Question

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 56.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question.

Answer 1

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =$\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =$\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H=$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =$\frac{0.422}{0.422}=1$

For H =$\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$.

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be=$4\times CH_2=C_4H_8$