[Sentinel]

Hi,

Please be patient...I'm tired and the obvious eludes me right now...

I have a molecule made of C, H and O with a mass of 4,376g

Products of combustion are 3,675g H20 and 12,82g CO2

Since it's combustion 02 was missing; I have 12,19g of O2

Hence, CxHyOz + O2 ---> H20 + CO2

So, I have 2,625 mol O2, 4,903 mol H2O and 3,433 mol of CO2.

How do I figure the empirical formula of the unknown? Steps, not the answer...

Thanks

[Mitch]

It helps to balnce the equation you posted. Always start by writing a balanced chemical equation.

[Sentinel]

It helps to balnce the equation you posted. Always start by writing a balanced chemical equation.

Sorry, CH20 + O2 ----> H20 + CO2

[Mitch]

You can't balance it with numbers yet, because you haven't figured out the problem. You have to balance the chemical equation you origonally posted with the x's,y's, and z's you used.

[Sentinel]

CxH2yOz + xO2 ---> yH2O + xCO2 ?

[Sentinel]

Do I ignore O and just find the weight of C and H in the products?

[Mitch]

I don't think you balanced it correctly.

[Sentinel]

Finally got it. My mol numbers were not only wrong (tired) but useless (so was the amount of O2). Sadly, I missed my class and this wasn't in my textbook...

Step 1: Find mass of C and H in products. Substract those to find O in unknown compound (since I had it's total weight).

Step 2: Convert mass of C, H and O to moles (correctly this time...).

Setp 3: Devide to ratio.

I got O=1 C=10 and H=14

So the empirical formula was C10H14O which is the answer I was looking for.

Thanks