# Question 030a2

Apr 22, 2017

We use the Ideal Gas equation to determine the molar quantity of dinitrogen.................I get approx. 12%................

#### Explanation:

$n = \frac{P V}{R T} = \frac{1 \cdot a t m \times 20.7 \times {10}^{-} 3 L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 288 \cdot K}$

$= 8.76 \times {10}^{-} 4 \cdot m o l$ with respect to $\text{dinitrogen}$.

And, this, in the initial mass of compound, there were $8.76 \times {10}^{-} 4 \cdot m o l \times 28.02 \cdot g \cdot m o {l}^{-} 1$ $= 0.0245 \cdot g$ of nitrogen BY MASS.

And thus...................$\text{% nitrogen}$.............

=(0.0245*g)/(0.20*g)xx100%=??%#