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 %

Step-by-step 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* 0.065L)=9.75* 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* 10^(-3)mol` of HCl will react with =
`(1)/(1)* 9.75* 10^(-3)mo=9.75* 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* 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* 10^(-3)mol`

Molar mass of nitrogen = 14 g/mol

Putting values in above equation, we get:

`9.75* 10^(-3)mol=\frac{\text{Mass of nitrogen}}{14g/mol}\\\\\text{Mass of nitrogen}=(9.75* 10^(-3)mol* 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}}* 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}=(0.136g)/(2.25g)* 100=6.04\%`

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