# Dr. Bob Chemistry question

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

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asked by LT
1. Answered below.

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posted by DrBob22
2. ok i did all the mols of those numbers and came out with
H2O = (20.51/18.051)=1.136
C2O4 = (52.60/88.016)=.5976
Fe 3+ = (11.55/55.845)= .2068
K + = (15.34/39.098)=.3923

Ive gotten that far... now im a little confused on how you do the rest of the question because they don't come out to whole numbers can you show me an example please

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posted by LT
3. Divide all the numbers by the smallest number, in this case that is 0.2068. Then round to the nearest whole number. If you have one that will round to 1/2, then multiply all by 2 to get rid of the 1/2. If you still don't get it, post your values for the division along with what the mols are for; i.e.,
H2O = 5.5 etc.
These numbers aren't coming out with anything I recognize; perhaps the teacher just made up a problem.

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posted by DrBob22
4. 1.136/.2068=5.5
.5976/.2068= 2.88 so 3
.2068/.2068 = 1
.3923/.2068 = 1.89 so 2

now do i do it to all four with different denominators like

1.136/1.136= 1
.5976/1.136 =
and so on or do i do it by just that one

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posted by LT
5. No, you're done, except for multiplying everything by 2 to get rid of the 1/2 in 5.5
So you have
H2O = 5.5 x 2 = 11
C2O4 = 3 x 2 = 6
Fe^+3= 1 x 2 = 2
K^+ = 4
So you have
K4Fe(C2O4)6*11H2O
I don't know what it is but something is wrong here BECAUSE Fe is NOT +3 in this formula, it is +4 and that's not possible. I don't know of ANY compound in which Fe is +4 and this one especially because it's listed in the problem as +3. The potassium ferrioxalate I know about is K3Fe(C2O4)3. Check your numbers in the orginal problemm Make sure your original post is correct.

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posted by DrBob22

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