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13 March, 23:49

Determine the empirical formula of a compound containing 48.38 grams of carbon, 8.12 grams of hydrogen, and 53.5 grams of oxygen.

In an experiment, the molar mass of the compound was determined to be 180.15 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations.

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  1. 14 March, 03:10
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    1. The empirical formula is CH2O

    2. The molecular formula is C6H12O6

    Explanation:

    Data obtained from the question include:

    Carbon (C) = 48.38g

    Hydrogen (H) = 8.12g

    Oxygen (O) = 53.5g

    Molar Mass of the compound = 180.15 g/mol

    1. The empirical formula can be obtained as follow:

    C = 48.38g

    H = 8.12g

    O = 53.5g

    Divide each by their molar mass

    C = 48.38/12 = 4.032

    H = 8.12/1 = 8.12

    O = 53.5/16 = 3.344

    Next, divide by the smallest

    C = 4.032/3.344 = 1

    H = 8.12/3.344 = 2

    O = 3.344/3.344 = 1

    Therefore, the empirical formula is CH2O.

    2. The molecular formula is mostly a multiple of the empirical i. e

    Molecular formula = > [CH2O]n

    Now, we need to find the value of 'n' in order to obtain the molecular formula.

    From the question given, we were told that the molar mass of the compound is 180.15 g/mol. With this information, the molecular formula can be obtained as follow:

    [CH2O]n = 180.15

    [12 + (2x1) + 16]n = 180.15

    [12 + 2 + 16]n = 180.15

    30n = 180.15

    Divide both side by the coefficient of n i. e 30

    n = 180.15/30

    n = 6

    Therefore, the Molecular formula is

    => [CH2O]n

    => [CH2O]6

    => C6H12O6
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