Chemical formulae, equations and calculations

What you'll learn

Write word equations and balanced symbol equations with state symbols.
Use $A_r$ , $M_r$ and moles to calculate masses and yields.
Find empirical and molecular formulae from experimental data.
Use concentration and gas volume calculations, including the Paper 2-only parts.

Equations: describing chemical reactions

A word equation uses the names of substances, for example:

magnesium + oxygen → magnesium oxide

A symbol equation uses chemical formulae. A balanced chemical equation has the same number of each type of atom on both sides, because atoms are rearranged in a reaction, not created or destroyed.

Definition

State symbols

State symbols show the physical state of each substance: (s) solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water.

For example, burning magnesium is:

$O_2(g) \to 2MgO(s)$

For unfamiliar reactions, use the information given to identify the reactants and products, write their formulae correctly, then balance the atoms.

Example

Balancing a symbol equation

Balance aluminium reacting with oxygen to form aluminium oxide, $Al_2O_3$ .

Start with the correct formulae: $O_2 \to Al_2O_3$ . Oxygen atoms are 2 on the left and 3 on the right, so use the lowest common multiple, 6: put 3 in front of $O_2$ and 2 in front of $Al_2O_3$ .
Now count aluminium atoms on the right: $2Al_2O_3$ contains 4 aluminium atoms, so put 4 in front of $A l$ .
Add state symbols: $3O_2(g) \to 2Al_2O_3(s)$ .

Common Mistake

Changing formulae while balancing

Do not change small numbers inside formulae, such as the 2 in $H_2O$ . Only change the big numbers in front, called coefficients.

Relative masses: $A_r$ and $M_r$

Definition

Relative atomic mass and relative formula mass

The relative atomic mass, $A_r$ , is the relative mass of an atom compared with carbon-12. The relative formula mass, $M_r$ , is the total of the $A_r$ values for all atoms in a formula; for molecular substances it may also be called relative molecular mass.

For example, $H_2O$ contains 2 hydrogen atoms and 1 oxygen atom, so its $M_r$ is $2 (1) + 16 = 18$ .

Example

Calculating relative formula mass and amount

Calculate the $M_r$ of $Ca(NO_3)_2$ and the amount in 16.4 g of it. Use $A_r$ : Ca = 40, N = 14, O = 16.

Interpret the formula carefully: $Ca(NO_3)_2$ contains 1 Ca atom, 2 N atoms and 6 O atoms.
Add the relative atomic masses: $M_r = 40 + 2(14) + 6(16) = 164$ .
Use $\frac{m}{M_r}$ , so $\frac{16.4}{164} = 0.100\ \text{mol}$ .

The mole and the big calculation map

Definition

The mole

The mole, symbol mol, is the unit for amount of substance. One mole of a substance has a mass equal to its $M_r$ in grams.

The most important mass relationship is:

\frac{m}{M_r}

where $n$ is amount in mol and $m$ is mass in g.

Most quantitative chemistry is about choosing the right route to moles, using the balanced equation ratio, then converting to the unit asked for.

Mole calculation map showing links between mass, amount, concentration, gas volume and equation ratios

Key Idea

Use moles as the bridge

Masses, solution concentrations and gas volumes all become easier once you convert them into moles.

Reacting masses and percentage yield

A balanced equation gives a mole ratio, not a mass ratio. For example:

$O_2(g) \to 2MgO(s)$

This means 2 mol of magnesium react to form 2 mol of magnesium oxide, so the mole ratio $M g : M g O$ is 1:1.

Definition

Percentage yield

The theoretical yield is the maximum product predicted by the equation. The actual yield is what is really obtained. Percentage yield compares them: $yield×100%\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$ .

Example

Calculating reacting mass and percentage yield

4.80 g of magnesium is burnt in excess oxygen. The actual mass of magnesium oxide collected is 7.50 g. Calculate the theoretical mass and percentage yield.

