Average isotope mass of X

= $${{200 \times 90 + 199 \times 8 + 202 \times 2} \over {90 + 8 + 2}}$$

= $${{18000 + 1592 + 404} \over {100}}$$

= 199.96 a.m.u $$ \approx $$ 200 a.m.u

At STP, 22.4 L H₂

= 6.023 $$ \times $$ 10²³ molecules

15 L H₂ = $${{6.023 \times {{10}^{23}} \times 15} \over {22.4}}$$ = $$4.033 \times {10^{23}}$$

5 L N₂ = $${{6.023 \times {{10}^{23}} \times 5} \over {22.4}}$$ = 1.344 $$ \times $$ 10²³

2 g of H₂ = 6.023 $$ \times $$ 10²³

0.5 g H₂ = $${{6.023 \times {{10}^{23}} \times 0.5} \over {2}}$$ = 1.505 $$ \times $$ 10²³

32 g O₂ = 6.023 $$ \times $$ 10²³

10 g O₂ = $${{6.023 \times {{10}^{23}} \times 10} \over {32}}$$ = 1.882 $$ \times $$ 10²³

	Element	%	Atomic mass	Relative no. of atoms	Simplest ratio of atoms
1	C	40	12	$${{40} \over {12}}$$ = 3.33	$${{3.33} \over {3.3}}$$ = 1
2	H	13.3	1	$${{13.3} \over 1}$$ = 13.3	$${{13.3} \over {3.3}}$$ = 4
2	N	46.7	14	$${{46.7} \over {14}}$$ = 3.3	$${{3.3} \over {3.3}}$$ = 1

$$ \therefore $$ The empirical formula is CH₄N.

1 mole = 6.023 $$ \times $$ 10²³ number of molecules.

1 g mole of O₂ = 32 g of O₂

$$ \Rightarrow $$ 16 g of O₂ = 0.5 g mole of O₂

1 g mole of N₂ = 28 g of N₂

$$ \Rightarrow $$ 7 g of N₂ = 0.25 g mole of N₂

1 g mole of H₂ = 2 g of H₂

$$ \Rightarrow $$ 2 g of H₂ = 1 g mole of H₂

1 g of NO₂ = 14 + 16 $$ \times $$ 2 = 46

$$ \Rightarrow $$ 16 g of NO₂ = 0.35 mole NO₂