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Question: A gaseous hydrocarbon gives upon combustion, 0.72g water and 3.08g of \[C{{O}_{2}}\]. The empirical ...

A gaseous hydrocarbon gives upon combustion, 0.72g water and 3.08g of CO2C{{O}_{2}}. The empirical formula of the hydrocarbon is:
(a) C6H5{{C}_{6}}{{H}_{5}}
(b) C7H8{{C}_{7}}{{H}_{8}}
(c) C2H4{{C}_{2}}{{H}_{4}}
(d) C3H4{{C}_{3}}{{H}_{4}}

Explanation

Solution

Hint: Empirical formula of a chemical compound is the simplest form of writing the chemical formula of the compound. It is the simplest positive integer ratio of the atoms that are present in the compound.

Complete step by step solution:
From a combustion reaction, it is possible to derive the empirical formula of the hydrocarbon, provided we know the number of moles of the products in the reaction. Let us assume the empirical formula as CxHy{{C}_{x}}{{H}_{y}}, where x is the number of carbon atoms and y is the number of hydrogen atoms. For a gaseous hydrocarbon combustion reaction can be written as
CxHy+(x+y4)O2xCO2+y2H2O{{C}_{x}}{{H}_{y}}+(x+\dfrac{y}{4}){{O}_{2}}\to xC{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O
(It is a general form of equation that has been calculated for combustion reaction.)
We know the number of moles is equal to weight divided by the molecular mass of the molecule.
Number of moles of H2O=0.7218=0.04{{H}_{2}}O=\dfrac{0.72}{18}=0.04 here, weight of water is 0.72g and molecular mass of water is 18g per mol.
Number of moles of CO2=3.0844=0.07C{{O}_{2}}=\dfrac{3.08}{44}=0.07here, weight of carbon dioxide is given 3.08g and molecular mass of carbon dioxide is 44g per mol.
Therefore, let’s take the coefficient of water and substitute it with its number of moles.
y2=0.04×100\dfrac{y}{2}=0.04\times 100 (y is multiplied with 100 so that we get a whole number)
y=8y=8
Now, let us take the coefficient of carbon dioxide x which is equal to the number of moles of carbon dioxide. We get the value of x.
x=0.07×100=7x=0.07\times 100=7(Here also we need to 0.07 is multiplied with 100 to get a whole number)
So, x=7 and y=8, substituting these values in CxHy{{C}_{x}}{{H}_{y}} we get C7H8{{C}_{7}}{{H}_{8}}.
Hence, the empirical formula is C7H8{{C}_{7}}{{H}_{8}}
Correct answer to the question is option (b).

Note: While solving this problem using this method, we should always convert the decimal we get for x and y into a whole number by multiplying it with 100. X and y are always whole numbers. This method is only applicable to simple hydrocarbons.