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.

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:
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

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

c=7.996
H=1.00
O=6.99

This looks like a ratio of 8:1:7

C8H107??

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.

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

To determine the empirical formula of a compound, we need to know the masses of the elements present in the compound. From the given information, we can calculate the masses of carbon, hydrogen, and oxygen in the unknown compound.

1. Start by calculating the mass of carbon in the compound. The mass of carbon dioxide (CO2) released during combustion corresponds to the mass of carbon in the compound. So, the mass of carbon is 3.451 grams.

2. Next, calculate the mass of hydrogen in the compound. The mass of water (H2O) released during combustion corresponds to the mass of hydrogen in the compound. So, the mass of hydrogen is 1.059 grams.

3. To find the mass of oxygen in the compound, subtract the sum of the masses of carbon and hydrogen from the total mass of the compound. In this case, the total mass of the compound is 2.000 grams. Therefore, the mass of oxygen is 2.000 - (3.451 + 1.059) grams.

Now, we can use these masses to determine the empirical formula.

4. Convert the masses of carbon, hydrogen, and oxygen to moles by dividing them by their respective molar masses:
- Carbon: 3.451 g / (12.01 g/mol) = 0.287 mol
- Hydrogen: 1.059 g / (1.008 g/mol) = 1.050 mol
- Oxygen: (2.000 - (3.451 + 1.059)) g / (16.00 g/mol) = 0.152 mol

5. Divide each of the mole values by the smallest number of moles present. In this case, the smallest number of moles is 0.152 mol.

- Carbon: 0.287 mol / 0.152 mol ≈ 1.89
- Hydrogen: 1.050 mol / 0.152 mol ≈ 6.91
- Oxygen: 0.152 mol / 0.152 mol = 1

6. Round off these values to the nearest whole numbers to obtain the mole ratios.

- Carbon: 2
- Hydrogen: 7
- Oxygen: 1

7. The empirical formula of the compound is therefore C2H7O.