# Help with two problems with empirical formula

1. Feb 27, 2004

### 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.

2. Feb 27, 2004

### thunderfvck

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

3. Feb 28, 2004

### himanshu121

HINT: Use Mole Concept

4. Feb 28, 2004

### 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' doin the first one right.

5. Feb 28, 2004

### 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.

Last edited: Feb 28, 2004
6. Feb 29, 2004

### Pepsi24chevy

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

7. Mar 16, 2004

### 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.

8. Mar 17, 2004

### 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.

9. Mar 17, 2004

### Bystander

Good, mass of carbon.
Waste of time.
Good, mass of hydrogen
WOT.

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