Convert magnesium mass into moles: $\frac{4.80}{24.0} = 0.200\ \text{mol}$ .
Use the equation ratio. Since $M g : M g O$ is 1:1, moles of $0.200\ \text{mol}$ .
Convert moles of $M g O$ into mass: $M_r(MgO) = 24 + 16 = 40$ , so $\times 40 = 8.00\ \text{g}$ .
Calculate percentage yield: $7.508.00×100%=93.8%\frac{7.50}{8.00} \times 100\% = 93.8\%$ .

Yields are often below 100% because some reactant may not react, some product may be lost during handling, or side reactions may occur.

Empirical and molecular formulae

Definition

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms or ions in a compound. The molecular formula gives the actual number of each type of atom in one molecule.

To find an empirical formula from data:

Find the mass of each element.
Convert each mass into moles.
Divide all mole values by the smallest.
Multiply if needed to get whole numbers.

Example

Finding empirical and molecular formulae

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Its $M_r$ is 180. Find its empirical and molecular formulae.

Assume 100 g, so the masses are 40.0 g C, 6.7 g H and 53.3 g O. Convert to moles: C is $40.012=3.33\frac{40.0}{12} = 3.33$ , H is $6.71=6.7\frac{6.7}{1} = 6.7$ , and O is $53.316=3.33\frac{53.3}{16} = 3.33$ .
Divide by the smallest value, 3.33: the ratio is C:H:O = 1:2:1, so the empirical formula is $CH_2O$ .
Find the multiplier: empirical formula mass is $12 + 2 (1) + 16 = 30$ , and $18030=6\frac{180}{30} = 6$ , so the molecular formula is $C_6H_{12}O_6$ .

Tip

Awkward ratios

If you get a ratio such as 1:1.5, multiply all parts by 2. If you get 1:1.33, multiply all parts by 3.

Finding formulae experimentally

Simple formulae can be found by measuring masses before and after a reaction, then converting mass changes into mole ratios.

For metal oxides, you can either burn a metal in oxygen or reduce a metal oxide. For water, you can form water from hydrogen and oxygen, then compare the masses of hydrogen and oxygen present. For salts containing water of crystallisation, heat the hydrated salt; the mass lost is water.

Definition

Water of crystallisation

Water of crystallisation is water chemically included in the crystal structure of a hydrated salt, for example in $CuSO4⋅5H2OCuSO_4 \cdot 5H_2O$ .

Example

Finding water of crystallisation

A 2.50 g sample of hydrated copper(II) sulfate is heated to constant mass. The anhydrous copper(II) sulfate left has mass 1.60 g. Find the formula $CuSO4⋅xH2OCuSO_4 \cdot xH_2O$ . Use $M_r(CuSO_4)=160$ and $M_r(H_2O)=18$ .

Find the mass of water lost: $0.90\ \text{g}$ .
Convert to moles: $moln(CuSO_4)=\frac{1.60}{160}=0.0100\ \text{mol}$ and $moln(H_2O)=\frac{0.90}{18}=0.0500\ \text{mol}$ .
Divide by the smaller amount: $CuSO_4:H_2O = 0.0100:0.0500 = 1:5$ , so the formula is $CuSO4⋅5H2OCuSO_4 \cdot 5H_2O$ .

Common Mistake

Heating hydrated salts

Heat gently and to constant mass. If the salt decomposes, the mass lost is not just water, so the calculated formula will be wrong.

Named practical: formula of a metal oxide

The two common practical routes are combustion of magnesium and reduction of copper(II) oxide.

Apparatus for determining metal oxide formula by heating magnesium in a crucible and reducing copper(II) oxide with hydrogen

Magnesium oxide by combustion

The reaction is:

$O_2(g) \to 2MgO(s)$

Method:

Weigh a clean, dry crucible and lid.
Add cleaned magnesium ribbon and reweigh.
Heat strongly with the lid slightly lifted at intervals to let oxygen in but reduce loss of white magnesium oxide smoke.
Cool, weigh, then reheat and reweigh until the mass is constant.

