A compound Y containing C,H and O only was burnt in a stream of pure oxygen. The carbon(iv)oxide and water produced were collected in pre-weighed absorption tubes and carefully re- weighed .
Readings obtained are:
Initial mass of Y=20.63mg
Mass of CO2 produced=57.94mg
Mass of H2O produced=11.85
What is the empirical formula
NOTE: I will round off to the nearest 0.1 to keep the numbers reasonable but this would not be allowed in a real lab experiment. But this problem can get away with it because we can round numbers. If you wish to carry out the numbers to four or more significant figures you may (and your teacher may require it) but rounding will not change the accuracy of the answer.
CxHzOw + O2 ==> CO2 + H2O
20.63 mg..............57.94 mg...11.85 mg
mg C = mg CO2 x (atomic C/molar mass CO2)=57.94 x (12/44) = 15.81 mg
g H = 11.85 x (2*atomic mass H/molar mass H2O) = 1.316 mg H
mg O = mg sample - mg C - mg H =20.63 -15.81 - 1.32 = 3.5 mg O
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mols C = 0.01581/12 = 0.00132
mols H = 0.001316/1 = 0.001316
mols O = 0.0035/16 = 0.000218
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Now you want to find the ratio of the values to one another with the smallest being no less than 1.00. The easy way to do that is to divide all of the numbers by the smallest; i.e.,
C = 0.00132/0.000218 = 6.05 which rounds to 6.0 as a whole number.
H = 0.001316/0.000218 = 6.04 which rounds to 6.0 as a whole number.
O = 0.000218/0.000218 = 1.00
So the empirical formula is C6H6O
Check all of those numbers and all of the math.