Calculations involving masses

What you'll learn

How to calculate relative formula mass and percentage composition.
How to find empirical and molecular formulae from data.
How conservation of mass explains mass changes in reactions.
How to calculate reacting masses, solution concentration, and — on Higher Tier — moles, particles and limiting reactants.

The building blocks: $A_r$ , formulae and balanced equations

Before the bigger calculations, you need three ideas.

Definition

Relative atomic mass, Ar

The relative atomic mass, $A_r$ , is the average mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom. In GCSE calculations, you usually use the $A_r$ values given in the periodic table.

A chemical formula tells you the elements present and how many atoms of each are in one particle or formula unit. For example, H₂O contains 2 hydrogen atoms and 1 oxygen atom.

A balanced equation shows the reacting ratio of substances. For example:

2Mg(s) + O₂(g) → 2MgO(s)

This means 2 magnesium atoms react with 1 oxygen molecule to form 2 formula units of magnesium oxide.

Relative formula mass, $M_r$

Definition

Relative formula mass

The relative formula mass, $M_r$ , is the total of the $A_r$ values for all the atoms shown in a formula. For simple molecules, it may also be called relative molecular mass.

For brackets, multiply everything inside the brackets by the number outside.

Example

Finding relative formula mass and percentage by mass

Find the $M_r$ of calcium hydroxide, Ca(OH)₂, and the percentage by mass of oxygen. Use $A_r$ : Ca = 40, O = 16, H = 1.

Count the atoms carefully: Ca(OH)₂ contains 1 Ca atom, 2 O atoms and 2 H atoms.
Add the relative masses: $Mr=40+(2×16)+(2×1)=74M_r = 40 + \left(2 \times 16\right) + \left(2 \times 1\right) = 74$ .
Find the oxygen part of the mass: oxygen contributes $\times 16 = 32$ .
Calculate the percentage: $3274×100=43.2%\frac{32}{74} \times 100 = 43.2\%$ .

Common Mistake

Forgetting the brackets

In Ca(OH)₂, the small 2 applies to both O and H. So there are 2 oxygen atoms and 2 hydrogen atoms, not just 1 oxygen and 2 hydrogen atoms.

Percentage by mass

Percentage by mass tells you what fraction of a compound’s mass comes from one element.

compound×100\% \text{ by mass} = \frac{\text{mass of the element in the formula}}{M_r \text{ of the compound}} \times 100

Key Idea

Percentage composition

A compound with a fixed formula always has the same percentage composition by mass, because its atoms are joined in a fixed ratio.

Empirical formulae

Definition

Empirical formula

An empirical formula is the simplest whole-number ratio of atoms of each element in a compound.

For example, the molecular formula C₆H₁₂O₆ simplifies to the empirical formula CH₂O, because 6:12:6 simplifies to 1:2:1.

When using reacting masses or percentage composition, convert each element’s mass to an amount ratio by dividing by its $A_r$ .

Example

Finding empirical and molecular formulae

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Its relative molecular mass is 180. Find its empirical and molecular formula. Use $A_r$ : C = 12, H = 1, O = 16.

Assume you have 100 g of the compound, so the masses are 40.0 g C, 6.7 g H and 53.3 g O.
Divide each mass by its $A_r$ : 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, to get the ratio C:H:O = 1:2:1, so the empirical formula is CH₂O.
Find the empirical formula mass: $\left(2 \times 1\right) + 16 = 30$ .
Compare with the molecular mass: $18030=6\frac{180}{30} = 6$ , so multiply CH₂O by 6 to get C₆H₁₂O₆.

Tip

If the ratio is not whole-number

If you get a ratio like 1:1.5, multiply all parts by 2 to make 2:3. If you get 1:1.33, multiply all parts by 3 to make 3:4.

Practical: empirical formula of magnesium oxide

You need to be able to describe an experiment to determine a simple empirical formula, such as magnesium oxide.

The idea is to heat magnesium so it reacts with oxygen from the air:

2Mg(s) + O₂(g) → 2MgO(s)

Diagram of the magnesium oxide empirical formula practical setup with crucible, lid, tripod and Bunsen burner

A sensible method is:

Weigh a clean, dry crucible and lid.
Add a coil of cleaned magnesium ribbon and weigh again.
Heat strongly with the lid slightly lifted sometimes to let oxygen in.
Keep the lid mostly on to reduce loss of white magnesium oxide smoke.
Cool, then reweigh the crucible, lid and product.
Repeat heating, cooling and weighing until the mass is constant.

