Combustion of 15 L of a gaseous compound of hydrogen, carbon, nitrogen gives 20g co2, 35 g h2o, 5g nitrogen at STP. What is the molecular formula of the compound?
Supposedly it's C5N2H2. I Can guess that it has something to do with the ideal gas law and stoichometry, but I can't figure out how to go about the problem. We didn't do anything like this is class.
Supposedly it's C5N2H2. I Can guess that it has something to do with the ideal gas law and stoichometry, but I can't figure out how to go about the problem. We didn't do anything like this is class.
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From the masses of CO2, H2O and N2, calculate the moles of C, H and N released from the compound:
20 g CO2 / 44 g/mol = 0.454 mol CO2 X (1 mol C/1 mol CO2) = 0.454 mol C
35 g H2O / 18 g/mol = 1.94 mol H2O X 2 mol H/1 mol H2O) = 3.89 mol H
5 g N2 / 28 g/mol = 0.178 mol N2 X (2 mol N/1 mol N2) = 0.357 mol N
Now, divide each of those moles by the smallest one:
C = 0.454/0.357 = 1.27
H = 3.89 / 0.357 = 10.9
N 0.357/0.357 = 1
I don't like how these ratios came out. Normally, these problems give "nicer" ratios than this. Please check the masses of CO2, H2O and N2 in the problem. If they are correct, then I'm sort of at a loss.
From this point, I might try multiplying everything by 4 to get the C to be about a whole number:
C = 1.27 X 4 = 5
H = 11 X 4 = 44
N = 4
Now, if your mass of water was actually 3.5 g (instead of 35),
moles H = 3.5 / 18 = 0.194 mol H2O X 2 H/H2O = 0.388 mol H
I still don't like the numbers but that would give you something like
C5H4N4 as the empirical formula.
Once you get the empirical formula, you can calculate the mass of the sample from the moles of C, H and N released. You can then calculate the moles of gas in the original sample using 15 L / 22.4 L/mol = 0.670 moles of sample. Dividing the mass of the sample by 0.67 moles will give you the molar mass of the compound. That will tell you if the empirical and molecular formulas are the same or if the molecular formula is a multiple of the empirical formula.
20 g CO2 / 44 g/mol = 0.454 mol CO2 X (1 mol C/1 mol CO2) = 0.454 mol C
35 g H2O / 18 g/mol = 1.94 mol H2O X 2 mol H/1 mol H2O) = 3.89 mol H
5 g N2 / 28 g/mol = 0.178 mol N2 X (2 mol N/1 mol N2) = 0.357 mol N
Now, divide each of those moles by the smallest one:
C = 0.454/0.357 = 1.27
H = 3.89 / 0.357 = 10.9
N 0.357/0.357 = 1
I don't like how these ratios came out. Normally, these problems give "nicer" ratios than this. Please check the masses of CO2, H2O and N2 in the problem. If they are correct, then I'm sort of at a loss.
From this point, I might try multiplying everything by 4 to get the C to be about a whole number:
C = 1.27 X 4 = 5
H = 11 X 4 = 44
N = 4
Now, if your mass of water was actually 3.5 g (instead of 35),
moles H = 3.5 / 18 = 0.194 mol H2O X 2 H/H2O = 0.388 mol H
I still don't like the numbers but that would give you something like
C5H4N4 as the empirical formula.
Once you get the empirical formula, you can calculate the mass of the sample from the moles of C, H and N released. You can then calculate the moles of gas in the original sample using 15 L / 22.4 L/mol = 0.670 moles of sample. Dividing the mass of the sample by 0.67 moles will give you the molar mass of the compound. That will tell you if the empirical and molecular formulas are the same or if the molecular formula is a multiple of the empirical formula.