Question

A gaseous hydrocarbon on complete combustion gives 3.38 g of CO2 and 0.690 g of H2O
and no other product. The empirical formula of the hydrocarbon is
a) CH
(6) CH₂
(c) CH₃
(d) data is incomplete​

Answer 1

If a gaseous hydrocarbon on complete combustion gives 3.38 g of CO2 and 0.690 g of H2O and no other product, then the empirical formula of the hydrocarbon is CH.

Explanation:

We are given that a gaseous hydrocarbon contains C and H.

So, let the empirical formula of the hydrocarbon be “CxHy”.

Now, in order to get the empirical formula of the hydrocarbon, we have to find the moles of C and H released in the form of CO2 & H2O respectively.

 

Step 1:

Mass of CO2 given is = 3.38 g

Molar mass of CO2 = 44 g/mol

Here we can see,

44 g of CO2 consists of 1 mole of C atom

∴ 3.38 g of CO2 will have = [1/44] * 3.38 = 0.0768 moles …. (i)

Step 2:

Mass of H2O given is = 0.690 g

Molar mass of H2O = 18 g/mol

Here we can see,

18 g of H2O consists of 2 moles of H atom

∴ 0.690 g of H2O will have = [2/18] * 0.690 = 0.0767 moles …. (ii)

Step 3:

Now, we will divide both numbers that we found in (i) & (ii) by the smallest one to get the mole ratio of C & H in the hydrocarbon.

For Carbon atom: x = 0.0768/0.0767 = 1

And  

For Hydrogen atom: y = 0.0767/0.0767 = 1

∴ CxHy = CH …… [putting x=1 & y=1]

Thus, the empirical formula of the hydrocarbon is option (1): CH.

Hope this is helpful!!!!