Percentage Composition, Empirical and Molecular Formulas Revision Notes for VCE SSCE Chemistry

Percentage Composition, Empirical and Molecular Formulas

Introduction

Compounds are made up of two or more different elements chemically bonded together. We can describe the composition of compounds in two main ways:

By mass: showing the percentage of each element's contribution to the total mass
By formula: showing either the ratio of atoms (empirical formula) or the actual number of atoms (molecular formula)

Understanding these representations is essential for working with chemical formulas and equations.

Percentage composition by mass

What is percentage composition?

Percentage composition by mass tells us what proportion of a compound's total mass comes from each element present. This is expressed as a percentage.

infoNote

For example, in ammonia ( $\text{NH}_3$ ), nitrogen accounts for 82% of the total mass, whilst hydrogen contributes 18%. This means that if you had 100 g of ammonia, 82 g would be nitrogen and 18 g would be hydrogen.

Calculating percentage composition

To calculate the percentage composition of an element in a compound, you need to know:

The molar mass of the compound
The mass of the element in one mole of the compound

Formula:

$\text{\% by mass of element} = \frac{\text{mass of the element in 1 mol of the compound}}{\text{molar mass of the compound}} \times 100$

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Worked Example: Percentage composition of carbon dioxide

Let's calculate the percentage by mass of carbon in carbon dioxide ( $\text{CO}_2$ ).

Step 1: Find the molar mass of $\text{CO}_2$

$M(\text{CO}_2) = 12.0 + (2 \times 16.0) = 44.0 \text{ g mol}^{-1}$

Step 2: Find the mass of carbon in one mole

$\text{mass of C in 1 mol} = 12.0 \text{ g}$

Step 3: Calculate the percentage

$\text{\% by mass of C} = \frac{12.0}{44.0} \times 100 = 27.3\%$

This tells us that carbon makes up 27.3% of the mass of carbon dioxide.

Empirical formula

What is an empirical formula?

The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound. It doesn't necessarily show the actual number of atoms in a molecule—just the simplest ratio.

chatImportant

Key points:

Ionic compounds always use empirical formulas because they exist as lattice structures, not discrete molecules
For molecular compounds, the empirical formula may be the same as the molecular formula, or it may be simplified

Examples of empirical formulas

Compound	Empirical formula	Simplest whole-number ratio
water ( $\text{H}_2\text{O}$ )	$\text{H}_2\text{O}$	H = 2:1
ethene ( $\text{C}_2\text{H}_4$ )	$\text{CH}_2$	C = 1:2
ethane ( $\text{C}_2\text{H}_6$ )	$\text{CH}_3$	C = 1:3
calcium carbonate	$\text{CaCO}_3$	Ca:C = 1:1:3

Determining empirical formulas from percentage composition

To find the empirical formula when you know the percentage composition, follow these steps:

Step 1: Obtain the mass of each element in the compound

If given percentages, assume you have 100 g of the compound
The percentages then become masses in grams

Step 2: Calculate the amount in moles of each element

Use the formula: $n = \frac{m}{M}$
Where $n$ is amount in moles, $m$ is mass, and $M$ is molar mass

Step 3: Convert to whole-number ratios

Divide each mole value by the smallest number of moles calculated
This gives you the simplest ratio

Step 4: Write the empirical formula

Use the whole-number ratios as subscripts in the formula

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Worked Example: Finding empirical formula from percentage composition

An organic compound contains 52.2% carbon, 13.0% hydrogen, and the remainder is oxygen. Let's find its empirical formula.

Step 1: Find the mass of each element in 100 g of compound

$m(\text{C}) = 52.2 \text{ g}$ $m(\text{H}) = 13.0 \text{ g}$ $m(\text{O}) = 100 - 52.2 - 13.0 = 34.8 \text{ g}$

Step 2: Calculate moles of each element

$n(\text{C}) = \frac{52.2}{12.0} = 4.35 \text{ mol}$

$n(\text{H}) = \frac{13.0}{1.0} = 13.0 \text{ mol}$

$n(\text{O}) = \frac{34.8}{16.0} = 2.18 \text{ mol}$

Step 3: Divide by the smallest value (2.18)

$\text{C} = \frac{4.35}{2.18} = 2.0$

$\text{H} = \frac{13.0}{2.18} = 6.0$

$\text{O} = \frac{2.18}{2.18} = 1.0$

Step 4: Write the empirical formula

The ratio is C:H

= 2:6:1

Therefore, the empirical formula is $\text{C}_2\text{H}_6\text{O}$

Elemental microanalysis: A case study

Analysing warfarin composition

Warfarin is an anticoagulant drug that prevents blood clots. It's also used as rat poison. The empirical formula of warfarin can be determined through elemental microanalysis.

