2) What is the empirical formula of a compound containing C, H, O if combustion of 1.23g of the compound yields 1.8g CO2 and .74g of H2O
3) What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10g of the compound yields 6.59g CO2 and 2.7g H20 and its molar mass is about 84g/mol
I really do not understand the steps to do this. I understand the standard way going from grams to moles then I do not know what to do with the values I calculate.
OK. Step by step #2 and I'll assume you know how to get from grams to mols.
6.59g CO2 = 0.150 mols CO2 = 0.150 mols C.
2.7 g H2O = 0.150 mols H2O = 0.300 mols H(atoms)
So now we have
0.150 g C
0.300 g H
We want to find the ratio, in small whole numbers, with no number being smaller than 1.00. The easy way to do that is to divide the smaller number by itself.
C = 0.150/0.150 = 1.00. That way assures us of getting 1.00 for the smaller number.
Then divide all of the other numbers (just one other one in this problem but many will have more than one).
0.300/0.150 = 2.00
So the ratio is C1.00H2.00 but we don't right the 1.00 and we would show CH2 for the empirical formula.
#1 you must convert mols CO2 and mols H2O to grams C, grams H and find O by total mass -mass C - mass H = mass O.
Then launch into the above to find the ratio. If you get stuck show your work and explain what you don't understand about the next step.posted by DrBob222