(Given : molar mass in $\text{g mol}^{-1} \text{ H} : 1, \text{ C} : 12, \text{ O} : 16, \text{ Br} : 80$)
1. Problem Analysis
The question describes the bromination of a cycloalkene (X). Key information provided:
- Reaction: Cycloalkene (X) + 1 mole $Br_2$ $\rightarrow$ Product (Y). This indicates (X) has one double bond.
- Product (Y): The ratio of Carbon (C) atoms to Bromine (Br) atoms is 3:1.
- Objective: Calculate the percentage of Bromine (% Br) in product (Y).
- Given Molar Masses: H = 1 g mol-1, C = 12 g mol-1, Br = 80 g mol-1.
2. Determining the Chemical Formula of Product (Y)
Bromination of alkenes involves the addition of $Br_2$ across the double bond.
Let the cycloalkene (X) have the formula $C_n H_{2n}$. Upon reaction with $Br_2$, the product (Y) will have the formula $C_n H_{2n} Br_2$.
The question states the C:Br ratio in product (Y) is 3:1. Let the formula of Y be $C_a H_b Br_c$. We have $a:c = 3:1$.
Since the reaction adds one molecule of $Br_2$, the product (Y) must contain 2 bromine atoms. Therefore, $c=2$.
Using the ratio $a:c = 3:1$, we find the number of carbon atoms ($a$): $a/2 = 3/1 \implies a = 3 \times 2 = 6$.
So, product (Y) contains 6 Carbon atoms and 2 Bromine atoms. This means $n=a=6$. The formula for product (Y) is $C_6 H_{12} Br_2$.
We can verify the C:Br ratio: In $C_6 H_{12} Br_2$, the ratio is 6:2, which simplifies to 3:1, matching the given condition.
3. Calculating Bromine Percentage in Product (Y)
The chemical formula for product (Y) is $C_6 H_{12} Br_2$. We need to calculate the percentage by mass of Bromine.
Step 1: Calculate the Molar Mass of (Y)
Molar Mass of Y = $(6 \times \text{Molar Mass of C}) + (12 \times \text{Molar Mass of H}) + (2 \times \text{Molar Mass of Br})$
Using the given values: $M(Y) = (6 \times 12 \text{ g mol}^{-1}) + (12 \times 1 \text{ g mol}^{-1}) + (2 \times 80 \text{ g mol}^{-1})$ $M(Y) = 72 + 12 + 160$ $M(Y) = 244 \text{ g mol}^{-1}$
Step 2: Calculate the Total Mass of Bromine in one mole of (Y)
Mass of Br = $2 \times \text{Molar Mass of Br}$ Mass of Br = $2 \times 80 \text{ g mol}^{-1}$ Mass of Br = $160 \text{ g mol}^{-1}$
Step 3: Calculate the Percentage of Bromine by Mass
Percentage Br $= \left( \frac{\text{Mass of Br}}{\text{Molar Mass of Y}} \right) \times 100\% $
Percentage Br $= \left( \frac{160 \text{ g mol}^{-1}}{244 \text{ g mol}^{-1}} \right) \times 100\% $
Percentage Br $\approx 0.6557377 \times 100\%$ Percentage Br $\approx 65.57\%$
Rounding to the nearest integer, the percentage of Bromine is 66%.
Similar Questions
- Dissociation of a gas $\text{A}_2$ takes place according to the following chemical reaction. At equilibrium, the total pressure is $1 \text{ bar}$ at $300\text{K}$.
$\text{A}_2\text{(g)} \rightleftharpoons 2\text{A(g)}$
The standard Gibbs energy of formation of the involved substances has been provided below:Substance $\Delta G_f^\circ / \text{kJ mol}^{-1}$ $\text{A}_2$ -100.00 A -50.832
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A. Hydrated salts can be used as primary standard.
B. Primary standard should not undergo any reaction with air.
C. Reactions of primary standard with another substance should be instantaneous and stoichiometric.
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Choose the correct answer from the options given below :- Consider $\text{A} \xrightarrow{k_1} \text{B}$ and $\text{C} \xrightarrow{k_2} \text{D}$ are two reactions. If the rate constant ($k_1$) of the $\text{A} \longrightarrow \text{B}$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T/K}$ and activation energy of $\text{C} \longrightarrow \text{D}$ reaction ($\text{Ea}_2$) is $\frac{1}{5}$th of the $\text{A} \longrightarrow \text{B}$ reaction ($\text{Ea}_1$), then the value of ($\text{Ea}_2$) is _________ $\text{kJ mol}^{-1}$. (Nearest Integer)
- Consider the following electrochemical cell :
$\text{Pt} \mid \text{O}_2\text{(g)}(1\text{bar}) \mid \text{HCl(aq)} \parallel \text{M}^{2+}\text{(aq, 1.0 M)} \mid \text{M(s)}$
The pH above which, oxygen gas would start to evolve at anode is _________ (nearest integer).
