Finding Molecular formula from a given molecular mass and equation.

  • #1
SaRaH...
7
0
The question is as follows:
A compound contains only carbon, hydrogen and oxygen. When 1g of the compound is reacted completely with oxygen, 2.2 g CO2 and 1.2 g H2O are obtained. The molar mass is close to 60. What is the molecular formula?

I first wrote the equation: compound + O2 = CO2 +H20 (is this right?)

Using the equation no.moles = no. grams/molar mass I found the number of moles of each to be:
compound: ~0.0167mol
O2: no. grams/32
CO2: 0.05mol
H2O: 0.067mol

Using these I found that (if the above equation is right) that the amount of the compound used is ~ 4.012*amount of H2O produced.

I also found the molucular mass of the compound CxHyOz to be: 12x+y+16z = 60

I'm not sure if everything I tried was necessary or correct however now I seem to be going around in circles with everything I do and can't seem to be able to reach an answer.

Any help in solving this would be very much appreciated.
Thanks,
Sarah.
 
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  • #2
SaRaH... said:
I first wrote the equation: compound + O2 = CO2 +H20 (is this right?)

Yes and no. Yes - that's a correct skeletal equation describing what is going on. No - it is not balanced (it can't be as long as you don't have your compound formula) and it doesn't help much.

compound: ~0.0167mol

That's not incorrect, but at this moment it is irrelevant.

O2: no. grams/32

You don't know number of grams of oxygen (yet) so while it is correct, it is not yet useful. Can you calculate how much oxygen was in the 1g of the compound?

CO2: 0.05mol
H2O: 0.067mol

That's the most important information at this stage. Molar ratio of these compound is linked to molar ratio of carbon and hydrogen in the original compound. Can you calculate how many atoms of hydrogen were there for each atom of carbon? For two atoms of carbon? For three?

I also found the molucular mass of the compound CxHyOz to be: 12x+y+16z = 60

That's correct, but not necesarilly required.

Check your textbook for empirical formula calculation.

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