Question

A 1.000 gram organic sample containing nitrogen is subjected to Kjeldahl analysis. The ammonia gas generated when ammonium is neutralized is distilled into 50.00 mL of 0.1000 M HCl to reform ammonium. If 17.29 mL of 0.1000 NaOH is required to titrate the unreacted HCl, calculate the percent by mass nitrogen in the original sample.

Answer 1

To determine percent mass of nitrogen in the original sample, we have to calculate moles of nitrogen = moles of ammonia.

moles of ammonia = moles of acid - moles of base ..............(1)

So in this problem moles of acid i.e. is moles of HCl and moles of base i.e. moles of NaOH

Hence, first we will calculate moles of HCl and NaOH from given data.

Moles of NaOH = molarity of NaOH volume added from burette

Moles of NaOH = 0.1000 17.29 ml

Moles of NaOH = 0.1000 0.1729 L

Moles of NaOH = 0.001792 moles .............(2)

Similarly,

Moles of HCl = molarity of HCl volume of HCl

Moles fo HCl = 0.1000 50 ml

Moles of HCL = 0.1000 0.05 L

Moles of HCl = 0.005 moles ...........(3)

Now, Substitute values of 2 and 3 in equation 1

moles of ammonia = moles of acid - moles of base = 0.005 - 0.001792

Moles of ammonia = 0.003208 moles .............(4)

As we know, moles of ammonia = Moles of Nitrogen

Moles of Nitrogen = 0.003208moles (from 4)

From 4 we can calculate Mass of Nitrogen in original sample

Mass of nitrogen = moles of nitrogen Atomic mass of nitrogen

Mass if nitrogen = 0.003208 14. 0067

Mass of nitrogen = 0.04493 gram ........ (5)

percent mass of nitrogen =  

percent mass of nitrogen = 0.04493 100

percent mass of nitrogen = 4.493 %