The magnesium changes into a white solid, magnesium oxide, and the mass increases because oxygen has been added. Calculate mass of magnesium from the starting data, mass of oxygen from the mass gain, then convert both to moles and find the simplest ratio.

Copper(II) oxide by reduction

The reaction with hydrogen is:

$H_2(g) \to Cu(s) + H_2O(l)$

Dry hydrogen is passed over heated copper(II) oxide. Black copper(II) oxide turns to pink-brown copper and water forms. The mass decreases because oxygen is removed from the oxide.

Key practical points:

Use excess oxygen for magnesium or excess hydrogen for copper(II) oxide.
Heat to constant mass to improve reliability.
Cool before weighing, because hot apparatus gives unreliable balance readings.
Main errors include loss of powder, incomplete reaction, wet apparatus, or re-oxidising hot copper in air.
Hydrogen is flammable, so air must be flushed out before lighting the excess hydrogen safely.

Solution concentration calculations

Some calculations involving solution concentration are Paper 2 only. Concentration in mol/dm³ means the amount of solute in 1 dm³ of solution.

n = c V

Here $n$ is amount in mol, $c$ is concentration in mol/dm³ and $V$ is volume in dm³.

Common Mistake

Using cm³ directly

Before using $n = c V$ , convert cm³ to dm³ by dividing by 1000.

Example

Using concentration to find mass

Calculate the mass of sodium hydroxide, $N a O H$ , needed to make 250 cm³ of 0.200 mol/dm³ solution.

Convert the volume: 250 cm³ = 0.250 dm³.
Calculate amount: $\times 0.250 = 0.0500\ \text{mol}$ .
Convert moles to mass: $M_r(NaOH)=23+16+1=40$ , so $\times 40 = 2.00\ \text{g}$ .

Gas volume calculations at rtp

Gas volume calculations are also Paper 2 only. At room temperature and pressure, rtp, 1 mol of any gas has volume 24 dm³, which is 24 000 cm³.

\frac{V_{\text{gas}}}{24}

Use this when the gas volume is in dm³. If the volume is in cm³, divide by 24 000 instead.

Example

Calculating gas volume at rtp

Calculate the volume of carbon dioxide formed at rtp when 5.00 g of calcium carbonate reacts with excess hydrochloric acid.

$CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)CaCO_3(s) + 2HCl(aq) \to CaCl_2(aq) + CO_2(g) + H_2O(l)$

Convert calcium carbonate mass into moles: $M_r(CaCO_3)=40+12+3(16)=100$ , so $moln=\frac{5.00}{100}=0.0500\ \text{mol}$ .
Use the equation ratio. $CaCO_3:CO_2$ is 1:1, so moles of $molCO_2 = 0.0500\ \text{mol}$ .
Convert moles of gas to volume: $\times 24 = 1.20\ \text{dm}^3$ , which is 1200 cm³.

Exam technique

In the exam

Balance the symbol equation first, then use the coefficients as mole ratios.
Convert the given quantity into moles: mass uses $n=mMrn=\frac{m}{M_r}$ , solution uses $n = c V$ , and gas at rtp uses 24 dm³ per mol.
Keep units with every calculation, especially converting cm³ to dm³ before concentration work.
For empirical formulae, divide all mole values by the smallest and only round to sensible whole-number ratios.

Self review

Check yourself

Why must you change coefficients, not formula subscripts, when balancing equations?
A hydrated salt loses mass when heated: how would you use the mass loss to find water of crystallisation?
In which calculations must you convert cm³ into dm³ before substituting into the formula?

What you'll learn

Write word equations and balanced symbol equations with state symbols.
Use $A_r$ , $M_r$ and moles to calculate masses and yields.
Find empirical and molecular formulae from experimental data.
Use concentration and gas volume calculations, including the Paper 2-only parts.

