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Chapter 3: Problem 151

A 0.755 -g sample of hydrated copper(II) sulfate $$ \mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O} $$ was heated carefully until it had changed completely to anhydrous copper(II) sulfate \(\left(\mathrm{CuSO}_{4}\right)\) with a mass of 0.483 g. Determine the value of \(x .[\text { This number is called the number of waters }\) of hydration of copper(Il) sulfate. It specifies the number of water molecules per formula unit of \(\mathrm{CuSO}_{4}\) in the hydrated crystal. \(]\)

Short Answer

Expert verified
The value of x, which represents the number of water molecules per formula unit of CuSO4 in the hydrated crystal, is approximately 5. Therefore, the formula of the hydrated copper(II) sulfate is given by: CuSO4 · 5H2O.

Step by step solution

01

1. Calculate the mass of water lost during the heating process

To find the mass of water lost during the heating process, subtract the mass of the anhydrous copper(II) sulfate from the mass of the hydrated copper(II) sulfate: Mass of water = Initial mass of hydrated copper(II) sulfate - Final mass of anhydrous copper(II) sulfate Mass of water = \(0.755 \, g - 0.483 \, g\) Mass of water = \(0.272 \, g\)
02

2. Determine the moles of anhydrous copper(II) sulfate and the moles of water

Next, we need to calculate the moles of anhydrous copper(II) sulfate and the moles of water. To do this, divide their respective masses by their molar masses. The molar mass of copper(II) sulfate (CuSO4) is approximately \(63.5 + 32 + (4 \times 16) \, g/mol = 159.5 \, g/mol\), and the molar mass of water (H2O) is approximately \(2 + 16 = 18 \, g/mol\). Moles of anhydrous copper(II) sulfate: Moles of CuSO4 = \( \frac{0.483 \, g}{159.5 \, g/mol} \) Moles of CuSO4 = \(0.00303 \, mol\) Moles of water: Moles of H2O = \( \frac{0.272 \, g}{18 \, g/mol} \) Moles of H2O = \( 0.0151 \, mol\)
03

3. Calculate the ratio between moles of water and moles of CuSO4

To find the value of x, we need to calculate the ratio between the moles of water and moles of anhydrous copper(II) sulfate: x = Moles of H2O / Moles of CuSO4 x = \( \frac{0.0151 \, mol}{0.00303 \, mol} \) x ≈ 5
04

Conclusion

The value of x, which represents the number of water molecules per formula unit of CuSO4 in the hydrated crystal, is approximately 5. Therefore, the formula of the hydrated copper(II) sulfate is given by: CuSO4 · 5H2O

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Most popular questions from this chapter

Bornite \(\left(\mathrm{Cu}_{3} \mathrm{Fe} \mathrm{S}_{3}\right)\) is a copper ore used in the production of copper. When heated, the following reaction occurs: $$ 2 \mathrm{Cu}_{3} \mathrm{FeS}_{3}(s)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{Cu}(s)+2 \mathrm{FeO}(s)+6 \mathrm{SO}_{2}(g) $$ If 2.50 metric tons of bornite is reacted with excess \(\mathrm{O}_{2}\) and the process has an 86.3\(\%\) yield of copper, what mass of copper is produced?

A 2.077 -g sample of an element, which has an atomic mass between 40 and \(55,\) reacts with oxygen to form 3.708 g of an oxide. Determine the formula of the oxide (and identify the element).

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What masses of iron(III) oxide and aluminum must be used to produce 15.0 \(\mathrm{g}\) iron? What is the maximum mass of aluminum oxide that could be produced?

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide $\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)$ could be produced? $\mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad$ (unbalanced)
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