Finding empirical and molecular formulae

Composition data tell you how much of each element a compound contains. To find its formula, first convert the element masses into amounts of atoms, then compare those amounts. To find a molecular formula, you also need the relative molecular mass.

A ratio is different from a molecule’s actual counts

An empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. A molecular formula gives the actual number of atoms of each element in one molecule.

A molecular formula applies to a substance made of separate molecules. Ionic solids and giant covalent structures have composition ratios rather than separate molecules. When writing an ionic formula from charges, preserve any polyatomic ions as groups; do not simplify their internal subscripts to make an empirical formula.

Find an empirical formula from element masses

Equal masses of different elements do not usually contain equal numbers of atoms. For example, 2.40 g of Mg atoms and 1.60 g of O atoms each represent 0.100 mol when their molar masses are 24 and 16 g mol⁻¹. Their atom ratio is 1:1, even though their masses differ.

1. If given mass percentages, choose a 100 g sample. A value of 40.0% then represents 40.0 g of that element.
2. For each element, calculate the amount of atoms using n = m/M. Use the numerical Aᵣ value as its atomic molar mass in g mol⁻¹ at this course’s precision. For oxygen atoms, use 16, not the molar mass of O₂ molecules.
3. Divide every amount by the smallest amount, using unrounded values from your calculator.
4. Express the ratios as the smallest whole numbers consistent with the precision of the data, then write them as subscripts.

Preserve fractions when simplifying

A small deviation caused by rounded composition data can be treated as an intended whole-number ratio: 1:2.01:1 can represent 1:2:1. A clear ratio such as 1:1.5 cannot be rounded to 1:2. Multiply every term by 2 to obtain 2:3 instead.

The multiplier must work for all terms. Recalculate the mass percentages from your proposed formula if you are unsure whether it fits the data.

Use relative molecular mass to find the actual counts

First find the empirical formula mass by adding the relative atomic mass contributions in the empirical formula. Then calculate the dimensionless multiplier k:

k = (Mᵣ of the molecular substance)/(empirical formula mass)

Multiply every empirical subscript by k, including an unwritten subscript of 1. For example, CH₂ has empirical formula mass 14. If the substance has Mᵣ = 56, then k = 4 and the molecular formula is C₄H₈.

This scales atom counts; it does not tell you how the atoms are connected. Different substances can share an empirical formula, so composition alone cannot determine the molecular formula or structure.

Check the quantities and the formula

• Convert element masses using their atomic molar masses, rather than the compound’s molar mass or a molecular element’s molar mass.
• Keep extra digits until you simplify the ratio. Scale every ratio term together.
• The molecular multiplier should be a positive whole number consistent with the data. If it is not, check the empirical formula, atom counts, relative masses and rounding before forcing an answer.
• Give a formula, not just a ratio. Check that its Mᵣ agrees with the supplied value when a molecular formula is requested.

For a reminder, see Relative atomic, molecular and formula mass and Moles, molar mass and particle counts.

Worked examples

Modelled example 1

Empirical formula from percentage composition

Core

Problem

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula. Use Aᵣ: C = 12, H = 1, O = 16.

Study the worked solution

1. Convert percentages to masses

   Method

   Assume a 100 g sample.

   Reason

   Each percentage then has the same numerical value as its mass in grams.

   Working

   m(C) = 40.0 g, m(H) = 6.7 g, m(O) = 53.3 g.

2. Convert each mass to moles

   Method

   Divide each element’s mass by its atomic molar mass.

   Reason

   Formula subscripts represent amount ratios, not mass ratios.

   Working

   The element amounts are shown below. Displayed values are rounded for readability; use the original quotients in the ratio calculation.

   Element	Mass in a 100 g sample / g	Atomic molar mass / g mol⁻¹	Amount of atoms / mol	Divide by 53.3/16
   C	40.0	12	3.3333…	approximately 1.0006
   H	6.7	1	6.7	approximately 2.0113
   O	53.3	16	3.33125	1

3. Find the simplest whole-number ratio

   Method

   Divide all amounts by the unrounded oxygen amount, 53.3/16 mol.

   Reason

   The ratios are close to 1:2:1; the small deviations are consistent with the rounded percentage data.

   Working

   C:H:O ≈ 1:2:1.

4. Write the empirical formula

   Method

   Use the ratio as subscripts.

   Reason

   A subscript of one is omitted.

   Working

   Empirical formula = CH₂O.

Guided practice 2

Empirical formula from element masses

About 6 min

Problem

2.40 g of magnesium combines with 1.60 g of oxygen to form an oxide. Find the empirical formula. Use Aᵣ: Mg = 24, O = 16.

Compare amounts rather than masses

Moles of Mg

Moles of O

Empirical formula

MgOMg2OMgO2

Check my attempt

Hints

Hint 1: convert both masses

Calculate 2.40/24 and 1.60/16.

Hint 2: simplify together

Divide both mole values by 0.100.

View solution step by step

1. Calculate element amounts

   Method

   Divide each mass by the corresponding atomic molar mass.

   Reason

   Atoms combine in mole ratios.

   Working

   n(Mg) = 2.40/24 = 0.100 mol; n(O) = 1.60/16 = 0.100 mol.

2. Simplify the ratio

   Method

   Divide both amounts by 0.100.

   Reason

   This gives the smallest whole-number ratio.

   Working

   Mg:O = 1:1.

3. Write the formula

   Working

   Empirical formula = MgO.

