Posted by college chem on Thursday, September 19, 2013 at 2:11pm.
Complete combustion of 4.80 g of a hydrocarbon produced 15.6 g of CO2 and 4.80 g of H2O. What is the empirical formula for the hydrocarbon?
- chemistry - DrBob222, Thursday, September 19, 2013 at 2:20pm
Convert to mols C and mols H.
15.6 x (12/44) = ?
4.80 x (2*1/18) = ?
Find the ratio of the two. The easy way to do that is to divide the smaller number by itself, then divide the other number by the same small number. I get 1:1.5 which in small whole numbers is 2:3
- chemistry - college chem, Thursday, September 19, 2013 at 3:52pm
How do you get a ratio?
- chemistry - DrBob222, Thursday, September 19, 2013 at 4:47pm
I slipped up by converting to C first. It can be done that way but it's longer. Let me start over.
Convert 15.6g CO2 to mols. mols = g/molar mass
15.6/44 = 0.354 mols CO2 = 0.354 mols C since there is 1 atom C in 1 molecule CO2.
4.80g H2O to mols.
4.80/18 = 0.2667 mols H2O and mols H will be twice that or 0.2667*2 = 0.533 mols H.
To find the ratio do what I suggested you do. Divide the smaller number by itself (that gives you a 1.00 every time), then divide the other number by the same smaller number.
We have 0.354 mols C
and 0.533 mols H.
0.354/0.354 = 1.00 C
0.533/0.354 = 1.51 H which rounds to 1.5 and that is a ratio of 2C for 3H or the empirical formula is C2H3.
However, let me point out that 4.80g C2H3 will give you 15.6 g CO2 but it will NOT give you 4.80g H2O so I think someone goofed when they constructed the problem. But the above procedure is how you work the problem.
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