Determine the empirical formula of the compound whose composition by mass is 42% nitrogen, 48% oxygen and 9% hydrogen. [N = 14, O = 16, H = 1]

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Solution

Element Atomic weight Percentage Relative number of moles Simplest molar ratio

N 14 42 $\frac{42}{14} = 3$ 1

O 16 48 $\frac{48}{16} = 3$ 1

H 1 9 $\frac{9}{1} = 9$ 3

Hence, the empirical formula of the compound is NOH₃ or NH₂OH.

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Determine the empirical formula of the compound whose composition by mass is 42% nitrogen, 48% oxygen and 9% hydrogen. [H = 1, N = 14, O = 16].

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Element	Atomic weight	Percentage	Relative number of moles	Simplest molar ratio
N	14	42	$\frac{42}{14} = 3$	1
O	16	48	$\frac{48}{16} = 3$	1
H	1	9	$\frac{9}{1} = 9$	3

Determine the empirical formula of the compound whose composition by mass is 42% nitrogen, 48% oxygen and 9% hydrogen. [N = 14, O = 16, H = 1]

Element Atomic weight Percentage Relative number of moles Simplest molar ratio N 14 42 4214=3 1 O 16 48 4816=3 1 H 1 9 91=9 3 Hence, the empirical formula of the compound is NOH3 or NH2OH.

Element Atomic weight Percentage Relative number of moles Simplest molar ratio

N 14 42 $\frac{42}{14} = 3$ 1

O 16 48 $\frac{48}{16} = 3$ 1

H 1 9 $\frac{9}{1} = 9$ 3

Hence, the empirical formula of the compound is NOH₃ or NH₂OH.