Question

How many grams of N are present in 3.67 grams of dinitrogen tetroxide

Answer 1

Percentage of Nitrogen by mass in $N_2 ​O_4$

​

$=\frac{2×14}{[2×14]+[4×16]}=30.43$

So, mass of Nitrogen in 3.67 g of $N_2 ​O_4$

​=[(% of Nitrogen)×(mass of sample)]÷(100}

$=\frac{30.43×3.67}{100}=1.117grams$