Equations: describing chemical reactions

A word equation uses the names of substances, for example:

magnesium + oxygen → magnesium oxide

Definition

State symbols

State symbols show the physical state of each substance: (s) solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water.

For example, burning magnesium is:

$O_2(g) \to 2MgO(s)$

For unfamiliar reactions, use the information given to identify the reactants and products, write their formulae correctly, then balance the atoms.

Example

Balancing a symbol equation

Balance aluminium reacting with oxygen to form aluminium oxide, $Al_2O_3$ .

Start with the correct formulae: $O_2 \to Al_2O_3$ . Oxygen atoms are 2 on the left and 3 on the right, so use the lowest common multiple, 6: put 3 in front of $O_2$ and 2 in front of $Al_2O_3$ .
Now count aluminium atoms on the right: $2Al_2O_3$ contains 4 aluminium atoms, so put 4 in front of $A l$ .
Add state symbols: $3O_2(g) \to 2Al_2O_3(s)$ .

Common Mistake

Changing formulae while balancing

Do not change small numbers inside formulae, such as the 2 in $H_2O$ . Only change the big numbers in front, called coefficients.

Relative masses: $A_r$ and $M_r$

Definition

Relative atomic mass and relative formula mass

For example, $H_2O$ contains 2 hydrogen atoms and 1 oxygen atom, so its $M_r$ is $2 (1) + 16 = 18$ .

Example

Calculating relative formula mass and amount

Calculate the $M_r$ of $Ca(NO_3)_2$ and the amount in 16.4 g of it. Use $A_r$ : Ca = 40, N = 14, O = 16.

Interpret the formula carefully: $Ca(NO_3)_2$ contains 1 Ca atom, 2 N atoms and 6 O atoms.
Add the relative atomic masses: $M_r = 40 + 2(14) + 6(16) = 164$ .
Use $\frac{m}{M_r}$ , so $\frac{16.4}{164} = 0.100\ \text{mol}$ .

The mole and the big calculation map

Definition

The mole

The mole, symbol mol, is the unit for amount of substance. One mole of a substance has a mass equal to its $M_r$ in grams.

The most important mass relationship is:

\frac{m}{M_r}

where $n$ is amount in mol and $m$ is mass in g.

Most quantitative chemistry is about choosing the right route to moles, using the balanced equation ratio, then converting to the unit asked for.

Mole calculation map showing links between mass, amount, concentration, gas volume and equation ratios

Key Idea

Use moles as the bridge

Masses, solution concentrations and gas volumes all become easier once you convert them into moles.

Reacting masses and percentage yield

A balanced equation gives a mole ratio, not a mass ratio. For example:

$O_2(g) \to 2MgO(s)$

This means 2 mol of magnesium react to form 2 mol of magnesium oxide, so the mole ratio $M g : M g O$ is 1:1.

Definition

Percentage yield

Example

Calculating reacting mass and percentage yield

4.80 g of magnesium is burnt in excess oxygen. The actual mass of magnesium oxide collected is 7.50 g. Calculate the theoretical mass and percentage yield.

Convert magnesium mass into moles: $\frac{4.80}{24.0} = 0.200\ \text{mol}$ .
Use the equation ratio. Since $M g : M g O$ is 1:1, moles of $0.200\ \text{mol}$ .
Convert moles of $M g O$ into mass: $M_r(MgO) = 24 + 16 = 40$ , so $\times 40 = 8.00\ \text{g}$ .
Calculate percentage yield: $7.508.00×100%=93.8%\frac{7.50}{8.00} \times 100\% = 93.8\%$ .

Yields are often below 100% because some reactant may not react, some product may be lost during handling, or side reactions may occur.

Empirical and molecular formulae

Definition

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms or ions in a compound. The molecular formula gives the actual number of each type of atom in one molecule.

To find an empirical formula from data:

Find the mass of each element.
Convert each mass into moles.
Divide all mole values by the smallest.
Multiply if needed to get whole numbers.

