Question

An analysis of a compound showed it contained 87.4% nitrogen and 12.6% hydrogen. the molecular weight was determined to be 32.05 g/mol. what is the molecular formula for this compound?

Answer 1

Co2 is correct buddy

Answer 2

Answer: The molecular formula for the given compound is $N_2H_4$

Explanation:

We are given:

Percentage of H = 12.6 %

Percentage of N = 87.4 %

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

Mass of H = 12.6 g

Mass of N = 87.4 g

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

• Step 1: Converting the given masses into moles.

Moles of Nitrogen =$\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{87.4g}{14g/mole}=6.24moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{12.6g}{1g/mole}=12.6moles$

• 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 6.24 moles.

For Nitrogen = $\frac{6.24}{6.24}=1$

For Hydrogen = $\frac{12.6}{6.24}=2.019\approx 2$

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

The ratio of N : H = 1 : 2

The empirical formula for the given compound is $N_1H_{2}=NH_2$

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 = 32.05 g/mol

Mass of empirical formula = 16 g/mol

Putting values in above equation, we get:

$n=\frac{32.05g/mol}{16g/mol}=2$

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

$N_{(1\times 2)}H_{(2\times 2)}=N_2H_4$

Hence, the molecular formula for the given compound is $N_2H_4$