Chemisty
posted by LT .
Ok this is what i got for the grams part now....
Total mass is 20.14 + 51.64 + 11.34 + 15.06 = 98.18
%H2O = (20.14/98.18)*100 = 20.51%
%C2O4 = (51.64/98.18)*100 = 52.60%
% Fe 3+ = (11.34/98.15) * 100= 11.55%
% K + = (15.06/98.15) * 100 = 15.43 %
20.51 g H2O, 52.60 g C2O4 11.55 g Fe 3+, 15.43 g K+
you said redo the mols and percent... can you give me an example of how please

Take a 100 g sample; therefore, we will have the following:
20.51 g H2O.
52.60 g C2O4.
11.55 g Fe^+3.Note: you typed 98.15 but apparently divided by 98.18).
15.43 g K^+. Note: you typed 98.15, it should be 98.15which gives the same numberBUT, I think you transposed the numbers. It should be 15.34. The reason I spotted this is that the percentages didn't add to 100%. I ALWAYS check to make sure they do before I go to the next part of the problem.
Convert these to mols.
mols H2O = 20.51/molar mass = 20.51/18.015 = ??
Find mols of the other entities, then find the ratio of the mols of each to each other. The easy way to do this is to divide the smallest number by itself, then divide all the other numbers by the same small number, to keep the equality. If the answers are close to whole numbers, round them to whole numbers and uou have the formula of the compound. I think you will get some numbers close to 1/2 and if so you can't throw those away by rounding. In that case, just multiply everything through by 2 to get the whole number.
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Ok this is what i got for the grams part now.... Total mass is 20.14 + 51.64 + 11.34 + 15.06 = 98.18 %H2O = (20.14/98.18)*100 = 20.51% %C2O4 = (51.64/98.18)*100 = 52.60% % Fe 3+ = (11.34/98.15) * 100= 11.55% % K + = (15.06/98.15) * … 
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