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11g of a hydrocarbon was completely burnt in air and 18g of water was produced. What is the empirical formula for the hydrocarbon?

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2 Answers

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A comment to add to Blackhitman's good answer.

AT the stage of 1:2.66 the multiplication by 3 can be awkward to determine.

Here, it is worth remembering that numbers ending in .333 or .666 can be nultiplied by 3 to give an integer as they are 1/3 and 2/3, respectively, as fractions.

Also, note that for .14 endings (which are rare in questions), multiplication by 7 is effective as .14 is approx. 1/7.

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Really good tip/insight! Knowing my decimal to fraction approximations really helped me out when I did VCE Chemistry. – Collin Li♦ Feb 14 at 22:07
Thanks a lot! really good tip, I never thought of that. – Blakhitman Feb 15 at 4:55
Colin and Blakhitman, thanks for the positive feedback, the fraction-decimal interconversion can be very helpful when approximating answers in questions-particularly some multiple choice questions where not using calculator may save time. – Hugh Feb 16 at 2:03
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O.k firstly, because it is completely burnt, that means that the masses are the same on each side of the equation, i.e g(H) in the reactants is equal to g(H) in the products.

First step is to work out the mass of Hydrogen.

n(H2O)=18/18= 1 mol, remember n(H)= 2 x n(H2O) hence n(H)= 2 mol

Therefore g(h)= 2 g

Then we can find the mass of Carbon in the hydrocarbon

m(C)= 11-2= 9g, therefore n(C)= 9/12= .75 mol

Then we use ratios

n(C) : n(H)

.75 : 2

divide by smallest number and you get

1 : 2.66 multiply it by 3 to get 3 : 8,

E.F should be C3H8.

Do you have solutions?

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Yes, it was C3H8 :) It was a multiple choice question and C: C3H8 was the correct answer. Thanks heaps – Dan Feb 18 at 10:36

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