Percentage by mass, finding an empirical formula from masses or percentages, going from empirical to molecular formula, and the magnesium oxide practical.
Objectives
Calculate the percentage by mass of an element in a compound.
Work out an empirical formula from reacting masses or percentage composition.
Work out a molecular formula from an empirical formula and the relative formula mass.
Describe and explain the magnesium oxide experiment used to find an empirical formula.
Empirical formulae and percentage composition explained
Start here
A formula tells you which atoms are in a compound and in what ratio. Sometimes you know the masses of the elements in a compound, but not its formula. This lesson shows how to work the formula out.
This matters because chemists find the formula of new compounds by weighing how much of each element they contain. It also gives you practice with relative atomic mass and moles, which are used in many other calculations.
By the end you will be able to calculate percentage by mass, find an empirical formula, find a molecular formula, and explain the magnesium oxide experiment.
Percentage by mass
The percentage by mass of an element in a compound is the part of the total mass that is that element. Percentage by mass = (Ar × number of atoms of the element ÷ Mr of the compound) × 100.
For MgO, Mr = 24 + 16 = 40. The percentage of magnesium is 24 ÷ 40 × 100 = 60%. The percentage of oxygen is 16 ÷ 40 × 100 = 40%. The percentages add up to 100%.
For a formula such as Fe₂O₃ remember to multiply the Ar by the number of atoms: Mr = (2 × 56) + (3 × 16) = 160, and iron is 112 ÷ 160 × 100 = 70%.
What an empirical formula is
The empirical formula is the simplest whole number ratio of atoms of each element in a compound. For glucose, C₆H₁₂O₆, the atoms are in the ratio 6 : 12 : 6, which simplifies to 1 : 2 : 1. So the empirical formula is CH₂O.
To go from a molecular formula to an empirical formula, divide every number by the largest number that goes into all of them. C₂H₆ becomes CH₃. H₂O₂ becomes HO. C₄H₁₀ becomes C₂H₅.
Some formulae are already as simple as they can be. H₂O, CH₄ and MgO have the same empirical formula as their molecular or ionic formula. Ionic compounds are always written as their empirical formula.
Empirical formula from masses
Follow the same steps each time. First write the mass of each element. Divide each mass by its Ar to get the number of moles of atoms. Divide each answer by the smallest one to get the ratio. If the ratio is not whole numbers, multiply them all to make them whole, for example by 2 if you have 1.5. Then write the formula.
Example: a compound contains 2.4 g of carbon and 0.6 g of hydrogen. Carbon: 2.4 ÷ 12 = 0.2 mol. Hydrogen: 0.6 ÷ 1 = 0.6 mol. Divide by 0.2 to get 1 : 3. The empirical formula is CH₃.
A table helps you to keep the steps in order: element, mass, Ar, moles, ratio.
Empirical formula from percentages
If you are given percentages, imagine that you have 100 g of the compound. Then each percentage is the mass in grams. Carry on with the same steps.
Example: a compound is 85.7% carbon and 14.3% hydrogen. In 100 g there is 85.7 g of C and 14.3 g of H. Carbon: 85.7 ÷ 12 = 7.14 mol. Hydrogen: 14.3 ÷ 1 = 14.3 mol. Divide by 7.14 to get 1 : 2. The empirical formula is CH₂.
Check that your answer is a sensible formula and that the ratio is made of small whole numbers. If you get 1 : 1.98, round it to 1 : 2, but if you get 1 : 1.5 multiply by 2.
Molecular formula
The molecular formula shows the actual number of atoms in one molecule. It is a whole number multiple of the empirical formula. To find it, work out the Mr of the empirical formula and divide the Mr of the compound by it. That gives the multiple.
Example: the empirical formula is CH₂ and the Mr of the compound is 42. The empirical formula mass is 12 + 2 = 14. 42 ÷ 14 = 3. So multiply every number by 3: the molecular formula is C₃H₆.
Use the Mr in the question, not your own guess. If the Mr of the compound equals the empirical formula mass, the two formulae are the same.
Practical: the formula of magnesium oxide
Weigh a crucible with its lid. Add a piece of cleaned magnesium ribbon and weigh again. Heat the crucible strongly. Lift the lid a little from time to time to let oxygen in. Do not let any white smoke escape. When the magnesium has all burned to a white powder, let it cool and weigh again.
