[abhishek007]

hey guys just starting IB chemistry, can any1 help with this problem...

2.4 grams of a compound of carbon hydrogen and oxygen gave, on combustion, 3.52 g of CO2 and 1.44grams of H2). The relative molecular mass of the compound was found to be 60 grams.

a. what are the masses of carbon, hydrogen and oxygen in the 2.4g of the compound?

b. what are the empirical and molecular formulas of the compound?

for part A all i could do is: 0.08 moles of CO2 and 0.08 moles of H2O

mole ratio CO2:C 1:1

           H2O:H 1:2

mass =.08* 12g=.96g Carbon
     =.08*1g=.08g Hydrogen
     = .96g + .08g= 1.04g   2.4g-1.12g= 1.28g oxgen


plz help some chem pro

[Dan]

for part A all i could do is: 0.08 moles of CO2 and 0.08 moles of H2O
mole ratio CO2:C 1:1

           H2O:H 1:2

mass =.08* 12g=.96g Carbon
     =.08*1g=.08g Hydrogen
     

plz help some chem pro

Your mistake is taking 0.08 mol of water to contain 0.08 mol H.

[abhishek007]

oh....0.08 mol of water will have 0.16 mol of H???

[Dan]

Yes. Now work out the mass of that H and then determine the mass and moles of oxygen, just as you were before.

[abhishek007]

mass =.08* 12g=.96g Carbon
     =.16g*1g=.16g Hydrogen
     = .96g + .16g= 1.12g   2.4g-1.12g= 1.28g oxgen

[abhishek007]

1.28g/16g = .08 moles of oxygen

that gives a ratio of 1:1:1...doesn't it.....oh....i am confused

[abhishek007]

so the answers

part a.    .96g carbon                .16g hydrogen               1.28g oxygen


b. no idea..

[Dan]

1.28g/16g = .08 moles of oxygen

that gives a ratio of 1:1:1...doesn't it.....oh....i am confused

No, careful. You are a gnat's whisker from the answer! You can do this.

So far you have worked out previously that you have:

0.08 mol C
0.16 mol H
0.08 mol O

Look again, is that really a 1:1:1 ratio?

[abhishek007]

CH2O!!!!!!! is the empirical formula

and for molecular

30g (mass of CH2O)/60g= 2

C2H4O2?

YAHOO!!!

thanks so much i hope im right

[Dan]

Yep, looks good to me, nicely done.