An oxide of nitrogen contains 30.4\% nitrogen and 69.6\% oxygen by mass. If its relative m… — JAMB UTME

An oxide of nitrogen contains $30.4\%$ nitrogen and $69.6\%$ oxygen by mass. If its relative molar mass is $92\text{ g/mol}$ , what is the molecular formula of the compound? $[N = 14, O = 16]$

Answer: N2O4 / N₂O₄ / $N_2O_4$

Answer

The molecular formula of the oxide is

N_2O_4

First, the empirical formula is derived by dividing the percentage composition by relative atomic masses (

30.4/14 = 2.17

for N and

69.6/16 = 4.35

for O), which gives a simple whole number ratio of

1:2

(

NO_2

). The empirical formula mass of

NO_2

46\text{ g/mol}

. Dividing the molar mass (

92\text{ g/mol}

) by the empirical mass (

46\text{ g/mol}

) yields an integer factor of

2

. Multiplying the empirical formula by

2

gives the molecular formula

N_2O_4

Step-by-Step Solution

Calculate the relative number of moles of each element.

Moles of

N = \frac{30.4}{14} = 2.17

; Moles of

O = \frac{69.6}{16} = 4.35

Dividing mass percentage by atomic mass gives the mole ratio.

Determine the simplest whole number mole ratio.

Ratio

N : O = \frac{2.17}{2.17} : \frac{4.35}{2.17} = 1 : 2

. Empirical formula =

NO_2

Dividing by the smallest mole value yields the empirical formula subscripts.

Calculate the empirical formula mass and the multiplier integer

n

Empirical formula mass =

14 + 2(16) = 46\text{ g/mol}

n = \frac{92}{46} = 2

The integer multiplier

n

is the ratio of molar mass to empirical formula mass.

Multiply empirical formula subscripts by

n

Molecular formula =

(NO_2)_2 = N_2O_4

Applying the integer multiplier gives the actual number of atoms in the molecule.

Key Concept

Determining empirical and molecular formulas from elemental percentage composition and relative molar mass.

Estimated Time:1m 30s

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