%H2O = (g H2O/total mass)*100 =??
%C2O4 = (g C2O4/total mass)*100 = >>
etc.
Then take a 100 g sample. In a 100 g sample you will have that many g of H2O, C2O4, etc.
Change all grams (from the 100 g sample) to mols.
Find the ratio of moles to each other to obtain the emppirical formula. If you have trouble, post your work and explain what you don't understand.
How do I find the percent composition of
H2O: 20.14grams
C2O4: 51.64grams
Fe 3+: 11.34grams
K^+: 15.06 grams
and the empirical formula with these numbers
3 answers
ok well here is what i have gotten so far:
H2O: 1.118 mol
C2O4: .58681 mol
Fe 3+: .2025 mol
K^+: .3862 mol
That is what ive gotten doing 20.14 g(1 mol / 18 g) = 1.118 mol
and so on with the rest of them
Did i do that wrong
H2O: 1.118 mol
C2O4: .58681 mol
Fe 3+: .2025 mol
K^+: .3862 mol
That is what ive gotten doing 20.14 g(1 mol / 18 g) = 1.118 mol
and so on with the rest of them
Did i do that wrong
yes and no.
What did you find for the percents? Not 20.14%.
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%
etc.When you finish, you should have 100%. Then take a 100 g sample which will give you 20.51 g H2O, 52.60 g C2O4, etc.
You have the mols part correct if you had done the percentage correctly. Correct the percentages, then redo the mols, then determine the ratios.
What did you find for the percents? Not 20.14%.
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%
etc.When you finish, you should have 100%. Then take a 100 g sample which will give you 20.51 g H2O, 52.60 g C2O4, etc.
You have the mols part correct if you had done the percentage correctly. Correct the percentages, then redo the mols, then determine the ratios.
0 answers