The mass of magnesium is the second reading minus the first. The mass of oxygen is the final reading minus the second reading, because oxygen has joined the magnesium. For example, if 0.48 g of magnesium makes 0.80 g of magnesium oxide, the oxygen is 0.32 g. Magnesium: 0.48 ÷ 24 = 0.02 mol. Oxygen: 0.32 ÷ 16 = 0.02 mol. The ratio is 1 : 1, so the formula is MgO.
The mass goes up because the oxygen from the air joins the magnesium. If some of the magnesium oxide escapes as smoke, the final mass is too small, so the oxygen mass looks too small and the answer is wrong.
Recap
Percentage by mass = Ar × atoms ÷ Mr × 100. The empirical formula is the simplest whole number ratio of atoms.
Steps for an empirical formula: masses (or percentages treated as grams), divide by Ar, divide by the smallest, make whole numbers.
The molecular formula is the empirical formula multiplied by (Mr of compound ÷ empirical formula mass). In the magnesium oxide experiment the mass gain is the mass of oxygen.
Key points
Percentage by mass = Ar × number of atoms ÷ Mr × 100.
Empirical formula = simplest whole number ratio. Mass (or %) ÷ Ar → ÷ smallest → whole numbers.
Molecular formula = empirical formula × (Mr ÷ empirical formula mass).
Magnesium oxide experiment: mass gain = mass of oxygen. Lift the lid for oxygen. Do not lose smoke.
Worked examples
Example 1
Give the empirical formula of glucose, C₆H₁₂O₆.
The atoms are in the ratio C : H : O = 6 : 12 : 6.
The largest number that divides 6, 12 and 6 is 6.
6 ÷ 6 = 1, 12 ÷ 6 = 2, 6 ÷ 6 = 1.
The empirical formula is CH₂O.
Answer: CH₂O
Common slips
Slip: C₃H₆O₃ You can divide again. The numbers must be the simplest whole numbers.
Slip: CH₁₂O You must divide all of the numbers by the same number.
Slip: C₆H₁₂O₆ This is the molecular formula. The empirical formula is the simplest ratio.
Slip: CHO Hydrogen has 12 atoms, which is twice 6, so it is H₂.
Example 2
A compound contains 5.6 g of iron and 2.4 g of oxygen. Calculate its empirical formula. (Ar: Fe = 56, O = 16)
Divide by the smallest (0.10): iron = 1, oxygen = 1.5.
The ratio is not whole numbers, so multiply by 2: iron = 2, oxygen = 3.
The empirical formula is Fe₂O₃.
Answer: Fe₂O₃
Common slips
Slip: FeO You must multiply 1 : 1.5 by 2 to get whole numbers.
Slip: Fe₁.₅O A formula needs whole numbers. Multiply by 2.
Slip: Fe₃O₂ You swapped the numbers. Iron is 1 and oxygen is 1.5, so iron : oxygen = 2 : 3.
Slip: FeO₁.₅ You need whole numbers, so multiply by 2 to get Fe₂O₃.
Example 3
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its Mr is 180. Calculate its empirical formula and its molecular formula. (Ar: C = 12, H = 1, O = 16)
In 100 g: 40.0 g C, 6.7 g H, 53.3 g O.
Moles: C = 40.0 ÷ 12 = 3.33, H = 6.7 ÷ 1 = 6.7, O = 53.3 ÷ 16 = 3.33.
Divide by 3.33: C = 1, H = 2, O = 1. The empirical formula is CH₂O.
The empirical formula mass = 12 + 2 + 16 = 30. 180 ÷ 30 = 6.
The molecular formula is C₆H₁₂O₆.
Answer: Empirical formula CH₂O. Molecular formula C₆H₁₂O₆
Common slips
Slip: Empirical formula C₂H₂O Check the hydrogen. 6.7 ÷ 3.33 = 2.0, which is 2, and carbon is 1.
Slip: Molecular formula CH₂O The molecular formula must have Mr 180. CH₂O has Mr 30.
Slip: Molecular formula C₃H₆O₃ Divide 180 by 30 to get 6, not 3.
Slip: Empirical formula COH Hydrogen is twice carbon, so it is CH₂O.
Slip: Empirical formula C₄₀H₆.₇O₅₃.₃ The percentages are masses. Divide by Ar first to get moles.
