Lesson 5.1.2

5.1.2 Empirical and molecular formulae Quiz: Pearson Edexcel Chemistry, Unit 5

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 5.1.2, Empirical and molecular formulae: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 5: Formulae, Equations and Amounts of Substance, written with Revision Ninja.

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The 20 questions

  1. What is an empirical formula?

    • The actual number of atoms of each element in one molecule of the compound, which can be a large number
    • The simplest whole-number ratio of atoms of each element in a compound
    • The formula showing how the bonds between atoms are arranged in a drawn structure of the molecule
    • The formula of the ion formed when the compound dissolves in an aqueous solution of its own salt

  2. What is a molecular formula?

    • The actual number of atoms of each element in one molecule
    • The number of moles of the compound contained in one mole of the element in a reaction mixture
    • A formula drawn with all of the bonds between the atoms shown as lines in a structural diagram
    • The simplest whole-number ratio of atoms of each element in a compound, obtained from its percentage composition

  3. What is the empirical formula of glucose, C6H12O6?

    • C3H6O3
    • C6H12O6
    • C2H4O2
    • CH2O

  4. What is the empirical formula of ethene, C2H4?

    • CH2
    • C2H4
    • CH4
    • C3H6

  5. A compound has empirical formula CH2O and molar mass 180 g mol-1. What is its molecular formula? (Empirical mass 30)

    • C3H6O3, which has a molar mass of 90 g/mol and is therefore not the correct molecular formula here
    • CH2O, which is the empirical formula itself and has a molar mass of 30 g/mol in this case
    • C2H4O2, which has a molar mass of 60 g/mol and is the wrong multiple of the empirical formula
    • C6H12O6

  6. A compound contains 40.0% C, 6.7% H and 53.3% O by mass. What is its empirical formula? (C = 12, H = 1, O = 16)

    • CH2O
    • CH4O
    • C2H4O2
    • CHO

  7. A hydrocarbon is 85.7% C and 14.3% H by mass. What is its empirical formula? (C = 12, H = 1)

    • CH2
    • C2H
    • CH
    • CH3

  8. A compound is 82.8% C and 17.2% H by mass. What is its empirical formula? (C = 12, H = 1)

    • C2H5
    • CH2
    • CH4
    • C3H8

  9. A hydrocarbon has empirical formula CH2 and molar mass 56 g mol-1. What is its molecular formula?

    • C4H8
    • C5H10
    • C3H6
    • C2H4

  10. 1.20 g of magnesium burns completely to give 2.00 g of an oxide. What is the empirical formula? (Mg = 24, O = 16)

    • MgO2
    • MgO
    • Mg3O2
    • Mg2O

  11. A compound contains 2.4 g C, 0.4 g H and 3.2 g O. What is its empirical formula? (C = 12, H = 1, O = 16)

    • C2H4O
    • CH4O
    • CHO2
    • CH2O

  12. Which equation relates pressure, volume, amount and temperature for a gas?

    • pn = VRT
    • p/V = nRT
    • pV = nRT
    • pV = R/nT

  13. How many moles of gas occupy 0.0245 m^3 at 298 K and 101325 Pa? (R = 8.31)

    • 0.100 mol
    • 1.00 mol
    • 10.0 mol
    • 0.0245 mol

  14. What is 250 cm^3 expressed in m^3?

    • 2.5 x 10^-6 m^3
    • 0.25 m^3
    • 2.5 x 10^-4 m^3
    • 2.5 x 10^-3 m^3

  15. 0.60 g of a gas occupies 245 cm^3 at 101325 Pa and 298 K. What is its molar mass? (R = 8.31)

    • 120 g mol-1
    • 6 g mol-1
    • 30 g mol-1
    • 60 g mol-1

  16. Why must temperature be converted to kelvin when using pV = nRT?

    • The gas constant is defined only in kelvin for liquids, so it must be converted before it is used in the equation
    • Kelvin is the only unit that can be used for pressure in the gas equation, so the temperature is converted to match it
    • Kelvin removes the need to convert the volume to m^3 before the calculation can be carried out correctly
    • Kelvin is an absolute scale, so zero corresponds to no thermal motion and volume is proportional to temperature

  17. A compound has empirical formula CH2O and molar mass 90 g mol-1. What is its molecular formula? (Empirical mass 30)

    • C6H12O6
    • CH2O
    • C3H6O3
    • C2H4O2

  18. Why can combustion data give only the empirical formula directly?

    • Combustion gives the molar mass directly from the mass of carbon dioxide produced, so no ratio is needed
    • Combustion produces only one product, which means the mass data cannot show the ratio of the elements
    • Combustion gives the masses of elements in the sample, not its total molar mass, so only the simplest ratio can be found
    • Combustion always destroys part of each molecule, so the measured masses cannot relate to the original compound

  19. What is the percentage by mass of carbon in ethene, C2H4? (C = 12, H = 1)

    • 66.7%
    • 50.0%
    • 14.3%
    • 85.7%

  20. What is the first step in finding an empirical formula from percentage composition by mass?

    • Write the balanced equation for the reaction that produced the compound, which gives the ratio of the elements
    • Divide the mass of each element by the total mass of the sample, which gives the mole fraction of each element
    • Convert each percentage to moles using its relative atomic mass
    • Find the molar mass of the compound from its percentage composition, which is the first step in the method

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