Questions: Copper(II) sulfate forms several hydrates with the general formula CuSO4 · x H2O, where x is an integer. If the hydrate is heated, the water can be driven off, leaving pure CuSO4, behind. Suppose a sample of a certain hydrate is heated until all the water is removed, and it's found that the mass of the sample decreases by 18%. Which hydrate is it? That is, what is x?

Determine the value of \( x \) in the hydrate \(\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\) given that the mass of the sample decreases by \(18\%\) upon heating.

Define the initial and final masses.

Let the initial mass of the hydrate be \( m \). After heating, the mass decreases by \(18\%\), so the final mass is \( 0.82m \).

Express the mass of the hydrate and the anhydrous compound.

The mass of the hydrate is \( m = M_{\mathrm{CuSO}_{4}} + xM_{\mathrm{H}_{2} \mathrm{O}} \), where \( M_{\mathrm{CuSO}_{4}} \) is the molar mass of \(\mathrm{CuSO}_{4}\) and \( M_{\mathrm{H}_{2} \mathrm{O}} \) is the molar mass of water.

Calculate the molar masses.

The molar mass of \(\mathrm{CuSO}_{4}\) is \( 63.55 + 32.07 + 4 \times 16.00 = 159.62 \, \text{g/mol} \).

The molar mass of \(\mathrm{H}_{2} \mathrm{O}\) is \( 2 \times 1.01 + 16.00 = 18.02 \, \text{g/mol} \).

Set up the equation for the mass decrease.

The mass of the anhydrous \(\mathrm{CuSO}_{4}\) is \( 0.82m \).

Thus, \( 0.82m = M_{\mathrm{CuSO}_{4}} \).

Substituting \( m = M_{\mathrm{CuSO}_{4}} + xM_{\mathrm{H}_{2} \mathrm{O}} \):

\( 0.82(M_{\mathrm{CuSO}_{4}} + xM_{\mathrm{H}_{2} \mathrm{O}}) = M_{\mathrm{CuSO}_{4}} \).

Solve for \( x \).

Rearrange the equation:

\( 0.82M_{\mathrm{CuSO}_{4}} + 0.82xM_{\mathrm{H}_{2} \mathrm{O}} = M_{\mathrm{CuSO}_{4}} \)

\( 0.82xM_{\mathrm{H}_{2} \mathrm{O}} = M_{\mathrm{CuSO}_{4}} - 0.82M_{\mathrm{CuSO}_{4}} \)

\( 0.82xM_{\mathrm{H}_{2} \mathrm{O}} = 0.18M_{\mathrm{CuSO}_{4}} \)

\( x = \frac{0.18M_{\mathrm{CuSO}_{4}}}{0.82M_{\mathrm{H}_{2} \mathrm{O}}} \)

Substitute the molar masses:

\( x = \frac{0.18 \times 159.62}{0.82 \times 18.02} \)

\( x \approx 1.98 \)

Since \( x \) must be an integer, we round to the nearest whole number:

\( x = 2 \).

\(\boxed{x = 2}\)

\(\boxed{x = 2}\)