Ano organic compound contains 52.17%of carbon, 13% of hydrogen and the rest being oxygen. Given that the relative molecular mass is92. A, it's empirical formula is what?

Physics

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Answer

C2H6O\text{C}_2\text{H}_6\text{O}

Step-by-step solution

4 steps

  1. Determine the percentage of oxygen.

    The sum of percentages of all elements in a compound must be 100%.

    Percentage of Oxygen=100%−(Percentage of Carbon+Percentage of Hydrogen)\text{Percentage of Oxygen} = 100\% - (\text{Percentage of Carbon} + \text{Percentage of Hydrogen}) Percentage of Oxygen=100%−(52.17%+13%)\text{Percentage of Oxygen} = 100\% - (52.17\% + 13\%) Percentage of Oxygen=100%−65.17%\text{Percentage of Oxygen} = 100\% - 65.17\% Percentage of Oxygen=34.83%\text{Percentage of Oxygen} = 34.83\%

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

    In a 100 g sample:
    Mass of Carbon (C) = 52.17 g
    Mass of Hydrogen (H) = 13 g
    Mass of Oxygen (O) = 34.83 g

  3. Convert masses to moles using atomic masses.

    Atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol.

    Moles of Carbon:

    Moles of C=52.17 g12.01 g/mol≈4.344 mol\text{Moles of C} = \frac{52.17 \text{ g}}{12.01 \text{ g/mol}} \approx 4.344 \text{ mol}

    Moles of Hydrogen:

    Moles of H=13 g1.008 g/mol≈12.897 mol\text{Moles of H} = \frac{13 \text{ g}}{1.008 \text{ g/mol}} \approx 12.897 \text{ mol}

    Moles of Oxygen:

    Moles of O=34.83 g16.00 g/mol≈2.177 mol\text{Moles of O} = \frac{34.83 \text{ g}}{16.00 \text{ g/mol}} \approx 2.177 \text{ mol}

  4. Determine the simplest whole-number ratio by dividing by the smallest number of moles.

    The smallest number of moles is 2.177 mol (for Oxygen).

    Ratio for Carbon:

    4.344 mol2.177 mol≈1.995≈2\frac{4.344 \text{ mol}}{2.177 \text{ mol}} \approx 1.995 \approx 2

    Ratio for Hydrogen:

    12.897 mol2.177 mol≈5.924≈6\frac{12.897 \text{ mol}}{2.177 \text{ mol}} \approx 5.924 \approx 6

    Ratio for Oxygen:

    2.177 mol2.177 mol=1\frac{2.177 \text{ mol}}{2.177 \text{ mol}} = 1

    The simplest whole-number ratio of C:H:O is 2:6:1.

    The empirical formula is C2H6O\text{C}_2\text{H}_6\text{O}.

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