
A mixture of cuprous oxide ($C{{u}_{2}}O$) and cupric oxide ($Cu O$) on analysis was found to contain 88 % copper. Calculate the percentage amounts of $C{{u}_{2}}O$ and $CuO$ in the given mixture.(At. Wt. of Cu = 64)
Answer
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Hint: According to the law of definite composition which states that in a mixture , all the compounds are present in a definite proportion by mass and we can calculate the percentage of any substance/ material by dividing the mass of that very compound by the total mass of the compound and then multiplying it by 100.
Complete step by step answer:
By quantitative analysis, we mean to evaluate the mathematical data and give the result in the form of numbers.
Now, considering the numerical;
Molecular weight of copper=64 (given)
Then, The molecular weight of cuprous oxide is = $64 + 16 = 80g$
The molecular weight of cupric oxide is = $64 + 16 = 80g$
Now, if;
64 g of the copper is present in =144 g of cuprous oxide(given)
1 g of the copper is present in = $\dfrac{144}{64}$g of cuprous oxide
88 g of the copper is present in = $\dfrac{144\times 88}{64}$g of cuprous oxide
= 198 g of cuprous oxide
- similarly,
64 g of the copper is present in =80 g of cupric oxide(given)
1 g of the copper is present in = $\dfrac{144}{64}$g of cupric oxide
88 g of the copper is present in = $\dfrac{144\times 88}{64}$g of cupric oxide
= 110 g of cupric oxide
- Now, the mass of the given mixture of $C{{u}_{2}}O$ and $Cu O$=$110+198g=308g$
Now, calculating the percentage of $C{{u}_{2}}O$ and $Cu O $in the mixture as;
$\begin{align}
& percentage\text{ }of\text{ }C{{u}_{2}}O=\dfrac{mass\text{ }of\text{ }C{{u}_{2}}O}{total\text{ }mass\text{ }of\text{ }the\text{ }mixture}\times 100 \\
& \text{ =}\dfrac{198}{308}\times 100 \\
& \text{ =64}\text{.28 }\%\text{ } \\
\end{align}$
$\begin{align}
& percentage\text{ }of Cu O=\dfrac{mass\text{ }of\text{ }Cu O}{total\text{ }mass\text{ }of\text{ }the\text{ }mixture}\times 100 \\
& \text{ =}\dfrac{110}{308}\times 100 \\
& \text{ =35}\text{.17 }\%\text{ } \\
\end{align}$
Note: Composition in any substance tells us about how pure the sample is and it can be indicated either quantitatively (i.e. to express something in terms of quantity) or qualitatively (which tells us about the qualities of the substance).
Complete step by step answer:
By quantitative analysis, we mean to evaluate the mathematical data and give the result in the form of numbers.
Now, considering the numerical;
Molecular weight of copper=64 (given)
Then, The molecular weight of cuprous oxide is = $64 + 16 = 80g$
The molecular weight of cupric oxide is = $64 + 16 = 80g$
Now, if;
64 g of the copper is present in =144 g of cuprous oxide(given)
1 g of the copper is present in = $\dfrac{144}{64}$g of cuprous oxide
88 g of the copper is present in = $\dfrac{144\times 88}{64}$g of cuprous oxide
= 198 g of cuprous oxide
- similarly,
64 g of the copper is present in =80 g of cupric oxide(given)
1 g of the copper is present in = $\dfrac{144}{64}$g of cupric oxide
88 g of the copper is present in = $\dfrac{144\times 88}{64}$g of cupric oxide
= 110 g of cupric oxide
- Now, the mass of the given mixture of $C{{u}_{2}}O$ and $Cu O$=$110+198g=308g$
Now, calculating the percentage of $C{{u}_{2}}O$ and $Cu O $in the mixture as;
$\begin{align}
& percentage\text{ }of\text{ }C{{u}_{2}}O=\dfrac{mass\text{ }of\text{ }C{{u}_{2}}O}{total\text{ }mass\text{ }of\text{ }the\text{ }mixture}\times 100 \\
& \text{ =}\dfrac{198}{308}\times 100 \\
& \text{ =64}\text{.28 }\%\text{ } \\
\end{align}$
$\begin{align}
& percentage\text{ }of Cu O=\dfrac{mass\text{ }of\text{ }Cu O}{total\text{ }mass\text{ }of\text{ }the\text{ }mixture}\times 100 \\
& \text{ =}\dfrac{110}{308}\times 100 \\
& \text{ =35}\text{.17 }\%\text{ } \\
\end{align}$
Note: Composition in any substance tells us about how pure the sample is and it can be indicated either quantitatively (i.e. to express something in terms of quantity) or qualitatively (which tells us about the qualities of the substance).
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