An oxide of nitrogen was found by elemental analysis to contain 30.4% nitrogen and 69.6% oxygen. if 23.0 g of this gas were found to occupy 5.6 l at stp, what are the empirical and molecular formulas for this oxide of nitrogen?
express the empirical and molecular formulas for this oxide of nitrogen, separated by a comma.

The answer is NO₂-NO₂, N₂O₄

Explanation:

If 23.0 g of this gas were found to occupy 5.6 L at STP we can apply the Ideal gas Law to determinate the quantity of mols of the oxide.

As I have the mass, I can find out the molar weight.

STP are 1 atm and 273, 1 K so:

P . V = n . R . T

1 atm .  5.6L = n . 0.082 L.atm/mol.K . 273.1K

(1 atm .  5.6L)/ 0.082 mol.K/L.atm . 273.1K = n

0.250 mol = n

Molar weight = mass/mols

23 g /0.250 mol = 92 g/m

The percentages are:  30.4% N and 69.6%O

100% 92 g

30.4% (30.4 .92)/100 = 28 g → 2 mols of N

100% 92 g

69.6% (69.6 . 92)/ 100 = 64 g → 4 mols of O

Molecular formula N₂O₄

Empirical formula: NO₂-NO₂