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Chemical Calculations

Empirical and molecular formulae

  • 3 things to know
  • AS Unit 1

Overview

Recall questions on Empirical formulae and molecular formulae.

What you need to know

Answer each one before you reveal it.

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Numbers

Give the empirical formulae of ethane, C₂H₆, and of benzene, C₆H₆.

Key term

Heating 2.8 g of iron in oxygen produces 4.0 g of an oxide. Work out the empirical formula of the oxide.

Key term

Why does an empirical formula calculation often give a ratio like 1 : 1.5 rather than whole numbers, and what should you do?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    What is the difference between an empirical formula and a molecular formula?

    Model answer

    An empirical formula gives the simplest whole-number ratio of the atoms of each element present. A molecular formula gives the actual number of atoms of each element in one molecule, and is always a whole-number multiple of the empirical formula.

  2. 2

    Give the empirical formulae of ethane, C₂H₆, and of benzene, C₆H₆.

    Model answer

    Ethane has a carbon-to-hydrogen ratio of 2 : 6, which simplifies to 1 : 3, so its empirical formula is CH₃. Benzene has a ratio of 6 : 6, which simplifies to 1 : 1, so its empirical formula is CH.

  3. 3

    Why can two different compounds share the same empirical formula?

    Model answer

    The empirical formula records only the ratio of the atoms, not how many there are or how they are joined. Ethene, C₂H₄, and butene, C₄H₈, both reduce to CH₂, yet they are different substances with different molecular formulae.

  4. 4

    Outline the steps for finding an empirical formula from percentage composition by mass.

    Model answer

    Treat the percentages as masses in grams for a 100 g sample, divide each by the relative atomic mass of that element to get the amount in moles, divide all the results by the smallest of them, and then multiply through if needed to reach whole numbers.

  5. 5

    A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula.

    Model answer

    In 100 g there are 40.0 ÷ 12.0 = 3.33 mol C, 6.7 ÷ 1.0 = 6.7 mol H and 53.3 ÷ 16.0 = 3.33 mol O. Dividing by the smallest, 3.33, gives 1.00 : 2.01 : 1.00, so the empirical formula is CH₂O.

  6. 6

    A compound with the empirical formula CH₂O has a relative molecular mass of 180. What is its molecular formula?

    Model answer

    The empirical formula mass is 12.0 + 2.0 + 16.0 = 30.0. Dividing gives 180 ÷ 30.0 = 6, so every subscript is multiplied by six and the molecular formula is C₆H₁₂O₆.

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