Example

Using magnesium oxide results

A student heats 0.48 g of magnesium and obtains 0.80 g of magnesium oxide. Find the empirical formula. Use $A_r$ : Mg = 24, O = 16.

Find the mass of oxygen gained: $0.80 - 0.48 = 0.32$ g.
Convert masses to amount ratios: Mg is $0.4824=0.020\frac{0.48}{24} = 0.020$ , and O is $0.3216=0.020\frac{0.32}{16} = 0.020$ .
Divide by the smallest value: Mg:O = 1:1, so the empirical formula is MgO.

Common Mistake

Product escaping

If magnesium oxide smoke escapes, the measured product mass is too low, so the calculated oxygen mass will also be too low.

Conservation of mass

Definition

Law of conservation of mass

The law of conservation of mass says that atoms are not created or destroyed in a chemical reaction, so total mass is conserved.

In a closed system, no substances can enter or leave, so the measured mass stays the same. A stoppered flask is a closed system.

In an open flask, the balance reading can change if a gas enters or leaves. The atoms are still conserved, but not all the substances may remain on the balance.

Example

Explaining mass loss in an open flask

Marble chips react with hydrochloric acid in an open flask:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Identify the gas made: carbon dioxide, CO₂(g), is produced.
Because the flask is open, some CO₂(g) escapes into the air instead of staying on the balance.
The balance reading decreases, but the total mass of all reactants and products, including the escaped gas, is still conserved.

Reacting masses from balanced equations

A balanced equation gives a ratio. You can use that ratio to scale masses up or down.

Example

Calculating product mass from a balanced equation

Calcium carbonate decomposes when heated:

CaCO₃(s) → CaO(s) + CO₂(g)

Calculate the mass of carbon dioxide made from 25.0 g of calcium carbonate. Use $A_r$ : Ca = 40, C = 12, O = 16.

Find the relative formula masses: CaCO₃ has $Mr=40+12+(3×16)=100M_r = 40 + 12 + \left(3 \times 16\right) = 100$ , and CO₂ has $Mr=12+(2×16)=44M_r = 12 + \left(2 \times 16\right) = 44$ .
Use the equation ratio: 1 formula unit of CaCO₃ makes 1 molecule of CO₂, so 100 g of CaCO₃ makes 44 g of CO₂.
Scale to 25.0 g: $25.0100×44=11.0\frac{25.0}{100} \times 44 = 11.0$ g of CO₂.

Tip

Mass ratio method

For reacting-mass questions, calculate the formula masses, use the balanced equation ratio, then scale to the mass in the question.

Concentration in g dm⁻³

Definition

Concentration

The concentration of a solution is how much solute is dissolved in a certain volume of solution. In this section, concentration is measured in grams per cubic decimetre, g dm⁻³.

dm3\text{concentration in g dm}^{-3} = \frac{\text{mass of solute in g}}{\text{volume of solution in dm}^3}

Remember: 1000 cm³ = 1 dm³, so divide cm³ by 1000 to convert to dm³.

Example

Calculating concentration

5.00 g of sodium chloride is dissolved to make 250 cm³ of solution. Calculate the concentration in g dm⁻³.

Convert the volume: 250 cm³ = 0.250 dm³.
Substitute into the equation: concentration = $5.000.250\frac{5.00}{0.250}$ .
Calculate the answer: concentration = 20.0 g dm⁻³.

Common Mistake

Using cm³ directly

Do not put 250 straight into the equation if the answer needs g dm⁻³. Convert 250 cm³ to 0.250 dm³ first.

Higher Tier: moles and particles

This next part is Higher Tier only. It gives a powerful way to link mass, formula mass and number of particles.

Definition

One mole

One mole of a substance contains the Avogadro constant number of particles: $6.02 \times 10^{23}$ atoms, molecules, ions or formula units. One mole also has a mass equal to the relative particle mass in grams.

For example, carbon has $A_r = 12$ , so 1 mole of carbon atoms has a mass of 12 g. Water has $M_r = 18$ , so 1 mole of water molecules has a mass of 18 g.

