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An Organic Compound is composed of 62.1% Carbon, 10.18% Hydrogen, and 27.88% Oxygen. Determine the empirical formula of the Compound. Given that its relative molecular mass is 116, deduce its molecular formula.

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2 parts
i

C3H6O\text{C}_3\text{H}_6\text{O}

ii

C6H12O2\text{C}_6\text{H}_{12}\text{O}_2

Step-by-step solution

7 steps

Here's how to solve this problem:

Part i· 4 steps

Determine the empirical formula of the compound.

The compound is composed of:

  • Carbon (C): 62.1%
  • Hydrogen (H): 10.1%
  • Oxygen (O): 27.8%

  1. Assume a 100 g sample.

    This converts the percentages directly into masses:
    Mass of C = 62.1 g62.1 \text{ g}
    Mass of H = 10.1 g10.1 \text{ g}
    Mass of O = 27.8 g27.8 \text{ g}

  2. Convert the mass of each element to moles using their atomic masses (C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol).

    Moles of C:

    Moles of C=62.1 g12.01 g/mol=5.1707 mol\text{Moles of C} = \frac{62.1 \text{ g}}{12.01 \text{ g/mol}} = 5.1707 \text{ mol}

    Moles of H:

    Moles of H=10.1 g1.008 g/mol=10.0198 mol\text{Moles of H} = \frac{10.1 \text{ g}}{1.008 \text{ g/mol}} = 10.0198 \text{ mol}

    Moles of O:

    Moles of O=27.8 g16.00 g/mol=1.7375 mol\text{Moles of O} = \frac{27.8 \text{ g}}{16.00 \text{ g/mol}} = 1.7375 \text{ mol}

  3. Divide each mole value by the smallest number of moles to find the simplest mole ratio.

    The smallest number of moles is 1.7375 mol1.7375 \text{ mol} (for Oxygen).
    Ratio for C:

    5.1707 mol1.7375 mol≈2.976≈3\frac{5.1707 \text{ mol}}{1.7375 \text{ mol}} \approx 2.976 \approx 3

    Ratio for H:

    10.0198 mol1.7375 mol≈5.767\frac{10.0198 \text{ mol}}{1.7375 \text{ mol}} \approx 5.767

    Ratio for O:

    1.7375 mol1.7375 mol=1\frac{1.7375 \text{ mol}}{1.7375 \text{ mol}} = 1

    The ratio for hydrogen (5.767) is very close to 6. In empirical formula problems, small deviations from whole numbers are typically due to experimental error or rounding in the given percentages. We will round 5.767 to 6.

  4. Write the empirical formula using these whole-number ratios.

    The empirical formula is C3H6O\text{C}_3\text{H}_6\text{O}.

    The empirical formula of the compound is C3H6O\boxed{\text{C}_3\text{H}_6\text{O}}.

Part ii

C6H12O2\text{C}_6\text{H}_{12}\text{O}_2

Given that its relative molecular mass is 116, deduce its molecular formula.

  1. Calculate the empirical formula mass (EFM) of C3H6O\text{C}_3\text{H}_6\text{O}.

    EFM = (3×Atomic mass of C)+(6×Atomic mass of H)+(1×Atomic mass of O)(3 \times \text{Atomic mass of C}) + (6 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of O})
    EFM = (3×12.01)+(6×1.008)+(1×16.00)(3 \times 12.01) + (6 \times 1.008) + (1 \times 16.00)
    EFM = 36.03+6.048+16.00=58.078 g/mol36.03 + 6.048 + 16.00 = 58.078 \text{ g/mol}

  2. Determine the factor 'n' by dividing the given relative molecular mass by the EFM.

    Given relative molecular mass = 116.

    n=Relative Molecular MassEmpirical Formula Mass=11658.078≈1.997≈2n = \frac{\text{Relative Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{116}{58.078} \approx 1.997 \approx 2

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

    Molecular Formula = (C3H6O)2=C3×2H6×2O1×2=C6H12O2(\text{C}_3\text{H}_6\text{O})_2 = \text{C}_{3 \times 2}\text{H}_{6 \times 2}\text{O}_{1 \times 2} = \text{C}_6\text{H}_{12}\text{O}_2.

    The molecular formula of the compound is C6H12O2\boxed{\text{C}_6\text{H}_{12}\text{O}_2}.

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