Percentage Composition by Mass for the ESAT

Understanding Composition by Mass

In chemical analysis, we often need to know the mass contribution of each element within a compound. Because atoms are too small to weigh individually, we use the mole concept and relative masses to calculate these proportions. The mass of an element or compound is comprised entirely of the masses of its constituent atoms or ions. To find the percentage composition, we first determine the relative molar mass ($M_r$) of the substance by summing the relative atomic masses ($A_r$) of every atom in the formula.

For example, in one mole of calcium carbonate, $CaCO_3$, there is one mole of calcium ions, one mole of carbon atoms, and three moles of oxygen atoms. Using $A_r$ values ($Ca = 40, C = 12, O = 16$):

1. The mass contribution of calcium is $1 \times 40 = 40$ g.
2. The mass contribution of carbon is $1 \times 12 = 12$ g.
3. The mass contribution of oxygen is $3 \times 16 = 48$ g.

The total $M_r$ of $CaCO_3$ is $40 + 12 + 48 = 100$. From these totals, we can calculate the percentage of each element:

• Percentage $Ca = \frac{40}{100} \times 100 = 40\%$
• Percentage $C = \frac{12}{100} \times 100 = 12\%$
• Percentage $O = \frac{48}{100} \times 100 = 48\%$

The General Formula

To calculate the percentage composition by mass for any element in a compound, use the following expression:

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

Worked Example: Glucose

What is the percentage by mass of carbon in glucose, $C_6H_{12}O_6$? Use $A_r$ values: $H = 1.0, C = 12.0, O = 16.0$.

Step 1: Calculate the $M_r$ of glucose. $M_r = (6 \times 12.0) + (12 \times 1.0) + (6 \times 16.0) = 72 + 12 + 96 = 180$.

Step 2: Calculate the mass of carbon in the formula. There are 6 carbon atoms: $6 \times 12.0 = 72.0$.

Step 3: Apply the formula. Percentage $C = \frac{72.0}{180} \times 100 = 40\%$.

Percentage of Water in Hydrated Compounds

For hydrated crystals, we can calculate the percentage of water of crystallisation in a similar manner. The mass of the water molecules must be included in the total $M_r$ calculation.

Worked Example: Sodium Sulfide-9-Water

Calculate the percentage by mass of water in crystals of sodium sulfide-9-water, $Na_2S \cdot 9H_2O$. Given $M_r$ of $Na_2S \cdot 9H_2O = 240$ and $M_r$ of $H_2O = 18$.

Step 1: Calculate the total mass of water in the formula unit. There are 9 molecules of water: $9 \times 18 = 162$.

Step 2: Apply the formula. Percentage water = $\frac{162}{240} \times 100 = 67.5\%$.

Empirical and Molecular Formulae

The empirical formula is the simplest whole number ratio of atoms of each element in a compound. It can be determined directly from percentage composition data.

Example calculation for beryllium oxide: If a compound is 36% beryllium and 64% oxygen ($A_r: Be = 9.0, O = 16.0$):

1. Assume a 100 g sample: 36 g $Be$ and 64 g $O$.
2. Divide by $A_r$ to find moles: $36 / 9.0 = 4$ mol $Be$; $64 / 16.0 = 4$ mol $O$.
3. The ratio is 4:4, which simplifies to 1:1. The empirical formula is $BeO$.

The molecular formula is the actual number of atoms in one molecule. It is always a multiple of the empirical formula. If the empirical formula is $CHO_2$ (mass 45.0) and the relative molar mass is 90.0, the multiplier is $90.0 / 45.0 = 2$. The molecular formula is therefore $C_2H_2O_4$.