Percentage Composition and Calculating Formulae Revision Notes for HSC SSCE Chemistry

Percentage Composition and Calculating Formulae

Understanding percentage composition

Percentage composition tells us the proportion by mass of each element present in a chemical compound. This is different from the simple atom ratio shown in a chemical formula.

For example, in aluminium oxide ( $\text{Al}_2\text{O}_3$ ), the formula shows us that aluminium and oxygen atoms are present in a $2:3$ ratio. Alternatively, it tells us that one mole of $\text{Al}_2\text{O}_3$ contains two moles of aluminium and three moles of oxygen.

However, percentage composition tells us the ratio by mass of the elements. This information is particularly useful in real-world applications:

Note

Real-World Applications of Percentage Composition:

Industrialists can determine how much aluminium can be extracted from a tonne of aluminium oxide ore
Farmers can decide which fertiliser contains the most nitrogen per kilogram of product

Calculating percentage composition from a chemical formula

To find the percentage composition of an element in a compound, we use the following approach:

$\%A \text{ in a compound} = \frac{\text{mass of A in one mole of the compound}}{\text{mass of one mole of the compound}} \times 100$

For a compound with the general formula $\text{A}_w\text{B}_y\text{C}_z$ , this becomes:

$\%A = \frac{w \times (\text{relative atomic mass of A}) \times 100}{\text{relative molecular or formula mass of } \text{A}_w\text{B}_y\text{C}_z}$

Example

Worked Example: Calculating the Percentage of Iron in Haematite

Let's calculate the percentage of iron in the common ore haematite, $\text{Fe}_2\text{O}_3$ . The relative atomic masses are: Fe = $55.85$ , O = $16.00$ .

Step 1: Calculate the molar mass ( $MM$ ) of $\text{Fe}_2\text{O}_3$

$MM = 2 \times 55.85 + 3 \times 16.00 = 159.7 \text{ g mol}^{-1}$

Step 2: Calculate the mass of iron in one mole of $\text{Fe}_2\text{O}_3$

One formula unit of $\text{Fe}_2\text{O}_3$ contains two atoms of iron. Therefore, one mole of $\text{Fe}_2\text{O}_3$ contains two moles of iron.

$\text{Mass of Fe} = 2 \times 55.85 = 111.7 \text{ g}$

Step 3: Calculate the percentage

$\% \text{iron in } \text{Fe}_2\text{O}_3 = \frac{111.7}{159.7} \times 100 = 69.9\%$

Result: This means that haematite is approximately 70% iron by mass.

What is an empirical formula?

An empirical formula shows the simplest whole-number ratio of atoms present in a compound. This is different from a molecular formula, which shows the actual number of atoms of each type in one molecule.

For example, consider these carbon-based molecular compounds:

Ethylene: $\text{C}_2\text{H}_4$
Propylene: $\text{C}_3\text{H}_6$
Octene: $\text{C}_8\text{H}_{16}$
Cyclohexane: $\text{C}_6\text{H}_{12}$

In all of these compounds, carbon and hydrogen atoms are present in a 1:2 ratio. Therefore, they all share the same empirical formula: $\text{CH}_2$ .

Note

When we analyse a compound experimentally, we determine its empirical formula. To find the molecular formula, we need additional information (such as the molar mass of the compound).

Calculating empirical formulae from experimental data

To determine the empirical formula of a compound, chemists perform experiments that provide either:

The mass of each element in a known mass of compound, or
The percentage composition of the compound

The five-step method

Note

Systematic Approach to Finding Empirical Formulae:

Write down the masses of all elements present in the sample (if given percentages, treat them as masses in $100$ g of compound)
Convert masses to moles by dividing each mass by the element's molar mass (relative atomic mass in grams)
Divide by the smallest number of moles to get a simple ratio
If needed, multiply all numbers by a suitable factor to get whole numbers (e.g., multiply $1.3:2.7:1$ by $3$ to get $3.9:8.1:3$ )
Round off to get whole numbers and write the empirical formula (experimental errors of up to $10\%$ are common, so round sensibly)

Example

Worked Example: Determining the Formula of Lithium Oxide

When $4.97$ g of lithium metal reacted with excess oxygen, it was completely converted to lithium oxide. The mass of lithium oxide formed was $10.89$ g. Let's calculate the formula.

Step 1: Calculate the mass of oxygen

$\text{Mass of O} = 10.89 - 4.97 = 5.92 \text{ g}$

Step 2: Convert masses to moles

Using molar masses: Li = $6.941$ g mol $^{-1}$ , O = $16.00$ g mol $^{-1}$

$\text{Moles of Li} = \frac{4.97}{6.941} = 0.716 \text{ mol}$

$\text{Moles of O} = \frac{5.92}{16.00} = 0.370 \text{ mol}$

Step 3: Divide by the smallest number

$\frac{0.716}{0.370} : \frac{0.370}{0.370} = 1.94 : 1.0$

Step 4: Round to whole numbers

$1.94 : 1.0 \approx 2.0 : 1.0$

Result: The empirical formula is $\text{Li}_2\text{O}$ .