The combustion analysis method

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Elemental microanalysis involves burning an organic compound in oxygen. The combustion products are then absorbed and weighed:

Carbon dioxide is absorbed by sodium hydroxide solution
Water is absorbed by a drying agent such as magnesium perchlorate, $\text{Mg(ClO}_4)_2$

Calculating masses from combustion data

For carbon:

From the mass of $\text{CO}_2$ produced:

$n(\text{CO}_2) = \frac{m}{44.0}$

Since all carbon in the original compound ends up in $\text{CO}_2$ :

$n(\text{C}) = n(\text{CO}_2)$

$m(\text{C}) = n \times 12.0$

For hydrogen:

From the mass of $\text{H}_2\text{O}$ produced:

$n(\text{H}_2\text{O}) = \frac{m}{18.0}$

Since each water molecule contains two hydrogen atoms:

$n(\text{H}) = 2 \times n(\text{H}_2\text{O})$

$m(\text{H}) = n \times 1.0$

For oxygen:

If the original mass of the compound is known, and we've calculated the masses of carbon and hydrogen:

$m(\text{O}) = m(\text{total}) - m(\text{C}) - m(\text{H})$

chatImportant

Exam tip: In combustion analysis questions, always remember that each water molecule contains two hydrogen atoms, so you must multiply the moles of water by 2 to get the moles of hydrogen.

Molecular formula

What is a molecular formula?

The molecular formula shows the actual number of atoms of each element present in a molecule. This is different from the empirical formula, which only shows the simplest ratio.

chatImportant

Ionic compounds do not have molecular formulas because they don't exist as discrete molecules—they form crystal lattices instead.

Comparing empirical and molecular formulas

Molecule	Molecular formula	Empirical formula
water	$\text{H}_2\text{O}$	$\text{H}_2\text{O}$
ethane	$\text{C}_2\text{H}_6$	$\text{CH}_3$
carbon dioxide	$\text{CO}_2$	$\text{CO}_2$
glucose	$\text{C}_6\text{H}_{12}\text{O}_2$	$\text{CH}_2\text{O}$

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Notice that:

Sometimes the molecular and empirical formulas are identical (water, carbon dioxide)
Sometimes the molecular formula is a whole-number multiple of the empirical formula (ethane, glucose)

Determining molecular formulas

To find the molecular formula, you need:

The empirical formula
The molar mass of the compound

The molecular formula is always a whole-number multiple of the empirical formula.

Formula:

$\text{Number of empirical formula units} = \frac{\text{molar mass of the compound}}{\text{molar mass of one empirical formula unit}}$

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Worked Example: Finding molecular formula

A compound has the empirical formula $\text{CH}$ . Its molar mass is 78 g mol $^{-1}$ . What is the molecular formula?

Step 1: Calculate the molar mass of one empirical formula unit

$M(\text{CH}) = 12.0 + 1.0 = 13.0 \text{ g mol}^{-1}$

Step 2: Determine how many empirical formula units are in the molecule

$\text{Number of CH units} = \frac{78}{13.0} = 6$

Step 3: Write the molecular formula

$\text{Molecular formula} = 6 \times \text{CH} = \text{C}_6\text{H}_6$

This compound is benzene.

chatImportant

Exam tip: Always check that your molecular formula makes chemical sense. The molecular formula must always be a whole-number multiple of the empirical formula.

Remember!

bookmarkSummary

Key Points to Remember:

Percentage composition shows what fraction of a compound's mass comes from each element, calculated using: $\text{\% by mass} = \frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100$
The empirical formula gives the simplest whole-number ratio of atoms in a compound. For ionic compounds, this is the only formula type used.
The molecular formula shows the actual number of atoms in a molecule and is always a whole-number multiple of the empirical formula.
To find empirical formulas: convert masses to moles ( $n = \frac{m}{M}$ ), divide by the smallest value to get ratios, then write the formula.
To find molecular formulas: divide the compound's molar mass by the empirical formula's molar mass, then multiply the empirical formula by this factor.

Percentage Composition, Empirical and Molecular Formulas (VCE SSCE Chemistry): Revision Notes