Question

An unknown compound contains only C, H and O. Combustion of 5.60g of this compound produced 11.18g of CO2 and 4.58g of H2O. Find the Empirical formula.

Could you please show me, step by step on how to do it? and if possible each step on a new line, I get very confused when they're all lumped together :( Thanks!

Answer 1

ok so I'm guessing this is the way to do that:
11.18gCO2 x 12g/44g = 3.04gC
So 3.04C x I mol/12g = .254
Is that right so far? And I have the answer already in case you want to make sure you're getting it right it's C2H4O

Answer 2

I had to cheat, as I was a little thrown off by the inclustion of Oxygen, but this is what i found:
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Obtaining Empirical and Molecular Formulas from Combustion Data

Empirical and molecular formulas for compounds that contain only carbon and hydrogen (CaHb) or carbon, hydrogen, and oxygen (CaHbOc) can be determined with a process called combustion analysis.
[procedural details omitted]
Assume that all the carbon in the compound has been converted to CO2. Calculate the mass of carbon in the compound from the mass of carbon in the measured mass of CO2 formed.

Assume that all of the hydrogen in the compound has been converted to H2O. Calculate the mass of hydrogen in the compound from the mass of hydrogen in the measured mass of water.

If the compound contains oxygen as well as carbon and hydrogen, calculate the mass of the oxygen by subtracting the mass of carbon and hydrogen from the total mass of the original sample of compound.
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Source: http://preparatorychemistry.com/bishop_combustion_analysis.htm

Answer 3

Ok, great! Let's see if I can get the rest right, and yeah I get the subtracting in that case I'd be subtracting 3.04 and whatever I get for H from 5.60 to get O, right?

Answer 4

Wow... no edit button huh. You have to take the moles of CO2: C=11.18/(12.011+2x15.9994) and the moles of H2O : H=[4.58/(1.00794x2+15.9994)]/2, then subract: C-H=O (in moles already. Convenient.). And there you have it. CaHbOc

Answer 5

Wow... no edit button huh. You have to take the moles->mass of CO2: C=[11.18/(12.011+2x15.9994)]*12.011 and the moles->mass of H2O : H=[4.58/(1.00794x2+15.9994)]/2*1.00794, then subract: C-H=O (in grams.). O/15.9994=moles of O and there you have it. CaHbOc

Sorry about the confusion!! >.<

Answer 6

Oh my gosh ok I was trying to follow the first one and I was like what the... ok let me read your second one haha

Answer 7

Just keep in mind you will have to subtract the mass and recalculate C in moles because it won't be the same mass anymore.

The empirical formula will be a whole number ratio*, so you want to take the smallest molar component and divide the the other molar elements by that number to get that number. It may not be exact, so you can round.

*In reality, there are likely other products such as CO, but we will consider them negligible. But they have screwed up your empirical formula by introducing non-integer values of molar composition.
Combustion is actually a Very complex reaction,despite its introduction very early on in chemistry. We don't study it's mechanisms or look too deeply at it's balanced formula until much later. WE're just interested in stochiometry.

Answer 8

Thanks! :)

Answer 9

Lol apparently that's what i'm here for XD No one answered My question *sniff*

Answer 10

lol seems like you're smart enough to do fine on your own!