Questions and answers
Empirical formulae and percentage composition practice questions for GCSE Combined Science Chemistry. Try each one, then open the answer. Easy questions come first.
Question 1 Easy
Give the empirical formula of C₂H₆.
Show answer
Answer: CH₃
Divide both numbers by 2.
C₂H₆ → CH₃.
Question 2 Easy
Give the empirical formula of H₂O₂.
Show answer
Answer: HO
Divide both numbers by 2.
H₂O₂ → HO.
Question 3 Easy
Give the empirical formula of C₈H₁₈.
Show answer
Answer: C₄H₉
Divide both numbers by 2.
C₈H₁₈ → C₄H₉.
Question 4 Easy
Give the empirical formula of C₄H₁₀.
Show answer
Answer: C₂H₅
Divide both numbers by 2.
C₄H₁₀ → C₂H₅.
Question 5 Easy
Does the empirical formula show the simplest ratio or the actual number of atoms?
Show answer
Answer: the simplest ratio
Empirical means the simplest whole number ratio.
It does not show the actual number of atoms.
Question 6 Medium
The empirical formula of a compound is CH₂ and its Mr is 56. What is the molecular formula? (Ar: C = 12, H = 1)
Show answer
Answer: C₄H₈
CH₂ has Mr 14.
56 ÷ 14 = 4, so the formula is C₄H₈.
Question 7 Medium
How many moles are in 0.48 g of magnesium? (Ar: Mg = 24)
Show answer
Answer: 0.020 mol
Moles = mass ÷ Ar.
0.48 ÷ 24 = 0.020 mol.
Question 8 Medium
How many moles are in 0.32 g of oxygen atoms? (Ar: O = 16)
Show answer
Answer: 0.020 mol
Moles = mass ÷ Ar.
0.32 ÷ 16 = 0.020 mol.
Exam-style questions
Longer questions in the style of a written paper, with the marking points. These are Yumio practice questions.
Exam-style question 15 marks
A student heats magnesium in a crucible to find the formula of magnesium oxide. Crucible and lid: 25.00 g. Crucible, lid and magnesium: 25.48 g. Crucible, lid and magnesium oxide: 25.80 g. (a) Calculate the mass of magnesium. [1 mark] (b) Calculate the mass of oxygen that combined. [1 mark] (c) Calculate the empirical formula of magnesium oxide. Ar: Mg = 24, O = 16. [3 marks]
Show answer and marking points
Answer: (a) 0.48 g (b) 0.32 g (c) MgO
(a) 0.48 g (1)
(b) 0.32 g (1)
(c) 0.48 ÷ 24 and 0.32 ÷ 16 (0.02 and 0.02) (1)
(c) ratio 1 : 1 (1)
(c) MgO (1)
Exam-style question 24 marks
(a) A compound contains 80% carbon and 20% hydrogen by mass. Calculate its empirical formula. Ar: C = 12, H = 1. [3 marks] (b) The Mr of the compound is 30. Give its molecular formula. [1 mark]
Show answer and marking points
Answer: (a) CH₃ (b) C₂H₆
(a) 80 ÷ 12 and 20 ÷ 1 (1)
(a) ratio 1 : 3 (1)
(a) CH₃ (1)
(b) C₂H₆ (1)
Common mistakes
Dividing the masses by each other instead of by Ar
Divide each mass by its own Ar to get moles, then compare the moles.
Rounding 1 : 1.5 to 1 : 2
Multiply the whole ratio by 2 to get 2 : 3. Only round values very close to a whole number.
Forgetting that percentages are masses in 100 g
Treat the percentage as the number of grams in a 100 g sample.
Giving the molecular formula when asked for the empirical formula
The empirical formula is the simplest ratio. Divide the molecular formula.
Not multiplying the empirical formula by the right number
Divide Mr of the compound by the empirical formula mass to find the multiple.
Questions students ask
Is Empirical formulae and percentage composition on the Foundation or Higher tier?
Empirical formulae and percentage composition is on both Foundation and Higher papers for GCSE Combined Science Chemistry, aimed at around grade 6.
What is a common mistake on Empirical formulae and percentage composition?
Dividing the masses by each other instead of by Ar. Divide each mass by its own Ar to get moles, then compare the moles.
Where are the answers to the Empirical formulae and percentage composition questions?
Open Show answer under each question on this page. The questions are Yumio practice questions, not past exam questions.