Answer to Question #114837 in General Chemistry for maria

Question

An unknown compound has the formula CxHyOz. You burn 0.2246g of the compound and isolate 0.5484g of CO2 and 0.2245g of H2O. What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula? (Enter the elements in the order: C, H, O) This is a part answer. Empirical formula:______ and Molecular Formula:_________

Answer 1

C_xH_yO_z + (x+y/4-z/2)O2 ----> xCO₂ + y/2H2O

First, work out the mass of each element that was present in the original compound. Carbon is always present as CO₂ in the ratio (12.011 g / 44.0098 g), and hydrogen is always present as H₂O in the ratio (2.0158 g / 18.0152 g).

So you need to work out the mass of carbon in 0.5484 g of CO₂ and the mass of hydrogen in 0.2245 g of water.

Carbon: 0.5484 x (12.011 / 44.0098) = 0.14967 g.

Hydrogen: 0.2245 x (2.0158 / 18.0152) = 0.02512 g.

You can determine the mass of oxygen by difference 0.2246 - 0.14967 - 0.02512 = 0.04981 g.

Next, convert each of these to numbers of moles:

Carbon: 0.14967 / 12.011 = 0.01246 mol

Hydrogen: 0.02512 / 1.0079 = 0.02492 mol

Oxygen: 0.04981 / 15.994 = 0.003114 mol

Next, divide each number of moles by the lowest value, so that you get as close as possible to whole numbers.

Carbon: 0.01246 / 0.003114 = 4

Hydrogen: 0.02492 / 0.003114 = 8

Oxygen: 0.003114 / 0.003114 = 1

So the emprical formula is C₄H₈O

The molar mass is 72.1 g/mol - work out the mass of C₄H₈O - this is (12 x 4) + (8 x 1) + 16 = 72. Therefore the molecular formula is the same as the empirical.