A sample is to be analyzed for its chloride content by precipitating and weighing silver chloride. What weight (g) of sample would have to be taken so that the weight of precipitate is equal to 1/2 the percent chloride in the sample?

Answer 1

am = atomic mass
mm = molar mass
%Cl = [mass ppt AgCl x ( am Cl/mm AgCl)/mass sample]*100 = ?
The way I read it you want mass ppt = 1/2 the %Cl or 2*ppt = % so
just put in numbers to do that like this.
mass sample = [mass ppt AgCl x (amcl/mm AgCl)/%Cl]*100
mass sample = (12.5 x (35.453/143.32)/25]*100
Since my interpretation that you want 12.5 g ppt x 2 = 25% I put the 12.5 for mass ppt and 25 for the %Cl;i.e. just two numbers so that 2*ppt gives %Cl.
mass sample then = 1/2*(35.453/143.32)*100 = ?
In this scenario you weigh an unknown sample of 12.368 g and for the calculation of %Cl you just multiply mass ppt x 2.
Note that we are talking about major weighings here (25 g is a lot of unknown sample) so you convert to mg and weigh the ppt in mg.
If your interpretation is different than mine just plug in as I did and solve the equation to meet your thoughts. This is the way it was done in the "old days" in order to save time with calculations; however, in today's techy world a computer is hooked up to the procedure and it takes the mass of the sample, remembers it, factors in the conversion from AgCl to Cl and the mass of the ppt and rolls out the answers faster than ever before. When I was in school it took me more than 30 minutes to make a calculation. No calculator was available. We used a book of logarithm tables to do that. Talk about tedious. But it worked. Apparently I'm no worse because of it. :-)