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
