Dr. Bob Chemistry question

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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) * 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

• Dr. Bob Chemistry question - ,

• Dr. Bob Chemistry question - ,

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

• Dr. Bob Chemistry question - ,

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.

• Dr. Bob Chemistry question - ,

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

• Dr. Bob Chemistry question - ,

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.