# )A compound that contains just C, H and N is burned in oxygen. When 5.000 g of the compound is burned in excess oxygen, 6.019 g of CO_2, 4.313 g of H_2O, and 9.438g of NO_2 are the only products. What is the empirical formula of the compound?

Sep 14, 2017

The empirical formula is ${\text{C"_2"H"_7"N}}_{3}$.

#### Explanation:

We must calculate the masses of $\text{C, H}$, and $\text{N}$ from the masses of carbon dioxide, water, and nitrogen dioxide.

$\text{Mass of C" = 6.019 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "1.6425 g C}$

$\text{Mass of H" = 4.313 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.482 52 g H}$

$\text{Mass of N" = 9.438 color(red)(cancel(color(black)("g NO"_2))) × "14.01 g C"/(46.01 color(red)(cancel(color(black)("g NO"_2)))) = "2.8739 g N}$

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

$\underline{\boldsymbol{\text{Element"color(white)(X) "Mass/g"color(white)(X) "Amt/mol"color(white)(m) "Ratio"color(white)(ml)×2color(white)(m)"Integers}}}$
$\textcolor{w h i t e}{m m} \text{C"color(white)(mmml)1.6425color(white)(mml)"0.136 76} \textcolor{w h i t e}{X m} 1 \textcolor{w h i t e}{m m m l l} 2 \textcolor{w h i t e}{X m m m m} 2$
$\textcolor{w h i t e}{m m} \text{H" color(white)(XXXl)0.48252 color(white)(mll)"0.478 69} \textcolor{w h i t e}{m m} 3.5001 \textcolor{w h i t e}{m l} 7.0002 \textcolor{w h i t e}{X m l} 7$
$\textcolor{w h i t e}{m m} \text{N" color(white)(XXXl)2.8739 color(white)(mml)"0.205 13} \textcolor{w h i t e}{m m} 1.4999 \textcolor{w h i t e}{m l} 2.9997 \textcolor{w h i t e}{X m l} 3$

The empirical formula is ${\text{C"_2"H"_7"N}}_{3}$.