# If 4.04 g of N combine with 11.46 g O to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound?

May 29, 2016

Dinitrogen pentoxide, ${N}_{2} {O}_{5}$.

#### Explanation:

$\text{Moles of nitrogen}$ $=$ $\frac{4.04 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.289 \cdot m o l$

$\text{Moles of oxygen}$ $=$ $\frac{11.46 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.716 \cdot m o l$

We divide thru by the lowest molar quantity to give an empirical formula of $N {O}_{2.5}$ or ${N}_{2} {O}_{5}$, because the empirical formula should be integral.

Now the molecular formula is always a multiple of the the empirical formula:

So using the quoted molecular mass:

$108 \cdot g \cdot m o {l}^{-} 1$ $=$ $n \times \left(2 \times 14.01 + 5 \times 15.999\right) \cdot g \cdot m o {l}^{-} 1$

Clearly, $n = 1$, and the molecular formula is ${N}_{2} {O}_{5}$.