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

N_{2} O H_{3}

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N (O H)_{2}

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N O_{2} H_{3}

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N O H_{3}

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Solution

The correct option is C $N O H_{3}$
Element % ratio simplest ratio
N 42 3 1
O 48 3 1
H 9 9 3
Hence, empirical formula is $N O H_{3}$ .

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

The correct option is C NOH3Element % ratio simplest ratioN 42 3 1 O 48 3 1H 9 9 3Hence, empirical formula is NOH3.

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

The correct option is C $N O H_{3}$
Element % ratio simplest ratio
N 42 3 1
O 48 3 1
H 9 9 3
Hence, empirical formula is $N O H_{3}$ .