C₅H₁₂
Explanation:
To obtain the answer for this question we need to do a combustion analysis. When a hydrocarbon is heated, it means that it reacts with oxygen (O₂) to produce two known products which are carbon dioxide (CO₂) and water (H₂O), and by knowing the masses of these products, we can know the proportions of the elements that initially were part of the hydrocarbon, in this case, the C/H ratio.
First, we propose the next reaction, assuming that all the hydrocarbon sample was combusted:
CxHy(s) + O₂(g) → xCO₂(g) + yH₂O(g)
Now, with the provided masses of the carbon dioxide and the water, we can calculate the molar amounts of carbon and hydrogen in the sample.
First we calculate the molar masses:
C = 12.011 x 1 = 12.011 g/mol
O = 15.99 x 2 = 31.99 g/mol
CO₂ = 12.011 + 31.99 = 44.001 g/mol
H = 1.008 x 2 = 2.016 g/mol
O = 15.99 x 1 = 15.99 g/mol
H₂O = 2.01 + 15.99 = 18.006 g/mol
Now we obtain the molar amounts of C and H using the obtaines masses of carbon dioxide and water:
mol C = 30.50g CO₂ x (1mol CO₂)/(44.001 g/mol) x (1mol C)/(1mol CO₂) = 0.6931 mol C
mol H = 14.98g H₂O x (1mol H₂O)/(18.006 g/mol) x (2mol H)/(1mol H₂O) = 1.6638 mol H
Finally, we can obtain the H/C molar ratio by identifying the smaller whole-number ratio for these molar amounts. For this we can first divide each molar amount by the smaller amount:
mol C = 0.6931/0.6931 = 1
mol H = 1.6638/0.6931 = 2.4
As we are still getting a decimal amount for the hydrogen, what we can do is multiply both molar amounts by the smaller whole multiple that can give us a whole number for the hydrogen's molar amount, in this case, that multiple would be 5:
mol C = 0.6931/0.6931 = 1 x 5 = 5
mol H = 1.6638/0.6931 = 2.4 x 5 = 12
Now we can write the empirical formula for the hydrocarbon, which is:
C₅H₁₂
