A compound with a molecular weight of 110.1 g

[priscilla89]

Homework Statement

A compound with a molecular weight of 110.1 g that contained only C, H, and O was analyzed by combustion. the combustion of a 5.19 g sample of the compound resulted in the formation of 12.4 g of carbon dioxide and 2.55 g of water. What is the molecular formulas of this compound?

2. The attempt at a solution

CO2

12.44
------ = .2818
44

But I'm confused on how to do this for H20
Is it this below?

2.55
-----
18

= .1416

 

[Point Conception]

Thats correct for the moles of water, and for the hydrocarbon, 5.19g/110.1g/m = .0471 moles.
So now you can get the molar ratios for the reaction . And from there the formula for the hydrocarbon.

 

[Borek]

Moles of water is not moles of hydrogen. They are related, but not identical.

 

[Point Conception]

Were not looking for moles of hydrogen, we want to know how many hydrogens on the
reactant. So .1416 moles water/.0471 moles hydrocarbon = 3/1
So with three waters in the product we have 6 hydrogens on the hydrocarbon.

 

[Borek]

Sorry, my lack of clarity. I was not referring to the moles of molecular hydrogen, just the fact moles of water have to be converted to moles of atomic hydrogen. That means moles of water times two, that's what you did. We were talking the same all the time, obviously neither of us was entirely clear.

 

[priscilla89]

Were not looking for moles of hydrogen, we want to know how many hydrogens on the
reactant. So .1416 moles water/.0471 moles hydrocarbon = 3/1
So with three waters in the product we have 6 hydrogens on the hydrocarbon.

Right, now I understand it. You would've to do the same for CO2::

CO2

.2818 / .0471 = 6

C6 H6 O2

 

[priscilla89]

Thanks for the help. I really do appreciate this

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