Given :
$\begin{bmatrix} \text{E}^\circ_{\text{M}^{2+}/\text{M}} = 0.994 \text{ V} \\ \text{E}^\circ_{\text{O}_2/\text{H}_2\text{O}} = 1.23 \text{ V} \end{bmatrix} \text{standard reduction potential}$
and $\frac{\text{RT}}{\text{F}} (2.303) = 0.059 \text{ V}$ at the given condition - At $27^\circ\text{C}$ in presence of a catalyst, activation energy of a reaction is lowered by $10 \text{ kJ mol}^{-1}$. The logarithm of ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is....
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A solution containing 0.01 M $\text{X}^{2+}$ and 0.01 M $\text{Y}^{2+}$ is saturated with $\text{H}_2\text{S}$. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer)
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[Given:
$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \quad \text{E}^\circ_{\text{red}} = +0.34 \text{ V}$
$\text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \text{E}^\circ_{\text{red}} = +1.23 \text{ V}$
Molar mass of Cu = $63.54 \text{ g mol}^{-1}$
Molar mass of $\text{O}_2$ = $32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = 22.4 L] - Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and $15 \text{ kN m}^{-2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?
Important Questions from Physical Chemistry
- Dissociation of a gas $\text{A}_2$ takes place according to the following chemical reaction. At equilibrium, the total pressure is $1 \text{ bar}$ at $300\text{K}$.
$\text{A}_2\text{(g)} \rightleftharpoons 2\text{A(g)}$
The standard Gibbs energy of formation of the involved substances has been provided below:Substance $\Delta G_f^\circ / \text{kJ mol}^{-1}$ $\text{A}_2$ -100.00 A -50.832
The degree of dissociation of $\text{A}_2\text{(g)}$ is given by $(x \times 10^{-2})^{1/2}$ where $x =$ _________. (Nearest integer).
$[\text{Given: } \text{R} = 8 \text{ J mol}^{-1}\text{ K}^{-1}, \log 2 = 0.3010, \log 3 = 0.48]$
Assume degree of dissociation is not negligible. - Which of the following mixture gives a buffer solution with pH = 9.25 ?
Given : $\text{pK}_b (\text{NH}_4\text{OH}) = 4.75$ Identify the correct statements :
A. Hydrated salts can be used as primary standard.
B. Primary standard should not undergo any reaction with air.
C. Reactions of primary standard with another substance should be instantaneous and stoichiometric.
D. Primary standard should not be soluble in water.
E. Primary standard should have low relative molar mass.
Choose the correct answer from the options given below :- Consider $\text{A} \xrightarrow{k_1} \text{B}$ and $\text{C} \xrightarrow{k_2} \text{D}$ are two reactions. If the rate constant ($k_1$) of the $\text{A} \longrightarrow \text{B}$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T/K}$ and activation energy of $\text{C} \longrightarrow \text{D}$ reaction ($\text{Ea}_2$) is $\frac{1}{5}$th of the $\text{A} \longrightarrow \text{B}$ reaction ($\text{Ea}_1$), then the value of ($\text{Ea}_2$) is _________ $\text{kJ mol}^{-1}$. (Nearest Integer)
- Consider the following electrochemical cell :
$\text{Pt} \mid \text{O}_2\text{(g)}(1\text{bar}) \mid \text{HCl(aq)} \parallel \text{M}^{2+}\text{(aq, 1.0 M)} \mid \text{M(s)}$
The pH above which, oxygen gas would start to evolve at anode is _________ (nearest integer).
Given :
$\begin{bmatrix} \text{E}^\circ_{\text{M}^{2+}/\text{M}} = 0.994 \text{ V} \\ \text{E}^\circ_{\text{O}_2/\text{H}_2\text{O}} = 1.23 \text{ V} \end{bmatrix} \text{standard reduction potential}$
and $\frac{\text{RT}}{\text{F}} (2.303) = 0.059 \text{ V}$ at the given condition
