Help with two problems with empirical formula

[Pepsi24chevy]

Hey, i know how to do empirical formula regularly but nto with all this other stuff in these problems. COuld someone help me set these 2 problems up.

1.A compound contrains only C,H, and N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1mg H2O. What is the empirical formula of the compound?

2. A compound contrains only carbon, hydrogen, nitrogen, and oxygen. COmbustion of 0.157g of the compoudn produced 0.213 g CO2 and 0.0310g H2O. In another experiment, it is foudn that 0.013g of the compound produces 0.0230g NH3. What is the empirical formula of the compound? Combustion involves reacting with excess O2. Assume that all the carbon ends up in the CO2 and all the hydrogen ends up in H2O. Also assume that all the nitrogen ends up in the NH3 in the second experiment.

 

[thunderfvck]

Show us the work you've done, and I'll show you what I can do.

 

[himanshu121]

HINT: Use Mole Concept

 

[Pepsi24chevy]

Well i started out by doing 33.5* (1/44.0098)*12.001 = 9.14
33.5*(1/114.0098) * 31.9988 = 24.36
41.1 * (1/18.01534) * 2.01594= 4.60
41.1 * (1/18.01534) *15.9988= 36.50

Then i thought about dividing the numbers by teh smallest, but then i figured i wasn't doing it right. FOr the second one i didn't start it becasue i figured i wasnt' doing the first one right.

 

[thunderfvck]

Okay, that appears to be quite wrong. You do not multiply the moles of substance by the molar mass of carbon/oxygen. And you have 33.5 MILLIgrams, that's 0.00335 g. And in question 1 you write the compound contains C, H and N.
So. Now you find out how many moles 0.00335 g of CO2 is. Do the same for H2O. Now you can multiply it by a comon factor to get a nice number, if necessary. Goddammit these problems are bugging me so I'll just do the first one...
You find that the ratio is 3:1 H2O:CO2

x(C&H&N) + O2 --> 3H2O + CO2

This doesn't even make sense. Where the hell does the nitrogen go? It's not balanced and so it discourages my doing it.

 

[Pepsi24chevy]

I am still confused. The nitrogen is burned off maybe?

 

[jrwnov]

i don't know if i am right but

okay you start out with 35.0mg, but then on the other end of the equation, you have more that 35.0mg, doesn't that violate the mass conservation law?...oh well, just my imput.

 

[thunderfvck]

It's being combusted, ie. it's reacting with oxygen. That's where the missing mass comes from. I don't know how good you are with chemistry, and I appologize if I make you feel dumb, but conservation of mass really means the conservation of combined mass on reaction side = combined mass on products side.

 

[Bystander]

Originally posted by Pepsi24chevy
Well i started out by doing 33.5* (1/44.0098)*12.001 = 9.14

Good, mass of carbon.

33.5*(1/114.0098) * 31.9988 = 24.36

Waste of time.

41.1 * (1/18.01534) * 2.01594= 4.60

Good, mass of hydrogen

41.1 * (1/18.01534) *15.9988= 36.50

WOT.

Now, what's the mass of nitrogen in the sample? How many moles of carbon, hydrogen, and nitrgen? What are the mole ratios?