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An 8.4g sample of hydrogen containing carbon and hydrogen atoms only contains 7.2g of carbon (C). (Relative atomic of hydrogen (H) = 1 and carbon (C) = 12). a. What is the empirical formula of this compound? b. If this compound has a molar mass of 84, what is its molecular formula?

Chemistry

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2 parts
a

CH2\text{CH}_2

b

C6H12\text{C}_6\text{H}_{12}

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Q11. An 8.4g sample of hydrogen (H, containing carbon and hydrogen atoms only) contains 7.2g of carbon (C).
(Relative atomic mass of hydrogen (H) = 1 and carbon (C) = 12)

Part a· 4 steps

What is the empirical formula of this compound?

  1. Calculate the mass of hydrogen (H).

    The total mass of the compound is 8.4 g, and the mass of carbon is 7.2 g.

    Mass of H=Total mass of compound−Mass of CMass of H=8.4 g−7.2 gMass of H=1.2 g\text{Mass of H} = \text{Total mass of compound} - \text{Mass of C} \\ \text{Mass of H} = 8.4 \text{ g} - 7.2 \text{ g} \\ \text{Mass of H} = 1.2 \text{ g}

  2. Convert the mass of each element to moles.

    Moles of C=Mass of CRelative atomic mass of C=7.2 g12 g/mol=0.6 molMoles of H=Mass of HRelative atomic mass of H=1.2 g1 g/mol=1.2 mol\text{Moles of C} = \frac{\text{Mass of C}}{\text{Relative atomic mass of C}} = \frac{7.2 \text{ g}}{12 \text{ g/mol}} = 0.6 \text{ mol} \\ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Relative atomic mass of H}} = \frac{1.2 \text{ g}}{1 \text{ g/mol}} = 1.2 \text{ mol}

  3. Find the simplest whole number ratio of moles.

    Divide the number of moles of each element by the smallest number of moles (which is 0.6 mol).

    Ratio of C=0.6 mol0.6 mol=1Ratio of H=1.2 mol0.6 mol=2\text{Ratio of C} = \frac{0.6 \text{ mol}}{0.6 \text{ mol}} = 1 \\ \text{Ratio of H} = \frac{1.2 \text{ mol}}{0.6 \text{ mol}} = 2

    The simplest whole number ratio of C:H is 1:2.

  4. Write the empirical formula.

    The empirical formula is CH2\boxed{\text{CH}_2}.

Part b

C6H12\text{C}_6\text{H}_{12}

If this compound has a molar mass of 84, what is its molecular formula?

  1. Calculate the empirical formula mass.

    Empirical formula = CH2\text{CH}_2

    Empirical formula mass=(1×Relative atomic mass of C)+(2×Relative atomic mass of H)Empirical formula mass=(1×12)+(2×1)Empirical formula mass=12+2=14 g/mol\text{Empirical formula mass} = (1 \times \text{Relative atomic mass of C}) + (2 \times \text{Relative atomic mass of H}) \\ \text{Empirical formula mass} = (1 \times 12) + (2 \times 1) \\ \text{Empirical formula mass} = 12 + 2 = 14 \text{ g/mol}

  2. Determine the ratio (n) between the molar mass and the empirical formula mass.

    Given molar mass = 84 g/mol.

    n=Molar massEmpirical formula massn=84 g/mol14 g/moln=6n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} \\ n = \frac{84 \text{ g/mol}}{14 \text{ g/mol}} \\ n = 6

  3. Multiply the subscripts in the empirical formula by n to get the molecular formula.

    Molecular formula = (Empirical formula)n=(CH2)6(\text{Empirical formula})_n = (\text{CH}_2)_6

    Molecular formula=C6H12\text{Molecular formula} = \boxed{\text{C}_6\text{H}_{12}}
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