Question

If 65.0 mL of 0.150 M HCl ( aq ) is needed to neutralize all the NH 3 ( g ) from a 2.25 g sample of organic material, calculate the mass percentage of nitrogen in the sample.

Answer 1

Answer: The mass percentage of nitrogen in the sample is 6.04 %

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of HCl solution = 0.150 M

Volume of solution = 65.0 mL = 0.065 L    (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.150M=\frac{\text{Moles of HCl}}{0.065L}\\\\\text{Moles of HCl}=(0.150mol/L\times 0.065L)=9.75\times 10^{-3}mol$

The chemical equation for the reaction of HCl and ammonia follows:

$HCl+NH_3\rightarrow NH_4Cl$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, $9.75\times 10^{-3}mol$ of HCl will react with = $\frac{1}{1}\times 9.75\times 10^{-3}mo=9.75\times 10^{-3}mol$ of ammonia

1 mole of ammonia contains 1 mole of nitrogen and 3 moles of hydrogen element

Moles of nitrogen in ammonia = $9.75\times 10^{-3}mol$

• To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Moles of nitrogen = $9.75\times 10^{-3}mol$

Molar mass of nitrogen = 14 g/mol

Putting values in above equation, we get:

$9.75\times 10^{-3}mol=\frac{\text{Mass of nitrogen}}{14g/mol}\\\\\text{Mass of nitrogen}=(9.75\times 10^{-3}mol\times 14g/mol)=0.136g$

• To calculate the mass percentage of nitrogen in the sample, we use the equation:

$\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of sample}}\times 100$

Mass of sample= 2.25 g

Mass of nitrogen = 0.136 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen}=\frac{0.136g}{2.25g}\times 100=6.04\%$

Hence, the mass percentage of nitrogen in the sample is 6.04 %