Question

When 6.114 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 20.67 grams of CO2 and 4.231 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

Answer:

Molecular formula : C₂H₂

Empirical formula H-C≡C-H

Explanation:

First of all, think the equation

CₓHy  + O₂  →  CO₂  +  H₂O

We have produced grams, of products, so we can find those moles.

Mass / molar weight = 20.67 g/ 44 g/m = 0.469 mol CO₂

Mass / molar weight   = 4.231 g/ 18g/m = 0.235 mol  H₂O

6.411 of a hydrocargon / 26.04 g/m = 0.246 mol

Now that we have the moles, we can work by thinking the relation 1:1

0.246 mol CxHy ___ produce __ 0.469 mol

1 mol CxHy __ produce ___ (0.469 /0.246) = 1.90 mol (almost 2)

0.246 mol CxHy___produce __ 0.235 mol

1 mol CxHy __ produce ___ (0.235/0.246) = 0.95 (almost 1)

Molar weight C = 12g/m

Molar weight H = 1g/m

Now we can write the equation as this:

CₓHy  + O₂  →  2CO₂  + H₂O

We have 2 carbons on product side, 5 O and 2 H. So in reactant side, we can add 2C to the hydrocarbon, so if the molar mass is 26.04 g/m and we have 2 carbons (24 g/m), H must be 2. The molecular formula for the hydrocarbon must be C2H2. Hydrocarbon's name is acetylene

Finally we have to ballance the oxigen.

C₂H₂ + 5/2 O₂ →  2CO₂  + H₂O