Question

Ten millilitres of a gaseous hydrocarbon was burned completely in 80 ml of ${{O}_{2}}$ at STP. The Volume of the remaining gas is 70 ml. The volume became 50 ml on treatment with NaOH. The formula of the hydrocarbon is:
A.${{C}_{2}}{{H}_{6}}$
B.${{C}_{2}}{{H}_{4}}$
C.${{C}_{3}}{{H}_{8}}$
D.${{C}_{3}}{{H}_{6}}$

Answer 1

Hint: The concept of the combustion of carbon in presence of oxygen to yield carbon dioxide and water is to be used in this question. Two equations will be formed by using the data given in the question after which the number of carbon and hydrogen need to be calculated.

Complete Solution :
- In order to answer the question, we need to learn about the mole concept and Avogadro’s Law. Before the advent of mass spectrometry for determining the atomic masses accurately, scientists were determining mass of one atom relative to another by experimental means. Hydrogen, being the lightest atom, was arbitrarily assigned a mass of 1 (without any units) and other elements were assigned masses relative to it. In this system, C is assigned a mass of exactly 12 atomic mass units (amu) and masses of all other atoms are given relative to this standard. One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. Mole is simply a unit for counting entities at the microscopic level (e. atoms, molecules, particles, electrons loans, etc) just as we use one dozen for twelve objects and one score for twenty objects. In the SI system, mole (symbol mot) was introduced as severith base quantity for the amount of a substance, One mole is defined as the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g of the $^{12}C$ isotope. Molar mass is the mass of the substance that contains $6\times {{10}^{23}}$ atoms or molecules inside it. It is different for different substances. Now let us come to our question. We have the volume of hydrocarbon and volume of oxygen as 10mL and 80mL respectively. Now, we can assume the formula of hydrocarbon to be ${{C}_{x}}{{H}_{y}}$. The combustion reaction of carbon with oxygen can be represented as:
\[{{C}_{x}}{{H}_{y}}+(x+\dfrac{y}{4}){{O}_{2}}\to xC{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O\]

- Here, 1 mole of ${{C}_{x}}{{H}_{y}}$produces x mole of $C{{O}_{2}}$by reacting with $(x+\dfrac{y}{4})$ moles of ${{O}_{2}}$. So, in order to produce 10mL of $C{{O}_{2}}$ we have 10mL of ${{C}_{x}}{{H}_{y}}$. It is given that residual gas occupied 70mL. So the volume of $C{{O}_{2}}$ is 70 - 50 = 20 mL as KOH has volume 50mL and it absorbs the $C{{O}_{2}}$. So, we can have that $10\times x$mL of $C{{O}_{2}}$=20mL of $C{{O}_{2}}$.So, we have x = 2 on solving. So the volume of oxygen used is 80-50=30mL. Again, from the above reaction, we can write that:
\[\begin{align}
 & 10(x+\dfrac{y}{4})=30 \\
 & 2+\dfrac{y}{4}=3 \\
 & y=4 \\
\end{align}\]
So, x = 2 and y = 4. The compound is ${{C}_{2}}{{H}_{4}}$
So, the correct answer is “Option B”.

Note: It is to be noted that in this question the $C{{O}_{2}}$ was absorbed by the KOH, so we subtracted it by the volume of KOH. In other cases, the $C{{O}_{2}}$ might not get absorbed, so the answer would have been different.