Question

2.40 g of an explosive, j, contains 0.473 g of nitrogen. j also contains 33.8% carbon and 1.41% hydrogen by mass. the remainder of j is oxygen. what is the empirical formula of j?

Explanation:

Step1: Calculate mass of each element

Mass of carbon in 2.40 g of J: $m_{C}=2.40\times0.338 = 0.8112$ g
Mass of hydrogen in 2.40 g of J: $m_{H}=2.40\times0.0141 = 0.03384$ g
Mass of nitrogen is given as $m_{N}=0.473$ g
Mass of oxygen: $m_{O}=2.40-(0.8112 + 0.03384+0.473)=1.08196$ g

Step2: Calculate moles of each element

Molar - mass of $C = 12.01$ g/mol, $n_{C}=\frac{0.8112}{12.01}\approx0.0675$ mol
Molar - mass of $H = 1.008$ g/mol, $n_{H}=\frac{0.03384}{1.008}=0.0336$ mol
Molar - mass of $N = 14.01$ g/mol, $n_{N}=\frac{0.473}{14.01}\approx0.0338$ mol
Molar - mass of $O = 16.00$ g/mol, $n_{O}=\frac{1.08196}{16.00}\approx0.0676$ mol

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (0.0336 mol)
For C: $\frac{0.0675}{0.0336}\approx2$
For H: $\frac{0.0336}{0.0336}=1$
For N: $\frac{0.0338}{0.0336}\approx1$
For O: $\frac{0.0676}{0.0336}\approx2$

Answer:

B. $C_{2}HNO_{2}$