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
583.8k+ 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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

