last one, i PROMISE

Question

A compound containing only C, H, O and Cr was subjected to combustion analysis. A sample of 10.88 mg produced 22.43 mg of CO2 and 4.59 mg of H2O. The chromium in the sample was precipitated as 14.09 mg of Ag2CrO4. Determine the empirical formula of the compound and match each element with its correct subscript from the drop box.

Answer 1

%CO2 = (23.43/10.88)*100.
%H2O = (4.59/10.88)*100.
For Cr, convert 14.09 mg Ag2CrO4 to g, then to mols Ag2CrO4, then to mols Cr, then to %Cr using the same format as above.
%O = 100% - %CO2 - %H2O - %Cr.
Now take a 100 g sample which will give you xx g CO2, yy g H2O, zz O and ww g Cr.
Convert g to mols C, H, O, and Cr.
Then find the ratio of each. The easy way to do this is to divide the smallest number by itself, then divide all the other mols by that number. Post your work if you need more help.

This should get you started.
Determine

Answer 2

I hope you didn't spend much time on this for I am way off base, including some typos. You need to look up the actual molar masses. I have estimated some of them. I just don't know what I was thinking. (I wasn't thinking.)

%C =[22.42 x (1 mol C/1 mol CO2)/10.88]*100 = [22.42 x (12/44)/10.88]*100 = xx
%H = [4.59 x (2 mol H/1 mol H2O/10.88]*100 = [4.59 x (2/18)/10.88] = yy
%Cr = [14.08 x (1 mol Cr/1 mol Ag2CrO4)/10.88]*100 = [14.08 x (52/331.7)/10.88]*100 = zz
%O = %C - %H - %Cr = ww
Now take a 100 g sample which will give you xx g C, yy g H, zz g Cr and ww g O
Then find the ratio of each. The easy way to do this is to divide the smallest number by itself, then divide all the other mols by that number. Post your work if you need more help.

Answer 3

After you take the 100 g sample, you want to take the grams of each and convert to mols, then find the ratio of the mols.
So,
xx/12= mols C
yy/1 = mols H
zz/52 = mols Cr
ww/16 = mols O
Now go back and read about dividing the smallest number etc.