Question

A compound contains only nitrogen and oxygen and is 30.5% nitrogen by mass. A gaseous sample of the compound has a density of 3.61 g/L at 924 torr and 105°C. What is the molecular formula of the compound?

Answer 1

Answer : The molecular formula of the compound will be, $NO_4$

Explanation:

First we have to calculate the mass of nitrogen and oxygen.

As we are given that 30.5 % nitrogen by mass that means 30.5 g of nitrogen present in 100 g of solution.

Mass of nitrogen = 30.5 g

Mass of oxygen = Mass of solution - Mass of nitrogen

Mass of oxygen = 100 g - 30.5 g

Mass of oxygen = 69.5 g

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

Step 1: Converting the given masses into moles.

Moles of N =$\frac{\text{Given mass of N}}{\text{Molar mass of N}}=\frac{30.5g}{28g/mole}=1.09moles$

Moles of O = $\frac{\text{Given mass of O}}{\text{Molar mass of O}}=\frac{69.5g}{16g/mole}=4.34moles$

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 1.09 moles.

For N = $\frac{1.09}{1.09}=1$

For O = $\frac{4.34}{1.09}=3.98\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of N : O = 1 : 4

Hence, the empirical formula for the given compound is $NO_4$

The empirical mass of given compound is $NO_4$  = 1(28) + 4(16) = 92 g/eq

To calculate the molecular mass of compound we are using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\\times \frac{RT}{M}\\\\P=\rho \frac{RT}{M}$

where,

P = pressure of gas = 924 torr = 1.216 atm

Conversion used : (1 atm = 760 torr)

T = temperature of gas = $105^oC=273+105=378K$

R = gas constant = 0.0821 L.atm/mole.K

$\rho$ = density of gas = 3.61 g/L

M = molar mass of gas = ?

Now put all the given values in the ideal gas equation, we get:

$(1.216atm)=3.61g/L\times \frac{(0.0821L.atm/mole.K)\times (378K)}{M}$

$M=92.1g/mol$

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

The equation used to calculate the valency is :

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

$n=\frac{92.1}{92}=1$

Molecular formula = $(NO_4)_n=(NO_4)_1=NO_4$

Thus, the molecular formula of the compound will be, $NO_4$