# Combustion of a 0.9827 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of CO_2 and 1.070 g of H_2O. What is the empirical formula of the compound?

Mar 6, 2017

The empirical formula is ${\text{C"_4"H"_11"O}}_{2}$.

#### Explanation:

Here is the equation with masses:

${\text{0.9827 g C"_a"H"_b"O"_c + "(1.900 g + 1.070 g - 0.9827 g) O"_2 rarr "1.070 g H"_2"O" + "1.900 g CO}}_{2}$

Use 32.00 g/mol for the molar mass of ${\text{O}}_{2}$
Use 18.03 g/mol for the molar mass of $\text{H"_2"O}$
Use 44.01 g/mol for the molar mass of ${\text{CO}}_{2}$

We do not know the molar mass of the hydrocarbon so we allow $a , b$, and $c$ to be any positive real number.

"C"_a"H"_b"O"_c+ ("1.9873 g")/("32.00 g/mol") "O"_2 rarr ("1.070 g")/("18.03 g/mol") "H"_2"O" + ("1.900 g")/("44.01 g/mol") "CO"_2

Perform the division:

${\text{C"_a"H"_b"O"_c+ ("0.06210 mol") "O"_2 rarr ("0.05935 mol") "H"_2"O" + ("0.04317 mol") "CO}}_{2}$

Matching coefficients, we get:

$a = 0.04317$
$b = 2 \times 0.05935 = 0.1187$
$c = 0.05935 + 2 \times 0.04317 - 2 \times 0.06210 = 0.02149$

Divide every number by 0.02149:

$\frac{a}{0.02149} = 2.009$
$\frac{b}{0.02149} = 5.523$
$\frac{c}{0.02149} = 1$

Multiply every number by 2:

$2.009 \times 2 = 4.018$
$5.523 \times 2 = 11.05$
$1 \times 2 = 2$

Round off each number to the nearest integer.

4.018 ≈ color(white)(l)4
11.05 ≈ 11
$\textcolor{w h i t e}{l l} 2 \textcolor{w h i t e}{m l} = \textcolor{w h i t e}{l l} 2$

The empirical formula is ${\text{C"_4"H"_11"O}}_{2}$.

Mar 12, 2017

The empirical formula is ${\text{C"_4"H"_11"O}}_{2}$.

#### Explanation:

First, we calculate the masses of $\text{C}$ and $\text{H}$ from the masses of their oxides (${\text{CO}}_{2}$ and $\text{H"_2"O}$).

$\text{Mass of C" = 1.900 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.5185 g C}$

$\text{Mass of H" = 1.070 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1197 g H}$

$\text{Mass of C + Mass of H" = "0.5185 g + 0.1197 g" = "0.6382 g}$

This is less than the mass of the sample.

The missing mass must be caused by $\text{O}$.

$\text{Mass of O = 0.9827 g - 0.6382 g = 0.3445 g}$

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

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

$\boldsymbol{\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(mm) "Ratio"color(white)(ml)×2color(white)(mm)"Integers}}$
$\textcolor{w h i t e}{m l} \text{C} \textcolor{w h i t e}{X X X m} 0.5185 \textcolor{w h i t e}{m l l} 0.04317 \textcolor{w h i t e}{X m l l l} 2.005 \textcolor{w h i t e}{m} 4.010 \textcolor{w h i t e}{X m m m l} 4$
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{X X X m} 0.1197 \textcolor{w h i t e}{m l l} 0.1188 \textcolor{w h i t e}{m m m l} 5.518 \textcolor{w h i t e}{m} 11.04 \textcolor{w h i t e}{X m m m} 11$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{X X X m} 0.3445 \textcolor{w h i t e}{m l l} 0.02153 \textcolor{w h i t e}{m m m} 1 \textcolor{w h i t e}{m m m l} 2 \textcolor{w h i t e}{X m m m m m} 2$

The empirical formula is ${\text{C"_4"H"_11"O}}_{2}$.

Note: This is an impossible empirical formula.

The molecular formula must have an even number of $\text{H}$ atoms, e.g. ${\text{C"_8"H"_22"O}}_{4}$.

A compound with 8 $\text{C}$ atoms can contain no more than 18 $\text{H}$ atoms.