The map below links the key calculations.

Calculation map linking mass, amount in moles, number of particles and concentration

\frac{m}{M_r}

where $n$ is amount in mol, $m$ is mass in g, and $M_r$ is relative formula mass.

Example

Calculating moles and particles

Calculate the number of water molecules in 9.0 g of water, H₂O. Use $A_r$ : H = 1, O = 16.

Calculate the relative formula mass: $Mr=(2×1)+16=18M_r = \left(2 \times 1\right) + 16 = 18$ .
Calculate the amount in moles: $\frac{9.0}{18} = 0.50$ mol.
Convert moles to molecules: $0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ molecules.

Higher Tier: limiting reactants and stoichiometry

Definition

Limiting reactant

The limiting reactant is the reactant that is completely used up first. It controls the maximum mass of product that can be formed. A reactant in excess is left over.

The limiting reactant is not always the one with the smaller mass. You must compare the amounts using the balanced equation ratio.

Example

Identifying the limiting reactant

Magnesium reacts with oxygen:

2Mg(s) + O₂(g) → 2MgO(s)

A mixture contains 0.060 mol Mg and 0.020 mol O₂. Find the limiting reactant and the mass of MgO made. Use $M_r$ of MgO = 40.

Use the equation ratio: 1 mol O₂ needs 2 mol Mg, so 0.020 mol O₂ needs 0.040 mol Mg.
Compare with what is available: 0.060 mol Mg is available, so there is more than enough Mg. Oxygen is the limiting reactant.
Use the limiting reactant to find product amount: 0.020 mol O₂ makes 0.040 mol MgO.
Convert to mass: $\times 40 = 1.6$ g of MgO.

Definition

Stoichiometry

Stoichiometry means the reacting ratios shown by the numbers in a balanced chemical equation.

On Higher Tier, you may be asked to deduce these ratios from masses. Convert each mass to moles, then simplify the ratio.

Example

Deducing equation ratios from masses

Hydrogen and oxygen react to form water. In an experiment, 0.40 g H₂ reacts with 3.20 g O₂ to make 3.60 g H₂O. Deduce the balanced equation. Use $M_r$ : H₂ = 2, O₂ = 32, H₂O = 18.

Convert each mass to moles: H₂ is $0.402=0.20\frac{0.40}{2} = 0.20$ mol, O₂ is $3.2032=0.100\frac{3.20}{32} = 0.100$ mol, and H₂O is $3.6018=0.200\frac{3.60}{18} = 0.200$ mol.
Divide by the smallest amount, 0.100 mol, to get the ratio H₂:O₂:H₂O = 2:1:2.
Write the balanced equation with those numbers: 2H₂(g) + O₂(g) → 2H₂O(l).

Exam technique

In the exam

Write the formula or balanced equation first; most mass calculations depend on getting this correct.
Keep units with your numbers, especially g, cm³, dm³ and mol.
For empirical formulae and stoichiometry, divide by $A_r$ or $M_r$ , then simplify to a whole-number ratio.
For solution concentration in g dm⁻³, always convert cm³ to dm³ before substituting.
Give answers to a sensible number of significant figures, usually matching the data in the question.

Self review

Check yourself

How would you calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃?
Why might the mass decrease when a reaction producing carbon dioxide is carried out in an open flask?
Higher Tier: What is the difference between a limiting reactant and a reactant in excess?

What you'll learn

How to calculate relative formula mass and percentage composition.
How to find empirical and molecular formulae from data.
How conservation of mass explains mass changes in reactions.
How to calculate reacting masses, solution concentration, and — on Higher Tier — moles, particles and limiting reactants.

The building blocks: $A_r$ , formulae and balanced equations

Before the bigger calculations, you need three ideas.

Definition

Relative atomic mass, Ar

A chemical formula tells you the elements present and how many atoms of each are in one particle or formula unit. For example, H₂O contains 2 hydrogen atoms and 1 oxygen atom.

A balanced equation shows the reacting ratio of substances. For example:

2Mg(s) + O₂(g) → 2MgO(s)

This means 2 magnesium atoms react with 1 oxygen molecule to form 2 formula units of magnesium oxide.

Relative formula mass, $M_r$

Definition

Relative formula mass

The relative formula mass, $M_r$ , is the total of the $A_r$ values for all the atoms shown in a formula. For simple molecules, it may also be called relative molecular mass.

