A10.00 g sample of a hydrocarbon (which is a compound that contains only carbon and hydrogen) was burned in oxygen, and the carbon dioxide and water produced were carefully collected and weighed. the mass of the carbon dioxide was 30.50 g, and the mass of the water was 14.98 g. what was the empirical formula of the hydrocarbon?

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

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