In a gravimetric analysis, 0.1595g of a hydrocarbon upon combustion produces 0.5008g CO2 and 0.2041g H2O. Its molar mass is found in another experiment to be 70.13g/mol. Determine the empirical formula and it's molecular formula.
In a gravimetric analysis, 0.1595g of a hydrocarbon upon combustion produces 0.5008g CO2 and 0.2041g H2O. Its molar mass is found in another experiment to be 70.13g/mol.
am = atomic mass
mm = molar mass
CxHy ==> CO2 + H2O (not balanced)
0.5008 g CO2 x (am C/mm CO2) = approx 0.1367 g C.
0.2041 g H2O x (2*am H/mm H2O) = approx 0.02268 g H
mols C = 0.1367/12 = about 0.0114
mols H = 0.02268/1 = about 0.0223
Find the ratio by dividing both by the smaller number and rounding to a whole number.
C = 0.0114/0.0114 = 1
H = 0.0223/0.0114 = 1.96 = 2.00 rounded.
empirical formula is CH2. empirical mass is about 12 for C + 2 for H or 14.
molar mass = 70.13 = empirical mass x some whole number.
70.13 = 14 x
x = about 70/14 = 5 so the molecular formula is
(CH2)5 or C5H10.