posted by Patrick .
A certain compound contains carbon, hydrogen , and oxygen. When 1.00 grams of this compound are burned in oxygen gas 1.47 grams of CO2 and 0.600 grams of water are formed. When 2.75 grams of the compound are dissolved in 10.0 grams of water the resulting solution freezes at –2.84oC.
1.Determine the empirical formula of the compound.
2.Calculate the molecular weight of the compound.
3.Determine the molecular formula of the compound.
4.Write a balanced equation for the combustion reaction described above and calculate the volume of oxygen gas required to complete the combustion. Assume the gas is measured at STP.
I would convert 1.47 g CO2 to grams C, then to percent C. Same for H2O to hydrogen. Then add the two and subtract from 100 to obtain percent oxygen.
1.47 g CO2 x (atomic mass C/molar mass CO@) = 1.47 x (12/44) = about 40 but you should do it more accurately.
For H2O, that is
0.600 x (2/18) = about 6; again, confirm that.
then 100-6-40 = 54.
Now take a 100 g sample and you will have 40 g C, 6 g H, 54 g O. Convert those to moles.
40/12 = about 3.3
6/1 = about 6
54/16 = about 3.3
You can see the ratio of these elements is 1C, 2H, 1 O for the empirical formula of CH2O.
You should redo the math and get better numbers.
2. Use the freezing point data to determine the approximate molecular weight of the compound.
delta T = Kf*m
Solve for m
m = moles/kg solvent
Solve for moles.
moles = grams/molar mass
Solve for molar mass.
I get about 170 or so but you need to do it more accurately than that.
3. You want to see how many of the empirical units are in the molecular unit.
Empirical formula mass = CH2O = 12 + 2 + 16 = 30
Molecular weight from #2 = about 170
170/30 = 5.7 or so. Round to a whole number of 6 so the formula is
(CH2O)6 or you can re-write it as C6H12O6
I will leave #4 for you. If you have a problem, post your work and tell us what you don't understand about it.
In number 2.Is the delta Tf = to -2.84?
I came up with 180 but confused with how -2.84 is delta Tf
The chloride of an unknown metal is believed to have the formula . A 1.603 sample of the compound is found to contain .