Question

An unknown compound contains only C, H, and O. Combustion of 8.90 g of this compound produced 17.8 g of CO2 and 7.28 g of H2O.


What is the empirical formula of the unknown compound?

CHO

Answer 1

@DanJS Any idea how to do this? How do I get started?

thank you for any help.

Answer 2

Unknown + O2 -----> CO2 + H2O

Answer 3

Look at problem 1 here. It is a good example of how to do this ...
http://www.chemteam.info/Mole/CombustionAnalysis.html

Answer 4

You are asked to find the mass of C (carbon), H (hydrogen), and O(Oxygen) from the production of the compound: 17.8 g of CO2 and 7.28 g of H2O.
For carbon, to find the mass we will use the mass in grams of CO2 and multiply it by the molar mass of carbon and divide by the molar mass of the compound CO2 (carbon dioxide).
Molecular weight of carbon is 12, and CO2 is 44
\(17.8 g~CO2) ~*~\Large\frac{12~g~Carbon}{44g~CO2}\)=4.8 grams of C
do the same for H with H2O.

Answer 5

We only know that the unknown compound contains only C, H, and O and the unknown compound mass has 8.90 g . So find these elements masses first from the given mass in grams of the production of CO2 and H2O.
These will determine the molar ratio of C : H : O
We can easily find the masses of C and H. I found the mass of carbon, now find the mass of hydrogen.

Answer 6

thank you all for helping, thank you DanJS for the link it looks similar to this problem, and thank you Zale101, I will try to solve the rest.

Answer 7

No problem, good luck :)

Answer 8

ok I found the masses of Hydrogen and Oxygen

now I have
4.85g of C
0.405g of H
6.47g of O (I solved it the same way as I did with C and H)

Do I convert all of them to moles and "divide each molar amount by the lowest value" like the example provided in that link above?

Answer 9

@Zale101

Answer 10

my bad I got .165 Hydrogen ... Now how do I find oxygen?

Answer 11

i used 7.28g h20 to find hydrogen.

Answer 12

then I divide them by the lowest value, right? which is 0.404 grams of H

Answer 13

thanks

Answer 14

For Hydrogen, there is 2 mole of H in H2O, so 0.404 g H is 0.404*2=0.808 grams of Hydrogen.
By molar ratios:
\((7.28 g H2O) \Large *\frac { (1 mol H2O)}{18.0 g H2O)} *\frac{(2 mol H)}{ 1 mol H2O) }*\frac{ (1.01 g H)} {1 mol H)}\)=0.808 grams of H

Answer 15

Get the moles of each grams of C, H, and O. Member, we use the lowest mole to divide the moles of the rest of elements. And remember, a subscript determines the amount of atoms (moles) on an element.
Get those grams we calculated for C,H, and O and calculate the moles. And then divide by the lowest value.

Answer 16

also, when we calculated for oxygen, that would be wrong since we used the wrong mass of hydrogen produced in H2), we forgot about the molar ratio

Answer 17

Sorry about the confusion

Answer 18

so what would the empirical formula would be? If you know it , please let me know. I have more of these problems to do.

Answer 19

Weight of oxygen is
Mass of unknown - mass of C - mass of H
8.90 - 4.85-0.808= 3.24 g of O

Answer 20

We want to know how many moles these elements have and then find the lowest mole value of the element and divide by the other elements
Convert grams to mols by taking the given gram and divide by the molecular weight of the compound or element.

Weight of Oxygen is 16
\(3.24~ g~ of ~O∗\Large\frac{1 mol}{16.0~ g~ O}=\color{blue}{0.203~ mol ~O}\)
Weight of Hydrogen is 1.008 which is rounded to 1
\(0.808~ g ~of H∗\Large\frac{1 mol}{1.0 g H}=\color{blue}{0.808~ mol ~H}\)
Weight of Carbon is 12
\(4.85 g of C∗\Large\frac{1 mol}{12.0 g C}=\color{blue}{0.404 ~mol~ C}\)

Therefore, the lowest mole is Oxygen (O) it has 0.203 mols

\(\Large C_{\frac{0.404}{\color{red}{0.203}}}~:~H_{\frac{0.808}{\color{red}{0.203}}}~:~O_{\frac{0.203}{\color{red}{0.203}}}\)


=\( C_{1.9} ~: ~H_{3.98}~ :~ O_{1}\)

Therefore, empirical formula is \(\color{green}{C_2H_4O}\)

Answer 21

it worked, that is the right answer, thank you so much :)