An oxide of nitrogen was found by elemental analysis tocontain
30.4% nitrogen and 69.6% oxygen. if 23.0 g of thisgas were found to
occupy 5.6 l at stp, what are the empirical andmolecular formulas
for this oxide of nitrogen?

Emperical formula: NO2

Molecular formula: N2O4

Explanation:

Emperical formula

N. O

% by mass. 30.4. 69.6

mass fraction. 30.4/14 = 2.17

69.6/12 = 4.35

2.17/2.17 = 1 4.35/2.17 = 2

Emperical formula is NO2

Molecular formula

(NO2)n = molecular weight (M) of the oxide

M = mRT/PV

At STP, P = 1 atm, T = 273K

M = 23×0.082057×273/1×5.6 = 92g/mol

(NO2)n = 92

(14+32)n = 92

46n = 92

n = 92/46 = 2

Molecular formula = (NO2)2 = N2O4