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Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass.

Chemistry

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Answer

SiO2\text{SiO}_2

2)2=C8H10O2N4_2)_2 = \text{C}_8\text{H}_{10}\text{O}_2\text{N}_4

Step-by-step solution

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Q9. A compound contains 46.7% of silicon (Si) and 53.3% oxygen (O) by mass. Find the empirical formula of the compound, given the relative atomic mass of silicon (Si) = 28 and oxygen (O) = 16.

  1. Assume a 100 g sample.

    This means we have 46.7 g of Si and 53.3 g of O.

  2. Convert mass to moles for each element.

    Moles of Si=Mass of SiRelative atomic mass of Si=46.7 g28 g/mol≈1.6679 molMoles of O=Mass of ORelative atomic mass of O=53.3 g16 g/mol≈3.3313 mol\text{Moles of Si} = \frac{\text{Mass of Si}}{\text{Relative atomic mass of Si}} = \frac{46.7 \text{ g}}{28 \text{ g/mol}} \approx 1.6679 \text{ mol} \\ \text{Moles of O} = \frac{\text{Mass of O}}{\text{Relative atomic mass of O}} = \frac{53.3 \text{ g}}{16 \text{ g/mol}} \approx 3.3313 \text{ mol}

  3. Find the simplest whole number ratio of moles.

    Divide each number of moles by the smallest number of moles (1.6679 mol).

    Ratio of Si=1.6679 mol1.6679 mol=1Ratio of O=3.3313 mol1.6679 mol≈1.997≈2\text{Ratio of Si} = \frac{1.6679 \text{ mol}}{1.6679 \text{ mol}} = 1 \\ \text{Ratio of O} = \frac{3.3313 \text{ mol}}{1.6679 \text{ mol}} \approx 1.997 \approx 2

    The simplest whole number ratio of Si:O is 1:2.

  4. Write the empirical formula.

    The empirical formula is SiO2\boxed{\text{SiO}_2}.

    Q10. Caffeine is a compound found in coffee and tea. The percentage composition of caffeine is 49.1% carbon, 5.1% hydrogen, 16.5% oxygen and 28.9% nitrogen. The relative molecular mass of caffeine is 195. Find the molecular formula of caffeine, given the atomic mass of carbon (C) = 12, hydrogen (H) = 1, oxygen (O) = 16, and nitrogen (N) = 14.

  5. Assume a 100 g sample and convert percentages to masses.

    • Mass of C = 49.1 g
    • Mass of H = 5.1 g
    • Mass of O = 16.5 g
    • Mass of N = 28.9 g

  6. Convert mass to moles for each element.

    Moles of C=49.1 g12 g/mol≈4.0917 molMoles of H=5.1 g1 g/mol=5.1 molMoles of O=16.5 g16 g/mol≈1.0313 molMoles of N=28.9 g14 g/mol≈2.0643 mol\text{Moles of C} = \frac{49.1 \text{ g}}{12 \text{ g/mol}} \approx 4.0917 \text{ mol} \\ \text{Moles of H} = \frac{5.1 \text{ g}}{1 \text{ g/mol}} = 5.1 \text{ mol} \\ \text{Moles of O} = \frac{16.5 \text{ g}}{16 \text{ g/mol}} \approx 1.0313 \text{ mol} \\ \text{Moles of N} = \frac{28.9 \text{ g}}{14 \text{ g/mol}} \approx 2.0643 \text{ mol}

  7. Find the simplest whole number ratio of moles.

    Divide each number of moles by the smallest number of moles (1.0313 mol for O).

    Ratio of C=4.0917 mol1.0313 mol≈3.967≈4Ratio of H=5.1 mol1.0313 mol≈4.945≈5Ratio of O=1.0313 mol1.0313 mol=1Ratio of N=2.0643 mol1.0313 mol≈2.001≈2\text{Ratio of C} = \frac{4.0917 \text{ mol}}{1.0313 \text{ mol}} \approx 3.967 \approx 4 \\ \text{Ratio of H} = \frac{5.1 \text{ mol}}{1.0313 \text{ mol}} \approx 4.945 \approx 5 \\ \text{Ratio of O} = \frac{1.0313 \text{ mol}}{1.0313 \text{ mol}} = 1 \\ \text{Ratio of N} = \frac{2.0643 \text{ mol}}{1.0313 \text{ mol}} \approx 2.001 \approx 2

    The simplest whole number ratio of C:H:O:N is 4:5:1:2.

  8. Write the empirical formula.

    The empirical formula is C4H5ON2\text{C}_4\text{H}_5\text{ON}_2.

  9. Calculate the empirical formula mass.

    Empirical formula mass=(4×12)+(5×1)+(1×16)+(2×14)=48+5+16+28=97 g/mol\text{Empirical formula mass} = (4 \times 12) + (5 \times 1) + (1 \times 16) + (2 \times 14) \\ = 48 + 5 + 16 + 28 \\ = 97 \text{ g/mol}

  10. Determine the factor 'n'.

    Given relative molecular mass = 195.

    n=Relative molecular massEmpirical formula mass=19597≈2.01≈2n = \frac{\text{Relative molecular mass}}{\text{Empirical formula mass}} = \frac{195}{97} \approx 2.01 \approx 2

  11. Write the molecular formula.

    Multiply the subscripts in the empirical formula by n=2.

    Molecular formula=(C4H5ON2)2=C8H10O2N4\text{Molecular formula} = (\text{C}_4\text{H}_5\text{ON}_2)_2 = \boxed{\text{C}_8\text{H}_{10}\text{O}_2\text{N}_4}
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