Most tested C4.5

Empirical and Molecular Formulas

This topic covers the essential chemical skill of determining a compound's formula from its composition. You will learn to calculate the simplest whole-number ratio of atoms (empirical formula) from mass data, and then find the actual number of atoms in a molecule (molecular formula) using its relative molecular mass.

Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.

Key points

  • The core process is always Mass → Moles → Ratio. First, convert the mass (or percentage as mass) of each element into moles.
  • To find the simplest ratio, divide the moles of each element by the smallest number of moles calculated.
  • If the ratio is not in whole numbers after dividing (e.g., 1: 1.5), multiply all parts of the ratio by a small integer (e.g., 2) to get whole numbers (e.g., 2: 3).
  • The molecular formula is always a whole-number multiple of the empirical formula.
  • To find this multiple, divide the compound's actual Mr by the Mr of the empirical formula.

Formulae

moles = mass / Ar

To convert the mass of an element into the amount in moles, which is the first step in any empirical formula calculation.

Multiplier (n) = Mr(molecular) / Mr(empirical)

To find the integer used to scale up the empirical formula to the molecular formula, when the Mr of the compound is known.

Definitions

Empirical Formula
The simplest whole-number ratio of atoms of each element present in a compound. For example, the empirical formula of glucose (C6H12O6) is CH2O.
Molecular Formula
The actual number of atoms of each element in one molecule of a compound. For example, the molecular formula of glucose is C6H12O6.

Worked example

A hydrocarbon is found to contain 90% carbon by mass. Its relative molecular mass is 40.0. Determine its molecular formula. (Ar values: H = 1.0, C = 12.0)

  1. 1

    Step 1:

    Assume a 100 g sample.

    This means we have 90.0 g of Carbon and 100 - 90 = 10.0 g of Hydrogen.

  2. 2

    Step 2:

    Convert masses to moles.

    Moles C = 90.0 / 12.0 = 7.5 mol
    Moles H = 10.0 / 1.0 = 10.0 mol
  3. 3

    Step 3:

    Find the simplest ratio by dividing by the smallest number of moles (7.5).

    C ratio = 7.5 / 7.5 = 1
    H ratio = 10.0 / 7.5 = 100 / 75 = 4/3
  4. 4

    Step 4:

    Convert the ratio to whole numbers.

    The ratio is 1 :

    4/3.

    Multiply by 3 to get 3 :

    4.

    The empirical formula is C3H4.

  5. 5

    Step 5:

    Calculate the empirical formula mass.

    Mr(C3H4) = (3 × 12.0) + (4 × 1.0) = 36 + 4 = 40.0
  6. 6

    Step 6:

    Find the multiplier to get the molecular formula.

    Multiplier = Mr(molecular) / Mr(empirical) = 40.0 / 40.0 = 1
  7. 7

    Step 7:

    The multiplier is 1, so the molecular formula is the same as the empirical formula.

Answer: C3H4

Common mistakes

  • ×Arithmetic errors in the division steps are the most frequent mistake. When dividing mass by Ar or finding the simplest ratio, work carefully and double-check your calculations.
  • ×Stopping after finding a non-integer ratio. If you get a ratio like 1:2.5, you must multiply the entire ratio by 2 to get the simplest whole number ratio, 2:5.
  • ×Forgetting the final step. If the question provides the Mr of the compound, you must use it to check if the empirical formula needs to be scaled up to the molecular formula.

No-calculator tips

  • ✓When calculating ratios, convert decimals to simple fractions if possible. For example, a ratio of 7.5 to 10 is easier to simplify as (15/2) to 10, which becomes 15 to 20, simplifying to 3 to 4.
  • ✓Before dividing to find the mole ratio, look at the numbers. If you have 2.5 mol and 5 mol, you can see instantly it's a 1:2 ratio without needing to formally divide both by 2.5.
  • ✓Recognise common decimal-to-fraction conversions for ratios: 0.25 is 1/4 (multiply by 4), 0.33 is 1/3 (multiply by 3), 0.5 is 1/2 (multiply by 2), 0.67 is 2/3 (multiply by 3), 0.75 is 3/4 (multiply by 4).

Read this topic in the official UAT-UK ESAT guide →

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