A hydride of nitrogen contains 87.5 per cent by mass fo nitrogen. Determine the empirical formula of this compound.

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Solution

Constituents Symbol %age Atomic mass moles of the element = %age/at. mass (relative no. of moles) Simplest molar ratio Simplest whole no. ratio

Nitrogen N 87.5 14 $\frac{87.5}{14} = 6.25$ $\frac{6.25}{6.25} = 1$ 1

Hydrogen H 12.5 1 $\frac{12.5}{6.25} = 4.34$ $\frac{12.5}{6.25} = 2$ 2

Thus, empirical formula ( $N H_{2}$ ).

But nitrogen forms a trivalent compound so hydride of nitrogen by 87.5 by mass % is $N_{2} H_{4}$ ( dimer of $N H_{2}$ ).

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Constituents	Symbol	%age	Atomic mass	moles of the element = %age/at. mass (relative no. of moles)	Simplest molar ratio	Simplest whole no. ratio
Nitrogen	N	87.5	14	$\frac{87.5}{14} = 6.25$	$\frac{6.25}{6.25} = 1$	1
Hydrogen	H	12.5	1	$\frac{12.5}{6.25} = 4.34$	$\frac{12.5}{6.25} = 2$	2