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4 December, 15:45

The combustion of 3.42 gg of a compound known to contain only nitrogen and hydrogen gave 9.82g of NO2 and 3.85 g of water. Determine the empirical formula of this compound.

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  1. 4 December, 16:13
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    Let x be the amount of mass of N while y is the amount of mass of H. The combustion reactions are as follows:

    N + O₂ → NO₂

    2 H + 1/2 O₂ → H₂O

    The solution is as follows:

    x + y = 3.42 g - - > eqn 1

    9.82 g NO₂ (1 mol/46 g) (1 mol N/1 mol NO₂) (14 g N/mol) = x

    2.989 g N = x

    y = 3.42 g - 2.989 g = 0.4313 g H

    Mol N: 2.989 g/14 g/mol = 0.2135

    Mol H: 0.4313 g/1 g/mol = 0.4313

    Divide both by 0.2135,

    N: 0.2135/0.2135 = 1

    H: 0.4313/0.2135 = 2

    Thus, the empirical formula is NH₂.
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