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A hydrocarbon has 86% carbon and 14% hydrogen and a molecular mass of 70. Determine its empirical and molecular formulas.

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: A hydrocarbon containing 86% of carbon and 14% of hydrogen has a molecular mass of 70.

a) Determine its Empirical formula.

  1. Calculate the moles of each element in 100 g of the compound.

    Moles of C=86 g12 g/mol=7.167 mol\text{Moles of C} = \frac{86 \text{ g}}{12 \text{ g/mol}} = 7.167 \text{ mol} Moles of H=14 g1 g/mol=14 mol\text{Moles of H} = \frac{14 \text{ g}}{1 \text{ g/mol}} = 14 \text{ mol}

  2. Find the simplest whole number ratio of moles.

    Divide by the smallest number of moles (7.167 mol):

    Ratio C:H=7.1677.167:147.167≈1:1.95≈1:2\text{Ratio C:H} = \frac{7.167}{7.167} : \frac{14}{7.167} \approx 1 : 1.95 \approx 1:2

  3. Write the empirical formula.

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

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

    b) Determine its Molecular formula.

  4. Calculate the empirical formula mass.

    Empirical formula mass of CH2=(1×12)+(2×1)=14 g/mol\text{CH}_2 = (1 \times 12) + (2 \times 1) = 14 \text{ g/mol}.

  5. Determine the factor nn by which the empirical formula must be multiplied.

    n=Molecular massEmpirical formula mass=70 g/mol14 g/mol=5n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{70 \text{ g/mol}}{14 \text{ g/mol}} = 5

  6. Write the molecular formula.

    Molecular formula =(CH2)5=C5H10= (\text{CH}_2)_5 = \text{C}_5\text{H}_{10}.

    b) The molecular formula is C5H10\boxed{\text{C}_5\text{H}_{10}}.

    : On complete combustion in oxygen, 0.42g of a gaseous hydrocarbon (X) gave 1.32g of carbon dioxide and 0.54g of water.

    a) What is the empirical formula of X?

  7. Calculate the mass of carbon and hydrogen in the products.

    Mass of C in CO2=1.32 g CO2×12 g C44 g CO2=0.36 g C\text{CO}_2 = 1.32 \text{ g } \text{CO}_2 \times \frac{12 \text{ g C}}{44 \text{ g CO}_2} = 0.36 \text{ g C}
    Mass of H in H2O=0.54 g H2O×2×1 g H18 g H2O=0.06 g H\text{H}_2\text{O} = 0.54 \text{ g } \text{H}_2\text{O} \times \frac{2 \times 1 \text{ g H}}{18 \text{ g H}_2\text{O}} = 0.06 \text{ g H}

  8. Verify that the total mass of C and H equals the initial mass of X.

    Total mass of C and H =0.36 g+0.06 g=0.42 g= 0.36 \text{ g} + 0.06 \text{ g} = 0.42 \text{ g}. This matches the given mass of hydrocarbon X.

  9. Calculate the moles of each element.

    Moles of C=0.36 g12 g/mol=0.03 mol\text{Moles of C} = \frac{0.36 \text{ g}}{12 \text{ g/mol}} = 0.03 \text{ mol} Moles of H=0.06 g1 g/mol=0.06 mol\text{Moles of H} = \frac{0.06 \text{ g}}{1 \text{ g/mol}} = 0.06 \text{ mol}

  10. Find the simplest whole number ratio of moles.

    Divide by the smallest number of moles (0.03 mol):

    Ratio C:H=0.030.03:0.060.03=1:2\text{Ratio C:H} = \frac{0.03}{0.03} : \frac{0.06}{0.03} = 1 : 2

  11. Write the empirical formula.

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

    a) The empirical formula of X is CH2\boxed{\text{CH}_2}.

    b) 0.21g of X occupied a volume of 120cm³ at r.t.p, what is the relative molecular mass of X?

  12. Convert the volume to dm³ and use the molar volume at r.t.p.

    (24 dm³/mol) to find the moles of X.

    Volume of X=120 cm3=0.120 dm3\text{Volume of X} = 120 \text{ cm}^3 = 0.120 \text{ dm}^3 Moles of X=0.120 dm324 dm3/mol=0.005 mol\text{Moles of X} = \frac{0.120 \text{ dm}^3}{24 \text{ dm}^3/\text{mol}} = 0.005 \text{ mol}

  13. Calculate the relative molecular mass (M) of X.

    M=Mass of XMoles of X=0.21 g0.005 mol=42 g/molM = \frac{\text{Mass of X}}{\text{Moles of X}} = \frac{0.21 \text{ g}}{0.005 \text{ mol}} = 42 \text{ g/mol}

    b) The relative molecular mass of X is 42\boxed{42}.

    c) Hence, determine the molecular formula of X.

  14. Calculate the empirical formula mass from part (a).

    Empirical formula mass of CH2=(1×12)+(2×1)=14 g/mol\text{CH}_2 = (1 \times 12) + (2 \times 1) = 14 \text{ g/mol}.

  15. Determine the factor nn by which the empirical formula must be multiplied.

    n=Molecular massEmpirical formula mass=42 g/mol14 g/mol=3n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{42 \text{ g/mol}}{14 \text{ g/mol}} = 3

  16. Write the molecular formula.

    Molecular formula =(CH2)3=C3H6= (\text{CH}_2)_3 = \text{C}_3\text{H}_6.

    c) The molecular formula of X is C3H6\boxed{\text{C}_3\text{H}_6}.

    Drop

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