Each copper(II) sulfate unit is associated with five water molecules in crystalline copper(II) sulfate pentahydrate \(\left(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)\). When this compound is heated in air above \(100^{\circ} \mathrm{C},\) it loses the water \(\mathrm{mol}-\) ecules and also its blue color:$$\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CuSO}_{4}+5 \mathrm{H}_{2} \mathrm{O}$$ If \(9.60 \mathrm{~g}\) of \(\mathrm{CuSO}_{4}\) are left after heating \(15.01 \mathrm{~g}\) of the blue compound, calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) originally present in the compound.

Short Answer

Expert verified
There were initially 0.3005 moles of water (H2O) present in the compound.

Step by step solution

01

Calculate moles of CuSO4

First, you will need to calculate how many moles of copper(II) sulfate (CuSO4) are present after heating. To do this, use the given mass of CuSO4 (9.60 g) and the molar mass of CuSO4, which is approximately 159.609 g/mol. The calculation involves dividing the mass by the molar mass, which gives \[\frac{9.60 \, \text{g}}{159.609 \, \text{g/mol}} = 0.0601 \, \text{mol}\]So there are 0.0601 moles of CuSO4 present after heating.
02

Use stoichiometry to find moles of H2O

According to the balanced chemical equation, one mole of copper(II) sulfate pentahydrate produces one mole of CuSO4 and five moles of H2O. This means for every one mole of CuSO4, there are five moles of H2O. Therefore, you can multiply the number of moles of CuSO4 by 5 to get the number of moles of H2O:\[0.0601 \, \text{mol} \times 5 = 0.3005 \, \text{mol}\]So, there were originally 0.3005 moles of H2O in the compound.

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Most popular questions from this chapter

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a component of gasoline. Complete combustion of octane yields \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2} .\) Incomplete combustion produces \(\mathrm{H}_{2} \mathrm{O}\) and CO, which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gal of octane is burned in an engine. The total mass of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(11.53 \mathrm{~kg} .\) Calculate the efficiency of the process; that is, calculate the fraction of octane converted to \(\mathrm{CO}_{2}\). The density of octane is \(2.650 \mathrm{~kg} / \mathrm{gal}\)

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A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

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