Question

A 4.927 g sample of a compound was analyzed for nitrogen. In the procedure, all the nitrogen present was completely converted to ammonia (NH3). 1.369 grams of ammonia were obtained. The percent, by mass, of nitrogen in the compound is therefore:

36.71 %

61.52 %

27.79 %

38.48 %

22.85 %

Answer 1

the reaction of ammonia formation is as below,

N₂ + 3 H₂ --------------> 2 NH₃

now we have 1.369 gm of ammonia having molecular weight is 17.031

So, weight / molecular weight = Number of moles of ammonia

1.369/17= 0.08038 mole for 2 mole of ammonia,

So for 1 Mole of ammonia,The number of moles are

0.08038 mole / 2 = 0.0401928 mole this is for 1 mole of ammonia molecule.

Now, for nitrogen atom ,N₂ molecular weight is 28 ( N+N=14.0067+14.0067=28.0134)

so quantity of nitrogen in the sample is = Number of moles of NH3 * Molecular weight of nitrogen

= 0.0401928 mole x 28.0134 g/mole

= 1.1258 g

Now % of nitrogen atom are = (weight of nitrogen /weight of input sample)*100

=  (1.1258 g / 4.927 g ) x 100

= 22.85%