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16 April, 11:02

Ammonia react with phosphorus acid to form a compound that contains 28.2% of nitrogen 20.8%, phosphorus 8.1% of hydrogen 42.9%oxygen. Calculate the empirical formula of this compound

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  1. 16 April, 13:55
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    Empirical formula will be (NH₄) ₃PO₄, which matches the molecular formula

    Explanation:

    This is the reaction:

    NH₃ + H₃PO₄ → 28.2% N, 20.8% P, 8.1% H, 42.9% O

    In 100 g of compound we have:

    28.2 g N

    20.8 g of P

    8.1 g of H

    42.9 g of O

    Now we divide each between the molar mass:

    28.2 g / 14 g/mol = 2.01 mol

    20.8 g / 30.97 g/mol = 0.671 mol

    8.1 g / 1 g/mol = 8.1 mol

    42.9 g / 16 g/mol = 2.68 mol

    And we divide again between the lowest value of moles

    2.01 mol / 0.671 mol → 3

    0.671 mol / 0.671 mol → 1

    8.1 mol / 0.671 mol → 12

    2.68 mol / 0.671 mol → 4

    Molecular formula will be: N₃PH₁₂O₄ → (NH₄) ₃PO₄

    Empirical formula will be (NH₄) ₃PO₄, which matches the molecular formula
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