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October 20, 2014

October 20, 2014

Posted by **Jake** on Tuesday, February 5, 2008 at 8:35pm.

Given this information:

CxHyOz (s) + O2 (g) ----> CO2 + H20

given:

2.165 g of the unknown hydrocarbon

v = 1.868 L

P = 130.2 kPa

T = 65.0 degrees C

g of H2O = 1.818

1.18g/18.02 g/mol = .101 mol H20

.101/1.01 g/mol H2 = .1 X 2 = .2 g H

PV=nRT

n= PV/RT

(1.28 atm)(1.868 L)/(.0821 L atm/mol K)(338 K)

n= .0863 mols CO2

.0862 mols X 12.01 g/mol = 1.035g C

2.165 g -(.2 g + 1.035) = .93 g O

.2/ 1.01 = .198 mol

.93/16.00 = .581 mol

1.035/12.01 = .0862 mol

my mol ratios don't work out correctly

- AP Chemistry -
**drbob222**, Tuesday, February 5, 2008 at 10:13pm**Look for the BOLD below**

Given this information:

CxHyOz (s) + O2 (g) ----> CO2 + H20

given:

2.165 g of the unknown hydrocarbon

v = 1.868 L

P = 130.2 kPa

T = 65.0 degrees C

g of H2O = 1.818

a)find g H in hydrocarbon

1.18g/18.02 g/mol = .101 mol H20

.101/1.01 g/mol H2 = .1 X 2 = .2 g H

b)find mols CO2 in product then find g of C

PV=nRT

n= PV/RT

(1.28 atm)(1.868 L)/(.0821 L atm/mol K)(338 K)

n= .0863 mols CO2

.0862 mols X 12.01 g/mol = 1.035g C

c) find g O

2.165 g -(.2 g + 1.035) = .93 g O

d) find empirical formula of hydrocarbon

.2/ 1.01 = .198 mol

.93/16.00 = .581 mol**Here is your problem. This should be 0.0581. I didn't go through to see if this would clear up the problem but everything above is ok. good work and thanks for showing your work. It makes it easier to fine the error when you do it.**

1.035/12.01 = .0862 mol

my mol ratios don't work out correctly

- AP Chemistry -
**drbob222**, Tuesday, February 5, 2008 at 10:19pmI worked it through with the new numbers and I think the formula is C3H7O2 but check my work. The ratios seem to work ok with that one change. By the way, I would make a habit of placing the 0 in front of the decimal. It makes it unmistakable that .2 is actually 0.2 ir that 0.199 is correct. The decimal sometimes can get lost in .199.

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