[rycharles]

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 0.106 3 M HCl to reach a bromocresol green end point. Find the weight percent of NH3 (FM 17.031) in the cleaner.

This is my attempt. 

14.22mL HCl*(L/10³mL)(0.1063mol/L) = 0.0015116mol HCl

0.0015116mol HCl(1mol NH3/1mol HCl) = 0.0015116 mol NH3

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

any hints on how to continue?

[Borek]

So far so good. Now it is just about dilution.

What was the total mas of the solution prepared using window cleaner and water? What fraction of this solution was titrated? If this fraction contained 0.02574g of ammonia, how much ammonia was present in the whole solution?

[rycharles]

The percent weight of NH3 is

(0.025744g / 4.373g)(100%) = 0.58870

I need a dilution factor,

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

so,

(0.58870g * 4.8575) = 2.860%

is this correct?  how does the dilution factor work in this case?

[Borek]

Looks OK to me.

You titrated 4.373g of 10.231g+39.466g - or 4.373/(10.231+39.466) of total ammonia. So there was (10.231+39.466)/4.373*0.025744g of ammonia in 10.231g of the window cleaner.

[rycharles]

That makes sense.  Thanks again!