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`

Step-by-step 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,
`(12)/(44)* 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,
`(2)/(18)* 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}}=(1.507g)/(12g/mole)=0.126moles`

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

Moles of Oxygen =
`\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(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 =
`(0.126)/(0.0133)=9.5`

For Hydrogen =
`(0.172)/(0.0133)=12.9\approx 13`

For Oxygen =
`(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=(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`