A hydrate of copper(II) sulfate, when heated, goes through the succession of changes suggested by the photograph. In this photograph, (a) is the original fully hydrated copper(II) sulfate; (b) is the product obtained by heating the original hydrate to \(140^{\circ} \mathrm{C}\) (c) is the product obtained by further heating to \(400^{\circ} \mathrm{C}\) and (d) is the product obtained at \(1000^{\circ} \mathrm{C}\) A \(2.574 \mathrm{g}\) sample of \(\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\) was heated to \(140^{\circ} \mathrm{C},\) cooled, and reweighed. The resulting solid was reheated to \(400^{\circ} \mathrm{C},\) cooled, and reweighed. Finally, this solid was heated to \(1000^{\circ} \mathrm{C},\) cooled, and reweighed for the last time. $$ \text {Original sample } \quad \quad\quad\quad\quad \text {\(2.574 \mathrm{g}\) } $$ $$ \text {After heating to \(140^{\circ} \mathrm{C}\) } \quad \quad\quad\quad\quad \text {\(1.833 \mathrm{g}\) } $$ $$ \text {After reheating to \(400^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(1.647 \mathrm{g}\) } $$ $$ \text {After reheating to \(1000^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(0.812 \mathrm{g}\)} $$ (a) Assuming that all the water of hydration is driven off at \(400^{\circ} \mathrm{C},\) what is the formula of the original hydrate? (b) What is the formula of the hydrate obtained when the original hydrate is heated to only \(140^{\circ} \mathrm{C} ?\) (c) The black residue obtained at \(1000^{\circ} \mathrm{C}\) is an oxide of copper. What is its percent composition and empirical formula?

Step by step solution

01

Determine formula of original hydrate

First, find the weight loss after heating to 400° C, which corresponds to the mass of the water in the original hydrate. The mass of the water is \(2.574 g - 1.647 g = 0.927 g \). Once the weight is known, calculate the moles of water using its molar mass (18.015 g/mol). The moles of water is \(0.927 g / 18.015 g/mol = 0.051 mol \). Then find the moles of CuSO4 left after heating to 400°C using its molar mass (159.609 g/mol). The moles of CuSO4 is \(1.647 g / 159.609 g/mol = 0.0103 mol \). The hydrate formula has a ratio of moles of water to moles of CuSO4. Therefore, x in \(CuSO4 \cdot xH2O\) is \(0.051 / 0.0103 \approx 5\). Therefore, the formula of the original hydrate is \(CuSO4 \cdot 5H2O\).

02

Determine formula of the reduced hydrate

Repeat a similar process to Step 1 except use the mass left after heating to 140°C to calculate the moles of water lost. The difference in g between the original and after heating to 140°C is \(2.574 g - 1.833 g = 0.741 g \). Calculate the moles of water lost (0.741g/18.015 g/mole = 0.041 moles). Repeat the calculation for CuSO4 where the moles of CuSO4 are 1.833g/159.609 g/mole = 0.011 mol. The ratio of moles of water to CuSO4 gives the formula of the hydrate as \(CuSO4 \cdot 4H2O\).

03

Determine the empirical formula of copper oxide

The mass of the oxide produced at 1000° C is the mass of the sample after complete dehydration, which is 0.812 g. Subtracting this from the mass after heating at 400°C gives the mass of oxygen in the sample, \(1.647 g - 0.812 g = 0.835 g\). Convert each mass to moles using their molar masses (63.546 g/mol for Cu and 15.999 g/mol for O). The mole ratio simplifies to approximately \(1:1\), so the empirical formula for copper oxide is \(CuO\). The percent composition is then found by dividing the mass of each element by the total mass of the compound and multiplying by 100, giving roughly \(60.43\% Cu\) and \(39.57\% O\) respectively.

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