Question

Androstenedione, which contains only carbon, hydrogen, and oxygen, is a steroid hormone produced in the adrenal glands and the gonads. Combustion analysis of a 1.893-g sample of androstenedione produced 5.527g of CO2 and 1.548g H2O. The molar mass of androstenedione is 286.4g/mol .
Find the molecular formula for androstenedione

Please show work and steps so that I can understand myself as well. Thank you!

Answer 1

Answer: The molecular formula for androstenedione is, $C_{19}H_{26}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=5.527g$

Mass of $H_2O=1.548g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 5.527 g of carbon dioxide, $\frac{12}{44}\times 5.527=1.507g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.548 g of water, $\frac{2}{18}\times 1.548=0.172g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.893g)-[(1.507g)+(0.172g)]=0.214g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.507g}{12g/mole}=0.126moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.172g}{1g/mole}=0.172moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.214g}{16g/mole}=0.0133moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.0133$ moles.

For Carbon = $\frac{0.126}{0.0133}=9.5$

For Hydrogen  = $\frac{0.172}{0.0133}=12.9\approx 13$

For Oxygen  = $\frac{0.0133}{0.0133}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9.5 : 13 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 19 : 26 : 2

Thus, the empirical formula for the given compound is $C_{19}H_{26}O_2$

The empirical formula weight of $C_{19}H_{26}O_2$ = 19(12) + 26(1) + 2(16) = 286 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{286.4}{286}=1$

Molecular formula = $C_{19}H_{26}O_2$

Therefore, the molecular formula for androstenedione is, $C_{19}H_{26}O_2$