# A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO_2 and 41.1 mg H_2O. What is the empirical formula of the compound?

Feb 18, 2016

We can write an equation for the (complete) combustion of the substance as follows:

${C}_{x} {H}_{y} {N}_{z} + \frac{x + \frac{y}{2} + z}{2} {O}_{2} \to x C {O}_{2} + \frac{y}{2} {H}_{2} O + z N {O}_{2}$

That might look complicated, but all we're saying is that each mole of $C$ in the reactant will create $1$ mole of $C {O}_{2}$ as a product, and require $\frac{1}{2}$ mole of ${O}_{2}$ to do so.

The same kind of reasoning applies for $H$ to ${H}_{2} O$ and $N$ to $N {O}_{2}$.

(There are several oxides of nitrogen such as ${N}_{2} O$, but this one corresponds to complete combustion.)

We know the molar mass of $C {O}_{2}$ is $44$ $g m o {l}^{-} 1$ and the molar mass of ${H}_{2} O$ is $18$ $g m o {l}^{-} 1$.

Find the number of moles of $C {O}_{2}$ in the products:

$n = \frac{m}{M} = \frac{33.5 \times {10}^{-} 3}{44} = 7.6 \times {10}^{-} 4$ $m o l$

(note that we had mg of product and grams for molar mass)

This is simply the value of $x$, the number of moles of $C$ in the reactant.

Find the number of moles of $C {O}_{2}$ in the products:

$n = \frac{m}{M} = \frac{41.1 \times {10}^{-} 3}{18} = 2.3 \times {10}^{-} 3$ $m o l$

Now each mole of $H$ in the reactant produces only $\frac{1}{2}$ mole of ${H}_{2} O$, so we need to double this number to find $y$, the number of moles of $H$ in the reactant: $y = 4.6 \times {10}^{-} 3$ $m o l$

We can now find the mass of $C$ and the mass of $H$ in the reactant:

For $C$:

$m = n M = 7.6 \times {10}^{-} 4 \cdot 12 = 9.12 \times {10}^{-} 3$ $g$ = $9.12$ $m g$

For $H$:

$m = n M = 4.6 \times {10}^{-} 3 \cdot 1 = 4.6 \times {10}^{-} 3$ $g$ = $4.6$ $m g$

Now, if we add these we have $13.72$ $m g$, but we know the sample was $35.0$ $m g$, so the remainder must be $N$: $21.28$ $m g$.

Find the number of moles of $N$:

$n = \frac{m}{M} = \frac{21.28 \times {10}^{-} 3}{14} = 1.52 \times {10}^{-} 3$ $m o l$

Now we have:

$x = 7.6 \times {10}^{-} 4 = 0.76 \times {10}^{-} 3$ $m o l$
$y = 4.6 \times {10}^{-} 3$ $m o l$
$z = {1.52}^{-} 3$ $m o l$

These are not neat and tidy numbers, but they look close to $\frac{3}{4} , 4 \frac{1}{2} \mathmr{and} 1 \frac{1}{2}$. To get them to nice clean whole numbers, multiplying by 4000 seems to make sense:

$x \approx 3 , y \approx 18 , z \approx 6$

Hmm, close, but we can make it simpler by dividing by $3$:

$x \approx 1 , y \approx 6 , z \approx 2$

So our empirical formula, in its simplest form, is:

$C {H}_{6} {N}_{2}$

(we don't bother including a subscript of 1)