Example

Finding empirical and molecular formulae

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Its $M_r$ is 180. Find its empirical and molecular formulae.

Assume 100 g, so the masses are 40.0 g C, 6.7 g H and 53.3 g O. Convert to moles: C is $40.012=3.33\frac{40.0}{12} = 3.33$ , H is $6.71=6.7\frac{6.7}{1} = 6.7$ , and O is $53.316=3.33\frac{53.3}{16} = 3.33$ .
Divide by the smallest value, 3.33: the ratio is C:H:O = 1:2:1, so the empirical formula is $CH_2O$ .
Find the multiplier: empirical formula mass is $12 + 2 (1) + 16 = 30$ , and $18030=6\frac{180}{30} = 6$ , so the molecular formula is $C_6H_{12}O_6$ .

Tip

Awkward ratios

If you get a ratio such as 1:1.5, multiply all parts by 2. If you get 1:1.33, multiply all parts by 3.

Finding formulae experimentally

Simple formulae can be found by measuring masses before and after a reaction, then converting mass changes into mole ratios.

Definition

Water of crystallisation

Water of crystallisation is water chemically included in the crystal structure of a hydrated salt, for example in $CuSO4⋅5H2OCuSO_4 \cdot 5H_2O$ .

Example

Finding water of crystallisation

Find the mass of water lost: $0.90\ \text{g}$ .
Convert to moles: $moln(CuSO_4)=\frac{1.60}{160}=0.0100\ \text{mol}$ and $moln(H_2O)=\frac{0.90}{18}=0.0500\ \text{mol}$ .
Divide by the smaller amount: $CuSO_4:H_2O = 0.0100:0.0500 = 1:5$ , so the formula is $CuSO4⋅5H2OCuSO_4 \cdot 5H_2O$ .

Common Mistake

Heating hydrated salts

Heat gently and to constant mass. If the salt decomposes, the mass lost is not just water, so the calculated formula will be wrong.

Named practical: formula of a metal oxide

The two common practical routes are combustion of magnesium and reduction of copper(II) oxide.

Apparatus for determining metal oxide formula by heating magnesium in a crucible and reducing copper(II) oxide with hydrogen

Magnesium oxide by combustion

The reaction is:

$O_2(g) \to 2MgO(s)$

Method:

Weigh a clean, dry crucible and lid.
Add cleaned magnesium ribbon and reweigh.
Heat strongly with the lid slightly lifted at intervals to let oxygen in but reduce loss of white magnesium oxide smoke.
Cool, weigh, then reheat and reweigh until the mass is constant.

Copper(II) oxide by reduction

The reaction with hydrogen is:

$H_2(g) \to Cu(s) + H_2O(l)$

Dry hydrogen is passed over heated copper(II) oxide. Black copper(II) oxide turns to pink-brown copper and water forms. The mass decreases because oxygen is removed from the oxide.

Key practical points:

Use excess oxygen for magnesium or excess hydrogen for copper(II) oxide.
Heat to constant mass to improve reliability.
Cool before weighing, because hot apparatus gives unreliable balance readings.
Main errors include loss of powder, incomplete reaction, wet apparatus, or re-oxidising hot copper in air.
Hydrogen is flammable, so air must be flushed out before lighting the excess hydrogen safely.

Solution concentration calculations

Some calculations involving solution concentration are Paper 2 only. Concentration in mol/dm³ means the amount of solute in 1 dm³ of solution.

n = c V

Here $n$ is amount in mol, $c$ is concentration in mol/dm³ and $V$ is volume in dm³.

Common Mistake

Using cm³ directly

Before using $n = c V$ , convert cm³ to dm³ by dividing by 1000.

Example

Using concentration to find mass

Calculate the mass of sodium hydroxide, $N a O H$ , needed to make 250 cm³ of 0.200 mol/dm³ solution.

