Question

The ratio of mass percent of C and H of an organic compound $({C_x}{H_y}{O_z})$ is $6:1$ . If one molecule of the above compound $({C_x}{H_y}{O_z})$ contains half as much oxygen as required to burn one molecule of compound ${C_x}{H_y}$ completely to $C{O_2}$ and ${H_2}O$ .The empirical formula of compound ${C_x}{H_y}{O_z}$ is:
A. ${C_2}{H_4}O$
B. ${C_3}{H_4}{O_2}$
C. ${C_2}{H_4}{O_3}$
D. ${C_3}{H_6}{O_3}$

Answer 1

To solve this question, we need to first calculate the number of oxygen atoms used in the combustion. Then we need to find the ratio of the number of atoms of carbon and hydrogen in the given compound. Then, on the basis of the relation of the number of oxygen atoms used in the combustion and in the compound, we can find the number of oxygen atoms present. Then finally, we can attain whole number ratios to get the final empirical formula.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some
important basic concepts. Whenever we burn any hydrocarbon compound, it results in the formation of carbon dioxide and water. According to the given conditions, one mole of a hydrocarbon with the molecular formula ${C_x}{H_y}$ is completely burnt by oxygen. Let us consider the number of moles of oxygen required to be ‘a’. Then, we can say:
${C_x}{H_y} + a{O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O$
Hence, the number of moles of oxygen involved in this reaction can be calculated to be:
$a = 2x + \frac{y}{2}$ moles
Now, another condition in the question is given that the number of oxygen molecules present in the
compound ${C_x}{H_y}{O_z}$ is half the value of ‘a’. Also, the mass ratio of carbon to hydrogen is given to be $6:1$ . The molecular masses of carbon and hydrogen are 6 g/mol and 2g/mol respectively. Hence, for every 1 atom of carbon, the compound will have 2 atoms of hydrogen. The ratio of the number of carbon atoms to the number of hydrogen atoms is $C:H:1:2$
Hence, substituting the values of x and y in the value of a, we get:
$a = 2(1) + \dfrac{2}{2} = 3$
Hence, the number of oxygen atoms present in ${C_x}{H_y}{O_z}$ is 1.5 atoms.
Hence, the value of $x = 1,y = 2,z = 1.5$
In empirical formulae, the ratio of the atoms present in a molecule are whole numbers.
Hence, $x = 2,y = 4,z = 3$

Hence, Option C is the correct option

Note: The empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound. A simple example of this concept is that the empirical formula of carbon monoxide, or CO, would simply be CO, as is the empirical formula of ethene Dione, ${C_2}{O_2}$ .