A compound contains only $C$ $C$ , $H$ $H$ , and $N$ $N$ . Combustion of $35.0$ $35.0$ $m g$ $mg$ of the compound produces $33.5$ $33.5$ $m g$ $mg$ $C O_{2}$ $CO_2$ and $41.1$ $41.1$ $m g$ $mg$ $H_{2} O$ $H_2O$ . What is the empirical formula of the compound?

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A compound contains only $C$ $C$ , $H$ $H$ , and $N$ $N$ . Combustion of $35.0$ $35.0$ $m g$ $mg$ of the compound produces $33.5$ $33.5$ $m g$ $mg$ $C O_{2}$ $CO_2$ and $41.1$ $41.1$ $m g$ $mg$ $H_{2} O$ $H_2O$ . What is the empirical formula of the compound?

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Answer 1 · 2016-02-18T02:49:19

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

$C_{x} H_{y} N_{z} + \frac{x + \frac{y}{2} + z}{2} O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O + z N O_{2}$ $C_xH_yN_z + (x+y/2+z)/2O_2 to xCO_2+y/2H_2O+zNO_2$

That might look complicated, but all we're saying is that each mole of $C$ $C$ in the reactant will create $1$ $1$ mole of $C O_{2}$ $CO_2$ as a product, and require $\frac{1}{2}$ $1/2$ mole of $O_{2}$ $O_2$ to do so.

The same kind of reasoning applies for $H$ $H$ to $H_{2} O$ $H_2O$ and $N$ $N$ to $N O_{2}$ $NO_2$ .

(There are several oxides of nitrogen such as $N_{2} O$ $N_2O$ , but this one corresponds to complete combustion.)

We know the molar mass of $C O_{2}$ $CO_2$ is $44$ $44$ $g m o l^{- 1}$ $gmol^-1$ and the molar mass of $H_{2} O$ $H_2O$ is $18$ $18$ $g m o l^{- 1}$ $gmol^-1$ .

Find the number of moles of $C O_{2}$ $CO_2$ in the products:

$n = \frac{m}{M} = \frac{33.5 \times 10^{- 3}}{44} = 7.6 \times 10^{- 4}$ $n=m/M=(33.5xx10^-3)/44 = 7.6xx10^-4$ $m o l$ $mol$

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

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

Find the number of moles of $C O_{2}$ $CO_2$ in the products:

$n = \frac{m}{M} = \frac{41.1 \times 10^{- 3}}{18} = 2.3 \times 10^{- 3}$ $n=m/M=(41.1xx10^-3)/18 = 2.3xx10^-3$ $m o l$ $mol$

Now each mole of $H$ $H$ in the reactant produces only $\frac{1}{2}$ $1/2$ mole of $H_{2} O$ $H_2O$ , so we need to double this number to find $y$ $y$ , the number of moles of $H$ $H$ in the reactant: $y = 4.6 \times 10^{- 3}$ $y=4.6xx10^-3$ $m o l$ $mol$

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

For $C$ $C$ :

$m = n M = 7.6 \times 10^{- 4} \cdot 12 = 9.12 \times 10^{- 3}$ $m = nM = 7.6xx10^-4*12=9.12xx10^-3$ $g$ $g$ = $9.12$ $9.12$ $m g$ $mg$

For $H$ $H$ :

$m = n M = 4.6 \times 10^{- 3} \cdot 1 = 4.6 \times 10^{- 3}$ $m = nM = 4.6xx10^-3*1=4.6xx10^-3$ $g$ $g$ = $4.6$ $4.6$ $m g$ $mg$

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

Find the number of moles of $N$ $N$ :

$n = \frac{m}{M} = \frac{21.28 \times 10^{- 3}}{14} = 1.52 \times 10^{- 3}$ $n=m/M=(21.28xx10^-3)/14=1.52xx10^-3$ $m o l$ $mol$

Now we have:

$x = 7.6 \times 10^{- 4} = 0.76 \times 10^{- 3}$ $x= 7.6xx10^-4 =0.76xx10^-3$ $m o l$ $mol$
$y = 4.6 \times 10^{- 3}$ $y= 4.6xx10^-3$ $m o l$ $mol$
$z = {1.52}^{- 3}$ $z= 1.52^-3$ $m o l$ $mol$

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

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

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

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

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

$C H_{6} N_{2}$ $CH_6N_2$

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

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A compound contains only $C$ $C$ , $H$ $H$ , and $N$ $N$ . Combustion of $35.0$ $35.0$ $m g$ $mg$ of the compound produces $33.5$ $33.5$ $m g$ $mg$ $C O_{2}$ $CO_2$ and $41.1$ $41.1$ $m g$ $mg$ $H_{2} O$ $H_2O$ . What is the empirical formula of the compound?

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A compound contains only $C$ $C$ , $H$ $H$ , and $N$ $N$ . Combustion of $35.0$ $35.0$ $m g$ $mg$ of the compound produces $33.5$ $33.5$ $m g$ $mg$ $C O_{2}$ $CO_2$ and $41.1$ $41.1$ $m g$ $mg$ $H_{2} O$ $H_2O$ . What is the empirical formula of the compound?