For brackets, multiply everything inside the brackets by the number outside.

Example

Finding relative formula mass and percentage by mass

Find the $M_r$ of calcium hydroxide, Ca(OH)₂, and the percentage by mass of oxygen. Use $A_r$ : Ca = 40, O = 16, H = 1.

Count the atoms carefully: Ca(OH)₂ contains 1 Ca atom, 2 O atoms and 2 H atoms.
Add the relative masses: $Mr=40+(2×16)+(2×1)=74M_r = 40 + \left(2 \times 16\right) + \left(2 \times 1\right) = 74$ .
Find the oxygen part of the mass: oxygen contributes $\times 16 = 32$ .
Calculate the percentage: $3274×100=43.2%\frac{32}{74} \times 100 = 43.2\%$ .

Common Mistake

Forgetting the brackets

In Ca(OH)₂, the small 2 applies to both O and H. So there are 2 oxygen atoms and 2 hydrogen atoms, not just 1 oxygen and 2 hydrogen atoms.

Percentage by mass

Percentage by mass tells you what fraction of a compound’s mass comes from one element.

compound×100\% \text{ by mass} = \frac{\text{mass of the element in the formula}}{M_r \text{ of the compound}} \times 100

Key Idea

Percentage composition

A compound with a fixed formula always has the same percentage composition by mass, because its atoms are joined in a fixed ratio.

Empirical formulae

Definition

Empirical formula

An empirical formula is the simplest whole-number ratio of atoms of each element in a compound.

For example, the molecular formula C₆H₁₂O₆ simplifies to the empirical formula CH₂O, because 6:12:6 simplifies to 1:2:1.

When using reacting masses or percentage composition, convert each element’s mass to an amount ratio by dividing by its $A_r$ .

Example

Finding empirical and molecular formulae

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Its relative molecular mass is 180. Find its empirical and molecular formula. Use $A_r$ : C = 12, H = 1, O = 16.

Assume you have 100 g of the compound, so the masses are 40.0 g C, 6.7 g H and 53.3 g O.
Divide each mass by its $A_r$ : 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, to get the ratio C:H:O = 1:2:1, so the empirical formula is CH₂O.
Find the empirical formula mass: $\left(2 \times 1\right) + 16 = 30$ .
Compare with the molecular mass: $18030=6\frac{180}{30} = 6$ , so multiply CH₂O by 6 to get C₆H₁₂O₆.

Tip

If the ratio is not whole-number

If you get a ratio like 1:1.5, multiply all parts by 2 to make 2:3. If you get 1:1.33, multiply all parts by 3 to make 3:4.

Practical: empirical formula of magnesium oxide

You need to be able to describe an experiment to determine a simple empirical formula, such as magnesium oxide.

The idea is to heat magnesium so it reacts with oxygen from the air:

2Mg(s) + O₂(g) → 2MgO(s)

Diagram of the magnesium oxide empirical formula practical setup with crucible, lid, tripod and Bunsen burner

A sensible method is:

Weigh a clean, dry crucible and lid.
Add a coil of cleaned magnesium ribbon and weigh again.
Heat strongly with the lid slightly lifted sometimes to let oxygen in.
Keep the lid mostly on to reduce loss of white magnesium oxide smoke.
Cool, then reweigh the crucible, lid and product.
Repeat heating, cooling and weighing until the mass is constant.

Example

Using magnesium oxide results

A student heats 0.48 g of magnesium and obtains 0.80 g of magnesium oxide. Find the empirical formula. Use $A_r$ : Mg = 24, O = 16.

Find the mass of oxygen gained: $0.80 - 0.48 = 0.32$ g.
Convert masses to amount ratios: Mg is $0.4824=0.020\frac{0.48}{24} = 0.020$ , and O is $0.3216=0.020\frac{0.32}{16} = 0.020$ .
Divide by the smallest value: Mg:O = 1:1, so the empirical formula is MgO.

Common Mistake

Product escaping

If magnesium oxide smoke escapes, the measured product mass is too low, so the calculated oxygen mass will also be too low.

Conservation of mass

Definition

Law of conservation of mass

The law of conservation of mass says that atoms are not created or destroyed in a chemical reaction, so total mass is conserved.

