Combustion analysis of 0.150 g of an unknown compound containing carbon,hydrogen,and oxygen produced 0.2607 g of CO₂ and 0.1418 g of H₂O.What is the empirical formula of the compound? A)C₂H₅O B)C₂H₅O₂ C)C₂H₁₀O₃ D)C₃H₈O₂

Question

Quizplus · Accepted Answer

1) Calculate moles of CO: 
0.2607 g CO x (1 mol CO/57.97 g CO) = 0.004501 mol CO 
2) Calculate moles of H2O: 
0.1418 g H2O x (1 mol H2O/18.015 g H2O) = 0.007867 mol H2O 
3) Calculate moles of H: 
(2 mol H/1 mol H2O) x (0.007867 mol H2O) = 0.01573 mol H 
4) Calculate moles of C: 
(1 mol C/1 mol CO) x (0.004501 mol CO) = 0.004501 mol C 
5) Calculate moles of O: 
2 mol O/1 mol H2O x (0.007867 mol H2O) + 1 mol O/1 mol CO x (0.004501 mol CO) = 0.022235 mol O 
6) Divide by smallest number of moles (0.004501) to get empirical formula: 
H: (0.01573/0.004501) = 3.50 --> 4 
C: (0.004501/0.004501) = 1 
O: (0.022235/0.004501) = 4.93 --> 5 
Empirical formula: C1H4O5, which is equivalent to choice D.

Combustion Analysis of 0