Question

A cycloalkene (X) is treated with Br\(_2\) and compound (Y) is formed with C : Br ratio 3 : 1. One mole of X required 1 mol of Br\(_2\). Find the composition of 'Br' in Y compound (percentage).

Hint:

To find the percentage composition of an element in a compound, divide the number of atoms of the element by the total atoms, then multiply by 100.

Answer 1

Correct Answer: 66

The question involves the reaction of a cycloalkene (X) with Br\(_2\) to form compound (Y), where the C:Br ratio is 3:1. We're tasked with finding the percentage of Br in compound Y.
1. **Understand the Reaction:** The cycloalkene (X) reacts with Br\(_2\) to form an addition product, incorporating Br atoms across the double bond in the cycloalkene.
2. **Determine the Formula of Compound Y:** Given the C:Br ratio of 3:1, we can denote the empirical formula of Y as C\(_3\)Br.
3. **Calculate the Molar Mass of Y**: Knowing the atomic masses C = 12 g/mol and Br = 80 g/mol, compute the molar mass of Y:
\[Molar\ Mass\ of\ C_3Br = 3 \times 12\ +\ 1 \times 80 = 36 + 80 = 116\ g/mol\]
4. **Calculate Composition of Br in Y:** Use the fraction of Br's contribution to the total molar mass:
\[\%\ Br = \left(\frac{80}{116}\right) \times 100\]%
\[= 68.97\%\]
5. **Confirm Within Range:** The calculated percentage of Br (68.97%) falls within the specified range of 66% to 66,66%.
In conclusion, the percentage of Br in compound Y is approximately 69%.