In a closed system, no substances can enter or leave, so the measured mass stays the same. A stoppered flask is a closed system.

In an open flask, the balance reading can change if a gas enters or leaves. The atoms are still conserved, but not all the substances may remain on the balance.

Example

Explaining mass loss in an open flask

Marble chips react with hydrochloric acid in an open flask:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Identify the gas made: carbon dioxide, CO₂(g), is produced.
Because the flask is open, some CO₂(g) escapes into the air instead of staying on the balance.
The balance reading decreases, but the total mass of all reactants and products, including the escaped gas, is still conserved.

Reacting masses from balanced equations

A balanced equation gives a ratio. You can use that ratio to scale masses up or down.

Example

Calculating product mass from a balanced equation

Calcium carbonate decomposes when heated:

CaCO₃(s) → CaO(s) + CO₂(g)

Calculate the mass of carbon dioxide made from 25.0 g of calcium carbonate. Use $A_r$ : Ca = 40, C = 12, O = 16.

Find the relative formula masses: CaCO₃ has $Mr=40+12+(3×16)=100M_r = 40 + 12 + \left(3 \times 16\right) = 100$ , and CO₂ has $Mr=12+(2×16)=44M_r = 12 + \left(2 \times 16\right) = 44$ .
Use the equation ratio: 1 formula unit of CaCO₃ makes 1 molecule of CO₂, so 100 g of CaCO₃ makes 44 g of CO₂.
Scale to 25.0 g: $25.0100×44=11.0\frac{25.0}{100} \times 44 = 11.0$ g of CO₂.

Tip

Mass ratio method

For reacting-mass questions, calculate the formula masses, use the balanced equation ratio, then scale to the mass in the question.

Concentration in g dm⁻³

Definition

Concentration

The concentration of a solution is how much solute is dissolved in a certain volume of solution. In this section, concentration is measured in grams per cubic decimetre, g dm⁻³.

dm3\text{concentration in g dm}^{-3} = \frac{\text{mass of solute in g}}{\text{volume of solution in dm}^3}

Remember: 1000 cm³ = 1 dm³, so divide cm³ by 1000 to convert to dm³.

Example

Calculating concentration

5.00 g of sodium chloride is dissolved to make 250 cm³ of solution. Calculate the concentration in g dm⁻³.

Convert the volume: 250 cm³ = 0.250 dm³.
Substitute into the equation: concentration = $5.000.250\frac{5.00}{0.250}$ .
Calculate the answer: concentration = 20.0 g dm⁻³.

Common Mistake

Using cm³ directly

Do not put 250 straight into the equation if the answer needs g dm⁻³. Convert 250 cm³ to 0.250 dm³ first.

Higher Tier: moles and particles

This next part is Higher Tier only. It gives a powerful way to link mass, formula mass and number of particles.

Definition

One mole

For example, carbon has $A_r = 12$ , so 1 mole of carbon atoms has a mass of 12 g. Water has $M_r = 18$ , so 1 mole of water molecules has a mass of 18 g.

The map below links the key calculations.

Calculation map linking mass, amount in moles, number of particles and concentration

\frac{m}{M_r}

where $n$ is amount in mol, $m$ is mass in g, and $M_r$ is relative formula mass.

Example

Calculating moles and particles

Calculate the number of water molecules in 9.0 g of water, H₂O. Use $A_r$ : H = 1, O = 16.

Calculate the relative formula mass: $Mr=(2×1)+16=18M_r = \left(2 \times 1\right) + 16 = 18$ .
Calculate the amount in moles: $\frac{9.0}{18} = 0.50$ mol.
Convert moles to molecules: $0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ molecules.

Higher Tier: limiting reactants and stoichiometry

Definition

Limiting reactant

The limiting reactant is the reactant that is completely used up first. It controls the maximum mass of product that can be formed. A reactant in excess is left over.

The limiting reactant is not always the one with the smaller mass. You must compare the amounts using the balanced equation ratio.

Example

Identifying the limiting reactant

Magnesium reacts with oxygen:

2Mg(s) + O₂(g) → 2MgO(s)

A mixture contains 0.060 mol Mg and 0.020 mol O₂. Find the limiting reactant and the mass of MgO made. Use $M_r$ of MgO = 40.

