Question

A 2.832 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 6.439 grams of CO2 and 2.636 grams of H2O are produced. In a separate experiment, the molar mass is found to be 116.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H, O empirical formula = molecular formula =

Answer 1

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$  and $C_6H_{12}O_2$ respectively.

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 $CO_2=6.439g$

Mass of $H_2O=2.636g$

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

So, in 6.439 g of carbon dioxide, $\frac{12}{44}\times 6.439=1.756g$ of carbon will be contained.

• For calculating the mass of hydrogen:

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

So, in 2.636 g of water, $\frac{2}{18}\times 2.636=0.293g$ of hydrogen will be contained.

• Mass of oxygen in the compound = (2.832) - (1.756 + 0.293) = 0.783 g

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

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.756g}{12g/mole}=0.146moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.293g}{1g/mole}=0.293moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.783g}{16g/mole}=0.049moles$

• 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 0.049 moles.

For Carbon = $\frac{0.146}{0.049}=2.97\approx 3$

For Hydrogen  = $\frac{0.293}{0.049}=5.97\approx 6$

For Oxygen  = $\frac{0.049}{0.049}=1$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 116.2 g/mol

Mass of empirical formula = 58 g/mol

Putting values in above equation, we get:

$n=\frac{116.2g/mol}{58g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(6\times 2)}O_{(1\times 2)}=C_6H_{12}O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_6O$ and $C_6H_{12}O_2$ respectively.