Example

Worked Example: Finding the Empirical Formula of CFC-113

A $2.37$ g sample of CFC-113 (once used in refrigerators, now banned) was analysed and found to contain $1.36$ g chlorine and $0.71$ g fluorine, with the remainder being carbon. Let's find its empirical formula.

Step 1: Calculate the mass of carbon

$\text{Mass of C} = 2.37 - 1.36 - 0.71 = 0.30 \text{ g}$

Step 2: Set up in tabular form

Element	Chlorine	Fluorine	Carbon
Mass (g)	$1.36$	$0.71$	$0.30$

Step 3: Convert to moles

Using molar masses: Cl = $35.5$ g mol $^{-1}$ , F = $19.0$ g mol $^{-1}$ , C = $12.0$ g mol $^{-1}$

Element	Chlorine	Fluorine	Carbon
Moles	$\frac{1.36}{35.5} = 0.0383$	$\frac{0.71}{19.0} = 0.0374$	$\frac{0.30}{12.0} = 0.0250$

Step 4: Divide by smallest number ( $0.0250$ )

$\frac{0.0383}{0.0250} : \frac{0.0374}{0.0250} : \frac{0.0250}{0.0250}$

$= 1.52 : 1.49 : 1.00$

Step 5: Multiply by 2 to get whole numbers

$3.06 : 2.98 : 2.00 \approx 3 : 3 : 2$

Result: The empirical formula is $\text{C}_2\text{F}_3\text{Cl}_3$ .

Practical investigations

Investigation 7.3: Empirical formula for magnesium oxide

This investigation involves heating magnesium ribbon in a crucible to convert it to magnesium oxide.

Aim: Determine the empirical formula of magnesium oxide

Important

Key Safety Considerations:

The intense white light from burning magnesium can hurt your eyes - keep the crucible lid on during heating and don't look directly at the burning metal
The crucible and lid become very hot - always use tongs to handle them

Method overview:

Clean and obtain the constant mass of an empty crucible and lid
Add magnesium ribbon and record the mass
Heat strongly for at least 30 minutes, occasionally lifting the lid to allow oxygen in
Cool and reweigh
Calculate the empirical formula from the mass changes

Note

Analysis: From the masses before and after heating, you can calculate:

Mass of magnesium used
Mass of oxygen that combined with the magnesium
Moles of each element
The whole-number ratio (empirical formula)

Investigation 7.4: Empirical formula for hydrated copper sulfate

Some salts combine with a definite number of water molecules when they crystallise. Hydrated copper sulfate ( $\text{CuSO}_4 \cdot x\text{H}_2\text{O}$ ) is blue, while anhydrous (water-free) copper sulfate ( $\text{CuSO}_4$ ) is white.

Aim: Determine the value of $x$ in the formula $\text{CuSO}_4 \cdot x\text{H}_2\text{O}$

Method overview:

Obtain the constant mass of an empty crucible and lid
Add about 5 g of blue hydrated copper sulfate and record mass
Heat with the lid ajar until all the salt changes from blue to white
Cool and reweigh
Calculate the empirical formula from mass changes

Note

Analysis: The loss in mass represents water driven off by heating. Calculate:

Mass of water lost
Mass of anhydrous copper sulfate remaining
Moles of water and moles of copper sulfate
The whole-number ratio to find $x$

Summary

Exam Tips:

Always show your working when calculating percentage composition or empirical formulae - marks are awarded for method
Check your rounding - remember that experimental errors up to $10\%$ are common, so $1.94$ can sensibly be rounded to $2.0$
Units matter - include units in all calculations (g, mol, g mol $^{-1}$ )
In empirical formula questions, if you get non-whole numbers after dividing by the smallest, multiply all numbers by 2, 3, or 4 to get whole numbers
For hydrated salt questions, remember that the water is shown separately with a dot (e.g., $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ )

Summary

Key Points to Remember:

Percentage composition shows the proportion by mass of each element in a compound, calculated using: $\%A = \frac{w \times (\text{relative atomic mass of A}) \times 100}{\text{molar mass of compound}}$
An empirical formula shows the simplest whole-number ratio of atoms in a compound, while a molecular formula shows the actual number of atoms in one molecule
To calculate an empirical formula: convert masses to moles, divide by the smallest number, multiply if needed to get whole numbers, then round sensibly
Practical investigations can determine empirical formulae by measuring mass changes when compounds are heated or formed from their elements
Always work systematically through calculations, showing clear steps and including units throughout

Percentage Composition and Calculating Formulae (HSC SSCE Chemistry): Revision Notes