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31 March, 19:14

determine the empirical formula for a compound that has the following analysis: 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine.

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  1. 31 March, 19:41
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    C₂HO₂Br₃

    Explanation:

    Data given

    Carbon = 8.09%

    Hydrogen = 0.34%

    Oxygen = 10.78%

    bromine = 80.78%

    Empirical formula = ?

    Solution:

    First find the masses of each component

    Consider total compound is 100g

    As we now

    mass of element = % of component

    So,

    8.09 g of C = 8.09% of Carbon

    0.34 g og H = 0.34% of Hydrogen

    10.78 g of O = 10.78% of Oxygen

    80.78 g of bromine = 80.78% of bromine

    Now convert the masses to moles

    For Carbon

    Molar mass of C = 12 g/mol

    no. of mole = mass in g / molar mass

    Put value in above formula

    no. of mole = 8.09 g / 12 g/mol

    no. of mole = 0.674

    mole of C = 0.6742

    For Hydrogen

    Molar mass of H = 1 g/mol

    no. of mole = mass in g / molar mass

    Put value in above formula

    no. of mole = 0.34 g / 1 g/mol

    no. of mole = 0.34

    mole of H = 0.34 mole

    For Oxygen

    Molar mass of O = 16 g/mol

    no. of mole = mass in g / molar mass

    Put value in above formula

    no. of mole = 10.78 g / 16 g/mol

    no. of mole = 0.674

    mole of O = 0.674 mole

    For Br

    Molar mass of Br = 80 g/mol

    no. of mole = mass in g / molar mass

    Put value in above formula

    no. of mole = 80.78 g / 80 g/mol

    no. of mole = 1.0098

    mole of Br = 1.0098 moles

    Now we have values in moles as below

    C = 0.6742

    H = 0.34

    O = 0.674

    Br = 1.0098

    Divide the all values on the smallest values to get whole number ratio

    C = 0.6742 / 0.34 = 1.983 ≅ 2

    H = 0.34 / 0.34 = 1

    O = 0.674 / 0.34 = 1.983 ≅ 2

    Br = 1.0098 / 0.34 = 2.97 ≅ 3

    So all have round value 1 mole

    C = 2

    H = 1

    O = 2

    Br = 3

    So the empirical formula will be C₂HO₂Br₃ i. e. all 3 atoms in simplest small ratio
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