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30 December, 18:50

3. A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of

hydrogen, and 33.6 g of oxygen. What is the empirical formula of the compound?

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Answers (1)
  1. 30 December, 19:15
    0
    HNO₃

    Explanation:

    Data given

    Nitrogen = 9.8 g

    Hydrogen = 0.70 g

    Oxygen = 33.6 g

    Empirical formula = ?

    Solution:

    Convert the masses to moles

    For Nitrogen

    Molar mass of N = 14 g/mol

    no. of mole = mass in g / molar mass

    Put value in above formula

    no. of mole = 9.8 g / 14 g/mol

    no. of mole = 0.7

    mole of N = 0.7 mol

    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 = 0.70 g / 1 g/mol

    no. of mole = 0.7

    mole of H = 0.7 mol

    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 = 33.6 g / 16 g/mol

    no. of mole = 2.1

    mole of O = 2.1 mol

    Now we have values in moles as below

    N = 0.7

    H = 0.7

    O = 2.1

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

    N = 0.7 / 0.7 = 1

    H = 0.7 / 0.7 = 1

    O = 2.1 / 0.7 = 3

    So all have following values

    N = 1

    H = 1

    O = 3

    So the empirical formula will be HNO₃ i. e. all three atoms in simplest small ratio.
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