Quantitative Chemistry: Empirical and Molecular Formulae

Quantitative chemistry relies on the ability to determine the composition of substances by mass and to express this composition through chemical formulae. These formulae provide a bridge between the macroscopic mass of a sample and the microscopic arrangement of atoms or ions.

Percentage Composition by Mass

The mass of an element within a compound is determined by the relative atomic masses ($A_r$) of the atoms present and the total relative molar mass ($M_r$) of the substance. To find the percentage composition by mass of an element in a compound, we use the following expression:

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

For example, in one mole of calcium carbonate ($CaCO_3$):

1. One mole of calcium ions has a mass of $40\text{ g}$.
2. One mole of carbon atoms has a mass of $12\text{ g}$.
3. Three moles of oxygen atoms have a mass of $3 \times 16 = 48\text{ g}$.

The total $M_r$ of $CaCO_3$ is $40 + 12 + 48 = 100$.

The percentage of calcium is $\frac{40}{100} \times 100 = 40\%$. The percentage of carbon is $\frac{12}{100} \times 100 = 12\%$. The percentage of oxygen is $\frac{48}{100} \times 100 = 48\%$.

Worked Example: Carbon in Glucose To find the percentage by mass of carbon in glucose ($C_6H_{12}O_6$): $M_r(C_6H_{12}O_6) = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 180$. Percentage of Carbon = $\frac{12.0 \times 6}{180} \times 100 = \frac{72}{180} \times 100 = 40\%$.

This method can also be used for hydrated compounds. For sodium sulfide:9:water ($Na_2S \cdot 9H_2O$): Given $M_r$ of $Na_2S \cdot 9H_2O = 240$ and $M_r$ of $H_2O = 18$. Percentage of water = $\frac{9 \times 18}{240} \times 100 = \frac{162}{240} \times 100 = 67.5\%$.

Empirical Formula from Percentage Composition

The empirical formula represents the simplest whole number ratio of the atoms of each element in a compound. It can be derived from the percentage composition by mass.

Worked Example: Beryllium and Oxygen A compound consists of $36\%$ beryllium and $64\%$ oxygen. Using $A_r$ values ($Be = 9.0$, $O = 16.0$):

1. Assume a $100\text{ g}$ sample. This gives $36\text{ g}$ of $Be$ and $64\text{ g}$ of $O$.
2. Calculate the amount in moles: For $Be$: $\frac{36}{9.0} = 4\text{ mol}$. For $O$: $\frac{64}{16} = 4\text{ mol}$.
3. Find the simplest ratio by dividing by the smallest value ($4$): $Be = 1$, $O = 1$. The empirical formula is $BeO$.

Empirical Formula from Reacting Masses

If experimental data provides the mass of reactants or products, the empirical formula is found by calculating the molar ratio of the elements.

Worked Example: Iron Chloride When $5.60\text{ g}$ of iron reacts, it forms $16.25\text{ g}$ of iron chloride. Using $A_r$ values ($Fe = 56.0$, $Cl = 35.5$):

1. Calculate the mass of chlorine: $16.25 - 5.60 = 10.65\text{ g}$.
2. Calculate the moles of each element: For $Fe$: $\frac{5.60}{56.0} = 0.1\text{ mol}$. For $Cl$: $\frac{10.65}{35.5} = 0.3\text{ mol}$.
3. Determine the simplest ratio: $0.1:0.3$ is $1:3$. The empirical formula is $FeCl_3$.

Determining the Molecular Formula

The molecular formula gives the actual number of atoms of each element in a single molecule of the compound. It is always a multiple of the empirical formula. To find it, you must know the relative molar mass ($M_r$) of the substance.

Worked Example: Finding a Molecular Formula A compound has the empirical formula $CHO_2$ and an $M_r$ of $90.0$. ($A_r$ values: $H = 1.0, C = 12.0, O = 16.0$):

1. Calculate the empirical formula mass: $12.0 + 1.0 + (2 \times 16.0) = 45.0$.
2. Find the multiplier ($n$): $n = \frac{M_r}{\text{empirical mass}} = \frac{90.0}{45.0} = 2$.
3. Multiply the empirical formula by $n$: $(CHO_2) \times 2 = C_2H_2O_4$. The molecular formula is $C_2H_2O_4$.