Convert the volume: 250 cm³ = 0.250 dm³.
Calculate amount: $\times 0.250 = 0.0500\ \text{mol}$ .
Convert moles to mass: $M_r(NaOH)=23+16+1=40$ , so $\times 40 = 2.00\ \text{g}$ .

Gas volume calculations at rtp

Gas volume calculations are also Paper 2 only. At room temperature and pressure, rtp, 1 mol of any gas has volume 24 dm³, which is 24 000 cm³.

\frac{V_{\text{gas}}}{24}

Use this when the gas volume is in dm³. If the volume is in cm³, divide by 24 000 instead.

Example

Calculating gas volume at rtp

Calculate the volume of carbon dioxide formed at rtp when 5.00 g of calcium carbonate reacts with excess hydrochloric acid.

$CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)CaCO_3(s) + 2HCl(aq) \to CaCl_2(aq) + CO_2(g) + H_2O(l)$

Convert calcium carbonate mass into moles: $M_r(CaCO_3)=40+12+3(16)=100$ , so $moln=\frac{5.00}{100}=0.0500\ \text{mol}$ .
Use the equation ratio. $CaCO_3:CO_2$ is 1:1, so moles of $molCO_2 = 0.0500\ \text{mol}$ .
Convert moles of gas to volume: $\times 24 = 1.20\ \text{dm}^3$ , which is 1200 cm³.

Exam technique

In the exam

Balance the symbol equation first, then use the coefficients as mole ratios.
Convert the given quantity into moles: mass uses $n=mMrn=\frac{m}{M_r}$ , solution uses $n = c V$ , and gas at rtp uses 24 dm³ per mol.
Keep units with every calculation, especially converting cm³ to dm³ before concentration work.
For empirical formulae, divide all mole values by the smallest and only round to sensible whole-number ratios.

Self review

Check yourself

Why must you change coefficients, not formula subscripts, when balancing equations?
A hydrated salt loses mass when heated: how would you use the mass loss to find water of crystallisation?
In which calculations must you convert cm³ into dm³ before substituting into the formula?

What you'll learn

Equations: describing chemical reactions

Relative masses: $A_r$ and $M_r$

The mole and the big calculation map

Reacting masses and percentage yield

Empirical and molecular formulae

Finding formulae experimentally

Named practical: formula of a metal oxide

Magnesium oxide by combustion

Copper(II) oxide by reduction

Solution concentration calculations

Gas volume calculations at rtp

Review Chemical formulae, equations and calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical formulae, equations and calculations

What you'll learn

Equations: describing chemical reactions

Relative masses: $A_r$ and $M_r$

The mole and the big calculation map

Reacting masses and percentage yield

Empirical and molecular formulae

Finding formulae experimentally

Named practical: formula of a metal oxide

Magnesium oxide by combustion

Copper(II) oxide by reduction

Solution concentration calculations

Gas volume calculations at rtp

Review Chemical formulae, equations and calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical formulae, equations and calculations

What you'll learn

Equations: describing chemical reactions

Relative masses: ArA_rAr​ and MrM_rMr​

The mole and the big calculation map

Reacting masses and percentage yield

Empirical and molecular formulae

Finding formulae experimentally

Named practical: formula of a metal oxide

Magnesium oxide by combustion

Copper(II) oxide by reduction

Solution concentration calculations

Gas volume calculations at rtp

Review Chemical formulae, equations and calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical formulae, equations and calculations Revision Guide

Chemical formulae, equations and calculations

What you'll learn

Equations: describing chemical reactions

Relative masses: ArA_rAr​ and MrM_rMr​

The mole and the big calculation map

Reacting masses and percentage yield

Empirical and molecular formulae

Finding formulae experimentally

Named practical: formula of a metal oxide

Magnesium oxide by combustion

Copper(II) oxide by reduction

Solution concentration calculations

Gas volume calculations at rtp

Review Chemical formulae, equations and calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical formulae, equations and calculations Revision Guide

Relative masses: $A_r$ and $M_r$

Relative masses: $A_r$ and $M_r$