Chemistry empirical formula please help.?

Question

A 58.90 g sample of a mystery compound containing carbon, hydrogen, and oxygen is subjected to combustion analysis. 86.35 g CO2 and 35.35 g H20 are produced in the combustion reaction. What is the empirical formula of the mystery compound?

a) CHO

b) CH2O

c) CHO2

d) CH2O3

e) C2H2O

______________________________________...

I have no idea how to do this problem...

please explain if you know...

Answer 1

Sure no problem. The basic reaction for combustion is:
O2 + CHO  = CO2 + H2O.

Now calculate the moles of CO2 and H20 you have from the masses.
You should get 1.95 moles CO2 and 3.9 moles H2O. Let's be big and say 2 moles CO2 and 4 moles H2O. I like round numbers.

Since the hydrocarbon is the only source of carbon and hydrogen, therefore
the empirical ratio must be 1:2 carbon to hydrogen.

So you have two possibilities: CH2O3 and CH2O. The oxygen amount will decide.

So far our working reaction is:

2 O2 + 2 CH2Ox = 2CO2 + 2H2O (where x is the unknown)
In the products we have 6 moles oxygen.
Take away the 4 moles of oxygen in O2, leaves us with 2 to account for.
Therefore since the coefficient is 2, then x=1,.

So your answer is CH2O.

P.S. Respectfully I think the last sender is not considering that some of the oxygen moles are found in the compound and some in pure oxygen gas. We can't say that all 1.96 moles are in the compound. (Although luckily you still get that the right answer).

Answer 2

Sure no problem. The basic reaction for combustion is:
O2 + CHO  = CO2 + H2O.

Now calculate the moles of CO2 and H20 you have from the masses.
You should get 1.95 moles CO2 and 3.9 moles H2O. Let's be big and say 2 moles CO2 and 4 moles H2O. I like round numbers.

Since the hydrocarbon is the only source of carbon and hydrogen, therefore
the empirical ratio must be 1:2 carbon to hydrogen.

So you have two possibilities: CH2O3 and CH2O. The oxygen amount will decide.

So far our working reaction is:

2 O2 + 2 CH2Ox = 2CO2 + 2H2O (where x is the unknown)
In the products we have 6 moles oxygen.
Take away the 4 moles of oxygen in O2, leaves us with 2 to account for.
Therefore since the coefficient is 2, then x=1,.

So your answer is CH2O.

P.S. Respectfully I think the last sender is not considering that some of the oxygen moles are found in the compound and some in pure oxygen gas. We can't say that all 1.96 moles are in the compound. (Although luckily you still get that the right answer).