Question

A 10.231-g sample of window cleaner containing ammonia was diluted with 39.466 g of water. Then 4.373 g of solution was titrated with 14.22 mL of 0.1063 M HCl to reach a bromocresol green end point. a.) What fraction of the 10.231-g sample of window cleaner is contained in the 4.373 g that was analyzed? b) How many grams of NH3 (MW 17.031) were in the 4.373-g sample? c) Find the weight percent of NH3 in the cleaner.

Answer 1

a.

moles of Hcl used = 14.22mL HCl*(L/10³mL) *(0.1063mol/L)

= 0.0015116mol HCl

We know that moles of HCl = moles of NH3

therefore, moles of NH3 = 0.0015116mol

0.0015116mol NH3(17.031g/1mol) = 0.025744 g NH3

mass of NH3 in the sample =  0.025744 g NH3

Now, for the weight %

we know that

fraction of ammonia in 4.373g solution = (0.025744g / 4.373g) = 0.58870

also, as per the dilution we get

(10.231g + 39.466g)/10.231g = 4.8575

weight % of NH3 = (0.58870g * 4.8575) = 2.860%