Question

A mixture of $CuO$ and $C{u_2}O$ contains 88% of Cu. The percentage of $CuO$ in the mixture is (as nearest integer):

Answer 1

Hint: We have to calculate the mass percentage of $CuO$ in the mixture by using the moles of $C{u^{2 + }}$, atomic mass of copper and atomic mass of oxygen hence moles of $C{u^{2 + }}$ is calculated by using the moles of $C{u^ + }$ and moles of $Cu$.

Complete step by step answer:
Given data contains,
The mass percentage of copper present in the mixture is 88%.
Let us consider the mass of the mixture as $100g$.
The mass of copper in the mixture is $88g$.
We can calculate the mass of oxygen by subtracting the mass of mixture from the mass of copper in the mixture.
Mass of oxygen$ = {\text{Mass of mixture}} - {\text{Mass of copper}}$
Mass of oxygen$ = 100g - 88g$
Mass of oxygen$ = 12g$
The mass of oxygen is $12g$.
12g of oxygen is there for 88g of copper.
We can now calculate the moles of copper and oxygen as,
Molar mass of oxygen is $16g/mol$.
Molar mass of copper is $63.5g/mol$.
We can calculate the molar mass of oxygen as,
Moles of oxygen$ = \dfrac{{12g}}{{16g/mol}}$
Moles of oxygen$ = 0.75mol$
The moles of oxygen is $0.75mol$.
We can calculate the molar mass of copper as,
Moles of copper$ = \dfrac{{88g}}{{63.6g/mol}}$
Moles of copper = $1.38mol$
The moles of copper is $1.38mol$.
$0.75mol$ of oxygen is there for $1.38mol$ of copper.
Let us consider $x$ moles of $C{u^ + }$ and $\left( {1.38 - x} \right)$ moles of $C{u^{2 + }}$. We know that the total charge present on the mixture is zero.
Therefore, $x + 2\left( {1.38 - x} \right) - 2\left( {0.75} \right) = 0$
On solving, we get the value of $x$ as $1.26$.
The moles of $C{u^ + }$ are $1.26$ moles.
So, the moles of $C{u^{2 + }}$ are $\left( {1.38 - x} \right)$. We know the value of $x$ is $1.26$ moles.
Therefore, the moles of $C{u^{2 + }}$ are $\left( {1.38 - 1.26} \right) = 0.12moles$.
The moles of $C{u^{2 + }}$ are $0.12moles$.
There are \[1.26moles\]of $C{u^ + }$ and $0.12moles$ of $C{u^{2 + }}$.
From this, we can calculate the mass percentage of $CuO$ in the mixture.
Mass percentage of $CuO$ = ${\text{Moles of C}}{{\text{u}}^{2 + }}\left( {{\text{Molar mass of copper + Molar mass of oxygen}}} \right)$
Mass percentage of $CuO$ = $0.12mol\left( {63.6g/mol + 16g/mol} \right)$
Mass percentage of $CuO$ = $9.552\% $
The mass percentage of $CuO$ is $9.552\% $. Therefore, the nearest integer is $9\% $.

Note:
We can also this problem in an alternate way,
Let the mass of the mixture be 100g.
We can consider the mass of $CuO$ as x g.
We can consider the mass of $C{u_2}O$ as y g.
The molar mass of $CuO$ is $79.5g/mol$.
The molar mass of $C{u_2}O$ is $143g/mol$.
The mass of oxygen is calculated as $\dfrac{{16x}}{{79.5}} + \dfrac{{16y}}{{143}} = 12g$
The mass of copper is calculated as $\dfrac{{63.5x}}{{79.5}} + \dfrac{{127y}}{{143}} = 88g$
When we solve the value of x and y, we get the value of x as $8.65\% $ and the value of y as $91.35\% $. So, the nearest integer is $9\% $.