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.

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:

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:

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

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, of HCl will react with = of ammonia

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

Moles of nitrogen in ammonia =

Moles of nitrogen =

Molar mass of nitrogen = 14 g/mol

Putting values in above equation, we get:

Mass of sample= 2.25 g

Mass of nitrogen = 0.136 g

Putting values in above equation, we get:

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

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