The combustion of 135 mg of a hydrocarbon sample produces 440. mg of co2 and 135 mg h2o. the molar mass of the hydrocarbon sample is 270 g/mol. determine the molecular formula of the hydrocarbon.

C10H150

Explanation:

An hydrocarbon is an organic molecule containing only carbon and hydrogen. Hence, for this exercise, we know we are to work with getting the mass of carbon and hydrogen. Kindly note, for convenience sake, we shall be working with unit of mass in grammes. Hence, we must know that 1000mg is present in 1g. To convert the mg unit to g , we simply divide by 1000.

The mass of the hydrocarbon is thus 135/1000 = 0.135g , the mass of carbon iv oxide produced is 440/1000 = 0.44g while the mass of water produced is 135/1000 = 0.135g.

We now proceed to the second part of the question. We can get the mass of hydrogen produced from the mass of water while we can get the mass of carbon produced from the mass of carbon iv oxide.

Since we know the mass of water produced, we now get the number of moles of water produced. To get this, we simply divide the mass of water produced by the molar mass of the water produced. The molar mass of water is 18g/mol. Hence the number of moles is 0.135/18 = 0.0075 moles.

Now, we know that one mole of water contains two moles of hydrogen. Hence, the number of moles of water produced is 2 * 0.0075 = 0.015 mole

To get the number of moles of carbon produced, we simply use the mass of carbon iv oxide produced.

To get the number of moles of carbon iv oxide produced, we simply divide the mass of carbon iv oxide by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol. The number of moles thus produced is 0.44/44 = 0.01 mole

We now proceed to get the empirical formula of the compound. To do this, we divide each by the smaller number of moles which in this case is that of carbon I.e 0.01 mole

C = 0.01/0.01 = 1

H = 0.015/0.01 = 15

The empirical formula is CH15.

To get the molecular formula, we simply use the empirical formula. The mass of the empirical formula is 27g/mol

Now (CH15)n = 270g/mol

We now get n.

27n = 270

n = 270/27 = 10

The molecular formula is thus C10H150