ratio and thus the moles of the unknown reactant.
3. Calculating the concentration of the unknown solution using its measured volume.

Example: $25.0$ cm$^3$ of $NaOH$ is neutralised by $20.0$ cm$^3$ of $0.100$ mol dm$^{-3}$ $HCl$.

• Moles $HCl = 0.020 \times 0.100 = 0.002$ mol.
• Ratio is $1:1$, so moles $NaOH = 0.002$ mol.
• Concentration $NaOH = \frac{0.002}{0.025} = 0.080$ mol dm$^{-3}$.

Solubility and Yield

A saturated solution is one where no more solute can dissolve at a specific temperature. Solubility is usually measured in grams of solute per $100$ g of solvent. Solubility often changes with temperature, as seen in the following charts:

The percentage yield compares the mass of product actually obtained to the theoretical maximum:

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