Question

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?

Thank you very much.

Answer 1

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.

Answer 2

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:
C=.94118 gC
H=.1177 gH
O=2.00-3.451-1.059 = .392

if i divide everything by .1177, i get

c=7.996
H=1.00
O=3.3305

Now, I am stuck on how to get the ratios and final answer

Answer 3

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.

Answer 4

Okay, so my O=.82348

then I divide everything by .1177, and i get

c=7.996
H=1.00
O=6.99

This looks like a ratio of 8:1:7

C8H107??

Answer 5

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.
Then
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.

Answer 6

Thank you - you are an awesome teacher!!!!!