When 6.54 mg of an organic compound containing only carbon and hydrogen was burned, 20.5 mg of carbon dioxide was formed. The vapor density of the compound at 800 torr and 25 degree Celsius is 1.215g/L. Find the formula of the compound.

Question

wolfe8 · Answer

Can you use PV=nRT? You will then get the volume which can be used to get the mass. This is the mass for water which is produced as vapor.

anonymous · Answer

I've tried using MM=dRT/P to get the molar mass of water

anonymous · Answer

I was able to find out the hydrogen part of the formula, I was stuck on the carbon part

wolfe8 · Answer

Oh what did you get?

anonymous · Answer

MM = 1.215*0.0821*298/1.05 = 28.2g H2O

anonymous · Answer

nevermind, I made an mistake

anonymous · Answer

I believe there is another approach.

wolfe8 · Answer

I don't know how to do this yet so I'm trying to work with you. I can set up:
n(carbon compound) + nO2 --> nCO2 + 2nH2O

but I'm not sure how that can help.

wolfe8 · Answer

Ok scratch that. I forgot the carbon compound has O

anonymous · Answer

Using the formula MM=dRT/P MM=1.215*0.0821*298/(800/760)=28.2g
We know the molar mass is 28.2g
the mass of the organic compound is 6.54mg
the mass of CO2 is 20.5mg

anonymous · Answer

and we know that the all the carbon in CO2 came from the organic compound

anonymous · Answer

so if we can figure out how many carbon is produced, then we know how many carbon is in the organic compund

anonymous · Answer

The answer is C2H4...if that helps

wolfe8 · Answer

I thought of the carbon part, but I guess it's too late for my brain to think of the rest. Thanks for the answer :)

anonymous · Answer

I got it. First use the percentage composite to find the mass of carbon, use the total mass to subtract the mass of carbon to get the mass of hydrogen.

anonymous · Answer

convert them back to moles, find the empirical formula. then divide the molar mass of the compound by the molar mass of the empirical formula to find the molecular formula.

wolfe8 · Answer

@.@ I am too sleepy for that. Good job!

anonymous · Answer

attachment

anonymous · Answer

Me too, I just want to think this one through before my test tomorrow.

anonymous · Answer

Thanks for staying with me to work this out!!! =D

wolfe8 · Answer

You're welcome. If that will help you then I'll stay up with you for a bit.

anonymous · Answer

I am all good! Getting a good sleep is better than studying all night~

anonymous · Answer

Goodnight~~~ =D

wolfe8 · Answer

Sleep well and good luck tomorrow!

anonymous · Answer

Thanks!!