Chemistry
posted by Patrick on .
The molecular formula of a compound containg only carbon and hydrogen is to be determined. When a sample of the compound is burned on oxygen gas, 7.2 grams of water and 7.2 liters of carbon dioxide gas are produced(measured at STP). When 0.600grams of the compound are dissolved in 100.0 grams of CHCl3 the mixture frezes at
64.0degrees C. The normal freezing point of CHCl3 is 63.5 degrees C and its molal freezing point depression constant is 4.68C/m.
1) What is the empirical formula of the compound?
2) Calculate the molecular weight of the compound.
3) Determine the molecular formula of the compound.
4) Wtite a balance equation for the combustion of the compound and calculate the mass of oxygen gas that would be required for the combustion described above.

Convert grams H2O and L CO2 to moles H and moles C.
moles H2O = grams/molar mass = 7.2/18.015 = 0.40 and moles H = 2x that = 0.80
moles CO2 = 7.2L/22.4 = 0.32 and moles C = 0.32
Take the ratio of the two to each other. The easy way to do that is to divide the smaller number by itself, then divide the other number by the same small number.
That should give you 1 for C and 2.5 for H. Normally you would round to whole numbers but we can't throw away as much as 1/2 so we note that 1:2.5 is the same as 2.0 to 5.0 in whole numbers. The empirical formula is C2H5.
2). delta T = Kf*molality
You know delta T and Kf, solve for molality.
molality = moles/kg solvent
You know m and kg solvent, solve for moles.
moles = grams/molar mass
You know moles and grams, solve for molar mass. I get approximately 56 but you need to confirm that.
3) empirical formula mass of C2H5 is about 29. Confirm that.
molar mass is about 56 from the freezing point data. The question actually is, how many of the C2H5 units will fit into the 56.
56/29 = 1.93; round that to the nearest whole number or 2.0 so the molecular formula is (C2H5)2 which can be rewritten as C4H10.
I'll leave 4 for you to do. This is a simple stoichiometry problem. If you need help with that here is a solved stoichiometry example. Just follow the steps. Post your work if you get stuck.
http://www.jiskha.com/science/chemistry/stoichiometry.html