Use the equation ratio: 1 mol O₂ needs 2 mol Mg, so 0.020 mol O₂ needs 0.040 mol Mg.
Compare with what is available: 0.060 mol Mg is available, so there is more than enough Mg. Oxygen is the limiting reactant.
Use the limiting reactant to find product amount: 0.020 mol O₂ makes 0.040 mol MgO.
Convert to mass: $\times 40 = 1.6$ g of MgO.

Definition

Stoichiometry

Stoichiometry means the reacting ratios shown by the numbers in a balanced chemical equation.

On Higher Tier, you may be asked to deduce these ratios from masses. Convert each mass to moles, then simplify the ratio.

Example

Deducing equation ratios from masses

Hydrogen and oxygen react to form water. In an experiment, 0.40 g H₂ reacts with 3.20 g O₂ to make 3.60 g H₂O. Deduce the balanced equation. Use $M_r$ : H₂ = 2, O₂ = 32, H₂O = 18.

Convert each mass to moles: H₂ is $0.402=0.20\frac{0.40}{2} = 0.20$ mol, O₂ is $3.2032=0.100\frac{3.20}{32} = 0.100$ mol, and H₂O is $3.6018=0.200\frac{3.60}{18} = 0.200$ mol.
Divide by the smallest amount, 0.100 mol, to get the ratio H₂:O₂:H₂O = 2:1:2.
Write the balanced equation with those numbers: 2H₂(g) + O₂(g) → 2H₂O(l).

Exam technique

In the exam

Write the formula or balanced equation first; most mass calculations depend on getting this correct.
Keep units with your numbers, especially g, cm³, dm³ and mol.
For empirical formulae and stoichiometry, divide by $A_r$ or $M_r$ , then simplify to a whole-number ratio.
For solution concentration in g dm⁻³, always convert cm³ to dm³ before substituting.
Give answers to a sensible number of significant figures, usually matching the data in the question.

Self review

Check yourself

How would you calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃?
Why might the mass decrease when a reaction producing carbon dioxide is carried out in an open flask?
Higher Tier: What is the difference between a limiting reactant and a reactant in excess?

What you'll learn

The building blocks: $A_r$ , formulae and balanced equations

Relative formula mass, $M_r$

Percentage by mass

Empirical formulae

Practical: empirical formula of magnesium oxide

Conservation of mass

Reacting masses from balanced equations

Concentration in g dm⁻³

Higher Tier: moles and particles

Higher Tier: limiting reactants and stoichiometry

Recap questions

Review Calculations involving masses by teaching Genie

Remember key concepts with flashcards

Calculations involving masses

What you'll learn

The building blocks: $A_r$ , formulae and balanced equations

Relative formula mass, $M_r$

Percentage by mass

Empirical formulae

Practical: empirical formula of magnesium oxide

Conservation of mass

Reacting masses from balanced equations

Concentration in g dm⁻³

Higher Tier: moles and particles

Higher Tier: limiting reactants and stoichiometry

Recap questions

Review Calculations involving masses by teaching Genie

Remember key concepts with flashcards

Calculations involving masses

What you'll learn

The building blocks: ArA_rAr​, formulae and balanced equations

Relative formula mass, MrM_rMr​

Percentage by mass

Empirical formulae

Practical: empirical formula of magnesium oxide

Conservation of mass

Reacting masses from balanced equations

Concentration in g dm⁻³

Higher Tier: moles and particles

Higher Tier: limiting reactants and stoichiometry

Recap questions

Review Calculations involving masses by teaching Genie

Remember key concepts with flashcards

Calculations involving masses Revision Guide

Calculations involving masses

What you'll learn

The building blocks: ArA_rAr​, formulae and balanced equations

Relative formula mass, MrM_rMr​

Percentage by mass

Empirical formulae

Practical: empirical formula of magnesium oxide

Conservation of mass

Reacting masses from balanced equations

Concentration in g dm⁻³

Higher Tier: moles and particles

Higher Tier: limiting reactants and stoichiometry

Recap questions

Review Calculations involving masses by teaching Genie

Remember key concepts with flashcards

Calculations involving masses Revision Guide

The building blocks: $A_r$ , formulae and balanced equations

Relative formula mass, $M_r$

The building blocks: $A_r$ , formulae and balanced equations

Relative formula mass, $M_r$