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Question

An unknown compound contains only  C, H and  O. Combustion of 5g  of this compound produced9.99g   CO2 and  4.09g H2O . What is the empirical formula of the unknown compound? 

Answer 1

Hello Krisann

Steps for solving this problem.

1) Convert g of CO₂ to g of C using stoichiometry

9.99 g CO₂ x (1 mol CO₂ / 44 g CO₂) x (1 mol C / 1 mol CO₂) x (12 g C / 1 mol C) = 2.72 g C

2) Convert g of H₂O to g of H

4.09 g H₂O x (1 mol H₂O / 18 g H₂O) x (2 mol H / 1 mol H₂O) x (1 g H / 1 mol H) = 0.454 g H

3) Since the starting compound has only C, H and O, the combined mass of each element must equal 5.00 g from which, the mass of O can be determined.

2.72 g C + 0.454 g H + ? g O = 5.00 g

g O = 5.00 - (2.72 + 0.454) = 1.83 g O

4) Now that we have the masses of C, H and O, we need to first convert their mass to moles.

2.72 g C x (1 mol C / 12 g C) = 0.227 mol C

0.454 g H x (1 mol H / 1 g H) = 0.454 mol H

1.83 g O x (1 mol O / 16 g O) = 0.114 mol O

5) Divide the moles of each element by the lowest # of moles (0.114) to determine the mole ratio of each element in the original compound.

0.227 mol C / 0.114 = 1.99 mol C ≅ 2 mol C

0.454 mol H / 0.114 = 3.98 mol H ≅ 4 mol H

0.114 mol O / 0.114 = 1.00 mol O ≅ 1 mol O

The empirical formula is therefore, C₂H₄O.