The combustion of 3.42 gg of a compound known to contain only nitrogen and hydrogen gave 9.82g of no2 and 3.85 g of water. determine the empirical formula of this compound.

Let x be the amount of mass of N while y is the amount of mass of H. The combustion reactions are as follows:

N + O₂ → NO₂
2 H + 1/2 O₂ → H₂O

The solution is as follows:
x + y = 3.42 g  --> eqn 1

9.82 g NO₂(1 mol/46 g)(1 mol N/1 mol NO₂)(14 g N/mol) = x
2.989 g N = x
y = 3.42 g - 2.989 g = 0.4313 g H

Mol N: 2.989 g/14 g/mol = 0.2135
Mol H: 0.4313 g/1 g/mol = 0.4313
Divide both by 0.2135,
N: 0.2135/0.2135 = 1
H: 0.4313/0.2135 = 2

Thus, the empirical formula is NH₂.