
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of carbon dioxide The empirical formula of the hydrocarbon is:
A. ${C_2}{H_4}$
B. ${C_3}{H_4}$
C. ${C_6}{H_5}$
D. ${C_7}{H_8}$
Answer
553.8k+ views
Hint: To solve this question firstly we will write the combustion reaction of a hydrocarbon in the form of x and y where x and y are ration for number of atoms of carbon and hydrogen respectively.
After writing the reaction we will compare the moles of water and carbon dioxide.
And finally we get the empirical formula for that hydrocarbon.
Complete answer:
Now we will write the combustion reaction for hydrocarbon:
${C_x}{H_y} + (x + \dfrac{y}{2}){O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O$
Where, X and y will be the ratio of carbon and hydrogen atoms.
Now from the equation we can conclude that
If 1 mole of ${C_x}{H_y}$ burns in $(x + \dfrac{y}{2})$ moles of oxygen molecule .then it produces x moles of carbon dioxide and $\dfrac{y}{2}$moles of water.
Now from the given question we have 0.72 g of water produced and 3.08 g of carbon dioxide produced.
So now we will calculate the moles of water and carbon dioxide produced.
So, we all know that the molecular mass of water is 18 and the molecular mass of carbon dioxide is 44.
Hence, the number of moles of water will be: $\dfrac{{0.72}}{{18}}$.
The number of moles of carbon dioxide will be:$\dfrac{{3.08}}{{44}}$
Now we will compare number of moles of water produced in the hydrocarbon equation and from the given question
We will get,
$
\dfrac{y}{2} = \dfrac{{0.72}}{{18}} \\
y = \dfrac{{0.72}}{9} \\
y = 0.08 \\
$
Similarly we will compare for the carbon dioxide:
$
x = \dfrac{{3.08}}{{44}} \\
x = 0.07 \\
$
Now x:y=7:8
Which means the Mean number of carbon atoms will be 7 and the number of hydrogen atoms will be 8.
Hence ,compound will be ${C_7}{H_8}$
Therefore, from the above equation we can say that the correct answer is (D).
Note: Combustion or complete oxidation : hydrocarbons readily burn with non-luminous flame in the excess of the air(oxygen) to form carbon dioxide and water. The combustion of hydrocarbons is an exothermic reaction i.e. a large quantity of heat is evolved and heat of combustion will increases with the molecular weight.
After writing the reaction we will compare the moles of water and carbon dioxide.
And finally we get the empirical formula for that hydrocarbon.
Complete answer:
Now we will write the combustion reaction for hydrocarbon:
${C_x}{H_y} + (x + \dfrac{y}{2}){O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O$
Where, X and y will be the ratio of carbon and hydrogen atoms.
Now from the equation we can conclude that
If 1 mole of ${C_x}{H_y}$ burns in $(x + \dfrac{y}{2})$ moles of oxygen molecule .then it produces x moles of carbon dioxide and $\dfrac{y}{2}$moles of water.
Now from the given question we have 0.72 g of water produced and 3.08 g of carbon dioxide produced.
So now we will calculate the moles of water and carbon dioxide produced.
So, we all know that the molecular mass of water is 18 and the molecular mass of carbon dioxide is 44.
Hence, the number of moles of water will be: $\dfrac{{0.72}}{{18}}$.
The number of moles of carbon dioxide will be:$\dfrac{{3.08}}{{44}}$
Now we will compare number of moles of water produced in the hydrocarbon equation and from the given question
We will get,
$
\dfrac{y}{2} = \dfrac{{0.72}}{{18}} \\
y = \dfrac{{0.72}}{9} \\
y = 0.08 \\
$
Similarly we will compare for the carbon dioxide:
$
x = \dfrac{{3.08}}{{44}} \\
x = 0.07 \\
$
Now x:y=7:8
Which means the Mean number of carbon atoms will be 7 and the number of hydrogen atoms will be 8.
Hence ,compound will be ${C_7}{H_8}$
Therefore, from the above equation we can say that the correct answer is (D).
Note: Combustion or complete oxidation : hydrocarbons readily burn with non-luminous flame in the excess of the air(oxygen) to form carbon dioxide and water. The combustion of hydrocarbons is an exothermic reaction i.e. a large quantity of heat is evolved and heat of combustion will increases with the molecular weight.
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