An unknown compound contains only C , H , and O . Combustion of 5.00 g of this compound produced 9.99 g CO 2 and 4.09 g H 2 O . What is the empirical formula of the unknown compound? Insert subscripts as needed.

C₂H₄O

Explanation:

What we want to calculate when determining empirical formulas is to obtain the proportion of the coefficients of the atoms in the molecule. This proportion is the same as the ratio in terms of number of moles, and since we have the quantities in grams  after combustion , we can calculate the number of moles and thus the empirical formula.

Molar mass CO₂ = 44 gmol⁻¹

Molar mass H₂O = 18 gmol⁻¹

# mol CO₂ =  9.99 g / 44 gmol⁻¹ =     0.227 mol CO₂  ⇒  mol C = 0.227

# mol H₂O =4.09 g H₂O / 18 gmol⁻¹ = 0.227 mol H₂O ⇒ mol H = 0.454

Now to determine the mol of oxygen  we would have to use the fact that we have 5.00 g in the combustion, and since we know the mol C and H, it is easy to determine the mass of O. We can not do the relations we did for C and H, since we already have oxygen in the formula.

mass C = 0.227 mol x 12 gmol⁻¹ = 2.72 g C

mass H = 0.454 mol x 1 gmol⁻¹ = 0.454 g H

mass O present in compound = 5.00 g - 2.72 g - 0.454 g = 1.82 g O

mol O = 1.82 g/ 16 gmol⁻¹ = 0.114

Now the composition and proportions are:

mol C = 0.227 / 0.114 = 1.99

mol H = 0.454 / 0.114 = 3.98

mol O = 0.114 / 0.114 = 1

Rounding these numbers we have the empirical formula C₂H₄O

in theory, limestone is composed principally of the compound calcium carbonate - caco3, but in reality this compound is deposited in conditions involving a number of other compounds as impurities so it ends up being a mixture of many compounds.

hope this : -)

tony (lordcaos066)