: The empirical formula for the compound is C3H60 (see below)
CO2 is the only product containing C,
C produced = 145.0 mg CO2 x (1 g / 1000 mg) x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.00330 moles C.
H2O is the only product containing H,
H produced = 59.38 mg H2O x (1 g / 1000 mg) x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.00660 moles H.
Oxygen is in both and the unknown reacts with oxygen(in the air)
0.00330 moles C x (12.0 g C / 1 mole C) = 0.0396 g C = 39.6 mg C
0.00660 moles H x (1.01 g H / 1 mole H) = 0.00667 g H = 6.7 mg H
Because the unknown weighed 63.8 mg and consists off justC, H, and O, then
mass O = g unknown - g C - g H = 63.8 mg - 39.6 mg - 6.7 mg = 17.5 mg = 0.0175 g
0.0175 g O x (1 mole O / 16.0 g O) = 0.00109 moles O
The mole ratio of C:H:O is:
C = 0.00330
H = 0.00660
O = 0.00109
Divide by the smallest you get:
C = 0.00330 / 0.00109 = 3.03
H = 0.00660 / 0.00109 = 6.06
O = 0.00109 / 0.00109 = 1.00