Common misconception 3

A half-unit ratio must be scaled

Find and correct the mistake

Learner response

A compound has 0.020 mol of X and 0.030 mol of Y. A student obtains 1:1.5, rounds it to 1:2 and writes XY₂. Explain the error and find the empirical formula.

Preserve the ratio exactly

Correct action for 1:1.5

Multiply both terms by 2Round 1.5 up to 2Drop the decimal part

Correct formula

X2Y3XY2X3Y2

Check my attempt

View solution step by step

1. Form the ratio

   Method

   Divide both amounts by the smallest, 0.020 mol.

   Reason

   Formula subscripts must preserve the measured amount ratio.

   Working

   X:Y = (0.020/0.020):(0.030/0.020) = 1:1.5.

2. Remove the half without rounding

   Method

   Multiply every ratio term by two.

   Reason

   Scaling both terms preserves the ratio; rounding only one changes it.

   Working

   1:1.5 → 2:3.

3. Correct the formula

   Working

   Empirical formula = X₂Y₃.

Examiner practice 4

Molecular formula from an empirical formula and Mᵣ

3 marks

Examination question

The empirical formula of a compound is CH₂. Its Mᵣ is 56. Use Aᵣ(C) = 12 and Aᵣ(H) = 1 to find its molecular formula. [3 marks]

Show the empirical-mass multiplier

Write the calculation and formula

View solution step by step

1. Find empirical formula mass

   1 mark

   Method

   Add the mass represented by CH₂.

   Reason

   The molecular Mᵣ must be compared with one empirical unit.

   Working

   Empirical formula mass = 12 + 2(1) = 14.

2. Find the whole-number multiplier

   1 mark

   Method

   Divide molecular Mᵣ by empirical formula mass.

   Reason

   A molecular formula contains a whole number of empirical units.

   Working

   k = 56/14 = 4.

3. Scale every subscript

   1 mark

   Method

   Multiply both empirical subscripts by four.

   Reason

   Each empirical subscript is multiplied by four.

   Working

   Molecular formula = C₄H₈.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark empirical formula mass, integer multiplier and scaled formula.

Mark points

1. 1 mark: Calculates empirical formula mass 14.
2. 1 mark: Calculates 56/14 = 4.
3. 1 mark: Obtains C4H8 by scaling every subscript.

Calculate my mark

Challenge 5

Combine composition and relative molecular mass

Minimal support

Combined-data transfer

A compound contains 92.3% carbon and 7.7% hydrogen by mass. Its Mᵣ is 78. Find its molecular formula. Use Aᵣ: C = 12, H = 1.

Complete both formula stages

Empirical formula

CHCH2C2H

Molecular multiplier

Molecular formula

C6H6C6H12CH

Check my attempt

Hints

Hint 1: first find the empirical unit

Assume 100 g and compare 92.3/12 with 7.7/1.

Hint 2: then use Mr

Once the empirical formula is CH, compare its mass 13 with 78.

View solution step by step

1. Convert composition to amounts

   Method

   Assume 100 g and divide each element’s mass by its atomic molar mass.

   Reason

   Percentage data become gram masses in a 100 g sample.

   Working

   n(C) = 92.3/12 ≈ 7.6917 mol; n(H) = 7.7/1 = 7.7 mol. Keep the original quotients when finding the ratio.

2. Find empirical formula

   Method

   Divide by the smaller amount.

   Reason

   The values are equal within the precision of the data.

   Working

   C:H ≈ 1:1.001 ≈ 1:1, so empirical formula = CH.

3. Find molecular multiplier

   Method

   Divide the molecular Mᵣ by empirical formula mass.

   Reason

   The empirical unit CH has mass 13.

   Working

   k = 78/13 = 6.

4. Write molecular formula

   Method

   Multiply every empirical subscript by six.

   Reason

   Each molecular atom count is six times its empirical subscript.

   Working

   Molecular formula = C₆H₆.

Next step

Try Mind Stretcher 1 independently.

Try it yourself

Mind stretcher 1: Use a mass change to find an empirical formulaExtension

5.60 g of iron is converted completely into 8.00 g of a pure oxide containing only iron and oxygen. All of the product is collected. Find the empirical formula of the oxide. Use Aᵣ(Fe) = 56 and Aᵣ(O) = 16.

Show answer

First find the oxygen mass from the increase:

m(O) = 8.00-5.60 = 2.40 g

Then convert both element masses to moles:

n(Fe) = 5.60/56 = 0.100 mol; n(O) = 2.40/16 = 0.150 mol

The ratio is 0.100:0.150 = 1:1.5. Multiply both terms by 2 to obtain 2:3.

Final: the empirical formula is Fe₂O₃.

Mind stretcher 2: Use combustion products to find a formulaExtension

0.260 g of a hydrocarbon burns completely to form 0.88 g of CO₂ and 0.18 g of H₂O. The hydrocarbon has Mᵣ = 78. Find its molecular formula. (Use Aᵣ: C = 12, H = 1, O = 16.)

Show answer

From products:

n(CO₂) = 0.88/44 = 0.0200 ⇒ n(C) = 0.0200; n(H₂O) = 0.18/18 = 0.0100 ⇒ n(H) = 2 × 0.0100 = 0.0200

Ratio C:H = 0.0200 : 0.0200 = 1 : 1, so empirical formula is CH.

Empirical formula mass = 13, so k = 78/13 = 6.

Molecular formula: C₆H₆.

Practise and check