Convert g CO2 to g C and g H2O to g H. Add grams C and grams H and subtract from total mass to find grams oxygen.
Convert g C, g H, g O to moles.
moles = grams/molar mass
Then find the ratio of the elements to each other. I worked the Na2S2O3 problem for you earlier; this one is done the same way to find the empirical formula. Post your work if you get stuck.
Thank you Dr. Bob222
An unknown compound contains carbon, hydrogen, and oxygen. When burned, a 2.000 gram sample of this compound releases 3.451 g of carbon dioxide and 1.059 grams of water. What is the empirical formula of this compound?
Here is the problem I worked out:
O=2.00-3.451-1.059 = .392
if i divide everything by .1177, i get
Now, I am stuck on how to get the ratios and final answer
First, you made an error before you get to that part.
grams O = 2.00-grams C - grams H. You subtracted grams CO2 and grams H2O. Your grams C (and moles) is ok as is the grams H (and moles H) but the oxygen must be redone.
Okay, so my O=.82348
then I divide everything by .1177, and i get
This looks like a ratio of 8:1:7
No, it looks like 8:1:7 to me but I don't think that is right.
I have C = 0.9421g (close enough to your answer) and H = 0.1176g (close enough to your answer) so O is
2.000 - 0.9421 - 0.1176 = 0.9403 g O (the slight differences in our C and H won't account for this difference in your value and mine. I've checked mine; perhaps you should check yours.
0.9421/12 = 0.0785 moles C
0.1176/1 = 0.1176 moles H
0.9403/16 = 0.05877 moles O
0.05877/0.05877 = 1 for O
0.1176/0.05877 = 2 for H
0.0785/0.05877 = 1.33 for C and these aren't whole numbers. You can quickly see that multiplying all by 2 won't work because 1.33 becomes 2.66; however, multiplying by 3 should get it.
O = 1*3 = 3
H = 2*3 = 6
C = 1.33*3 = 3.99 which rounds to 4.00
I would write C4H6O3 for the empirical formula.
Check my work.
Thank you - you are an awesome teacher!!!!!