Question

A sample of $CuS{{O}_{4}}.5{{H}_{2}}O$ contains 3.782 g of Cu. How many grams of oxygen are in this sample? (Cu =63.5)
(A) 0.952g
(B) 3.80g
(C) 4.761g
(D) 8.576g

Answer 1

Hint: One mole of $CuS{{O}_{4}}.5{{H}_{2}}O$ will have 144 gram of oxygen and 63.5 gram of copper. In any mole of the given compound, the moles of oxygen and copper will be in the ratio of 9:1. Since copper pentahydrate is given in the question, we will add the weight of 5 molecules of water in molar mass of copper sulphate, to get the molar mass of copper pentahydrate.

Complete step by step solution:
The mole (symbol: mol) is the unit of measurement for the amount of substance in the International System of Units (SI). A mole of a substance or a mole of particles is defined as exactly $6.02214076\times {{10}^{23}}~$ particles, which may be atoms, molecules, ions, or electrons. In short, for particles $1 mol =$ $6.02214076\times {{10}^{23}}~$.
Given in the question is that a sample of $CuS{{O}_{4}}.5{{H}_{2}}O$ contains 3.782 g of Cu. Therefore, we have to tell how much ram of oxygen is present in this sample.
Now we can see that the moles of Copper and Oxygen in $CuS{{O}_{4}}.5{{H}_{2}}O$ are present in the ratio of 1:9, and this ratio will be same for any weight of compound to be taken.
Molar mass of $CuS{{O}_{4}}.5{{H}_{2}}O$ is $249.6$ $g/mol$. It contains $63.5g$ of copper and $144g$ of oxygen, therefore, it can be related as:
\[\begin{align}
& \dfrac{144g\text{ of O}}{63.5g\text{ of Cu}}=\dfrac{x\text{ g of O}}{3.782g\text{ of Cu}} \\
& x\text{ g of O}=\dfrac{144g\text{ of O}}{63.5g\text{ of Cu}}\times 3.782g\text{ of Cu} \\
& x\text{ g of O}=8.576g \\
\end{align}\]
Therefore, in compound $CuS{{O}_{4}}.5{{H}_{2}}O$ having $3.782$ gram of Copper, will have $8.576$ gram of Oxygen.

Hence the correct option is the (D) option.

Note: The molar mass of a substance is the mass of 1 mole of that substance, in multiples of the gram. The amount of substance is the number of moles in the sample. For most practical purposes, the magnitude of molar mass is numerically the same as that of the mean mass of one molecule, expressed in daltons.