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21 May, 23:08

15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula

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  1. 21 May, 23:16
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    OsCO or COOs

    Explanation:

    Data given

    Carbon = 15.89 %

    Oxygen = 21.18 %

    Osmium = 62.93%

    Empirical formula = ?

    Solution:

    First find the masses of each component

    Consider total compound is 100g

    As we now

    mass of element = % of component

    So,

    15.89 g of C = 15.89 % Carbon

    21.18 g of O = 21.18 % Oxygen

    62.93 g of Os = 62.93% Osmium

    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 = 15.89 g / 12 g/mol

    no. of mole = 1.3242

    mole of C = 1.3242

    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 = 21.18 g / 16 g/mol

    no. of mole =

    mole of O = 1.3238

    For Os

    Molar mass of Os = 190 g/mol

    no. of mole = mass in g / molar mass

    Put value in above formula

    no. of mole = 62.93 g / 190 g/mol

    no. of mole =

    mole of Os = 1.3312

    Now we have values in moles as below

    C = 1.3242

    O = 1.3238

    Os = 1.3312

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

    C = 1.3242 / 1.3238 = 1.0003

    O = 1.3238 / 1.3238 = 1

    Os = 1.3312 / 1.3238 = 1.0056

    So all have round value 1 mole

    C = 1

    O = 1

    Os = 1

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