Could you please review my work and see what i'm doing incorrectly? Thank you
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
1.035/12.01 = .0862 mol
my mol ratios don't work out correctly
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
I 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.
To review your work and identify any errors, let's go through each step one by one.
a) Finding the mass of hydrogen (H) in the hydrocarbon:
First, you correctly determined the number of moles of water (H2O) formed from the given mass of water (1.818g).
However, the calculation for the mass of hydrogen (H) in the hydrocarbon is incorrect. It should be:
0.101 mol H2O x (2 mol H / 1 mol H2O) x (1.008 g / mol H) = 0.203 g H
So, the correct value for the mass of hydrogen in the hydrocarbon is 0.203 g.
b) Finding the moles of carbon dioxide (CO2) and the mass of carbon (C):
You correctly used the ideal gas law equation, PV = nRT, to find the number of moles of CO2. However, the values used in the calculation are incorrect.
Using the given values:
P = 130.2 kPa
V = 1.868 L
T = 65.0 degrees C (which needs to be converted to Kelvin, so T = 65.0 + 273.15 = 338.15 K)
n = (130.2 kPa x 1.868 L) / (0.0821 L atm / mol K x 338.15 K)
n ≈ 11.207 mol CO2
To find the mass of carbon (C), you need to multiply the number of moles of CO2 by the molar mass of carbon:
11.207 mol CO2 x (12.01 g / mol) = 134.6 g C
Therefore, the correct mass of carbon in the hydrocarbon is approximately 134.6 g.
c) Finding the mass of oxygen (O):
To find the mass of oxygen in the hydrocarbon, subtract the masses of hydrogen (0.203 g) and carbon (134.6 g) from the total mass of the hydrocarbon (2.165 g):
2.165 g - (0.203 g + 134.6 g) = 1.827 g O
Therefore, the correct mass of oxygen in the hydrocarbon is 1.827 g.
d) Finding the empirical formula of the hydrocarbon:
To determine the empirical formula, you need to find the mole ratios of each element (H, C, and O). However, the calculated values for the moles of H, C, and O are incorrect.
The correct values are as follows:
Moles of H:
0.203 g H / (1.008 g/mol H) = 0.201 mol H
Moles of C:
134.6 g C / (12.01 g/mol C) = 11.21 mol C
Moles of O:
1.827 g O / (16.00 g/mol O) = 0.114 mol O
Now, divide each number of moles by the smallest value (0.114 mol) to get the simplest mole ratio:
H: 0.201 mol H / 0.114 mol = 1.76 ≈ 2
C: 11.21 mol C / 0.114 mol = 98.4 ≈ 98
O: 0.114 mol O / 0.114 mol = 1
Therefore, the empirical formula of the hydrocarbon is C98H2O.
It seems the incorrect calculations in steps a) and b) caused the subsequent errors in the mole ratios. By correcting the calculations, you should obtain the correct empirical formula of the hydrocarbon.