Question

Analysis of a sample of a gaseous compound shows that it contains 85.7% C and 14.3% H by mass. At standard conditions, 112 mL of the gaseous compound weighs 0.21 g. What is the molecular formula for the compound

Answer 1

Answer: The molecular formula for the given compound is $C_3H_6$

Explanation : Given,

Percentage of C = 85.7 %

Percentage of H = 14.3 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 85.7 g

Mass of H = 14.3 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{85.7g}{12g/mole}=7.14moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{14.3g}{1g/mole}=14.3moles$

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 7.14 moles.

For Carbon = $\frac{7.14}{7.14}=1$

For Hydrogen  = $\frac{14.3}{7.14}=2.00\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 2

The empirical formula for the given compound is $C_1H_2=CH_2$

Mass of empirical formula = $CH_2$  = 1(12) + 2(1) = 14 g/eq.

Now we have to determine the molar mass of compound.

As, 112 mL of volume contains 0.21 g of compound

So, 22400 mL of volume contains $\frac{22400}{112}\times 0.21=42g$ of compound

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 = 42 g/mol

Mass of empirical formula = 14 g/mol

Putting values in above equation, we get:

$n=\frac{42}{14}=3$

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

$CH_2=(CH_2)_n=(CH_2)_3=C_3H_6$

Thus, the molecular formula for the given compound is $C_3H_6$