ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: --whiteskies on February 21, 2014, 11:32:33 am
-
Please help and thank you so much!
Complete combustion of a 0.0100mol sample of a hydrocarbon gives 1.792L of CO2 at STP and 1.261g of H20.
a) What is the MF of the hydrocarbon?
b) What is the EF of the hydrocarbon?
I know the theory behind this question and how to work out the empirical formula in general, but I'm getting a weird mole ratio that doesn't seem right.
-
strange.. I'm getting C2H63 or something like that.
perhaps i'm completely wrong. lol
EDIT: completely screwed up
let me try again.
For CO2
PV=nRT V=1.792L R=8.31 At STP P= 101.3 kPa T= 273K
101.3 x 1.792 = n x 8.31 x 273
n(CO2)= 0.08001 mols
n(CO2) = n(C) from hydrocarbon
n(C) = 0.08001 mols
For H2O
1.261g of H2O
n=m/M
n= 1.261/18
n(H2O) = 0.07005 mols
n(H2O) x 2 = n(H)
n(H)= 0.1401111 mols
0.1401111/0.080017279 = 1.75
For every C there will be 1.75 H x 4 to get lowest whole number mol ratio
C4H7
-
Is it C8H14 and C4H7?
-
Is it C8H14 and C4H7?
I got the C4H7 aswell.
-
strange.. I'm getting C2H63 or something like that.
perhaps i'm completely wrong. lol
EDIT: completely screwed up
let me try again.
For CO2
PV=nRT V=1.792L R=8.31 At STP P= 101.3 kPa T= 273K
101.3 x 1.792 = n x 8.31 x 273
n(CO2)= 0.08001 mols
n(CO2) = n(C) from hydrocarbon
n(C) = 0.08001 mols
For H2O
1.261g of H2O
n=m/M
n= 1.261/18
n(H2O) = 0.07005 mols
n(H2O) x 2 = n(H)
n(H)= 0.1401111 mols
0.1401111/0.080017279 = 1.75
For every C there will be 1.75 H x 4 to get lowest whole number mol ratio
C4H7
Is it C8H14 and C4H7?
I got the C4H7 aswell.
yes guys yes it is and thank you so so much :)