Question

The crystalline salt \(\mathrm{Na}_{2} \mathrm{SO}_{4} \times \mathrm{xH}_{2} \mathrm{O}\) on heating loses \(55.9 \%\) of its weight. The formula of the crystalline salt is a. \(\mathrm{Na}_{2} \mathrm{SO}_{4}, 10 \mathrm{H}_{2} \mathrm{O}\) b. \(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{Na}_{2} \mathrm{SO}_{4} .7 \mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{Na}_{2} \mathrm{SO}_{4} .5 \mathrm{H}_{2} \mathrm{O}\) e. \(\mathrm{Na}_{2} \mathrm{SO}_{s} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

Short answer

The formula is \( \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O} \).

Step-by-step solution

Understand the problem

We need to determine the number of water molecules (x) in the hydrate of sodium sulfate, \( \mathrm{Na}_{2} \mathrm{SO}_{4} \times \mathrm{xH}_{2} \mathrm{O} \), by finding which formula results in a weight loss of 55.9% upon heating.

Calculate Molecular Weights

Calculate the molecular weights: \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) is \( 2 \times 23 + 32 + 4 \times 16 = 142 \) g/mol, and \( \mathrm{H}_{2} \mathrm{O} \) is \( 2 + 16 = 18 \) g/mol.

Set Up Weight Ratio Equation

Assume the initial molecular weight of the hydrate to be \( 142 + 18x \). When the water is lost: Final weight = \( 142 \). Weight loss = \( 142 + 18x - 142 = 18x \). Set up the equation for percentage weight loss: \[ \frac{18x}{142 + 18x} = 0.559 \]

Solve the Equation for x

Solve for \( x \) in \( \frac{18x}{142 + 18x} = 0.559 \). Cross-multiply to get \( 18x = 0.559(142 + 18x) \). Expand to \( 18x = 79.338 + 10.062x \). Simplify and solve to find \( x = 10 \).

Match with Given Options

The calculated \( x \) equals 10, thus the formula of the crystalline salt is \( \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O} \). Check this option against the given choices.

Key concepts

Molecular Weight Calculation

In chemistry, understanding the molecular weight of a compound is fundamental in solving problems, particularly in stoichiometry and determining chemical formulas.

The molecular weight (or molar mass) of a compound is the sum of the atomic weights of all atoms in a molecule. It is expressed in grams per mole (g/mol). For example, sodium sulfate (\(\text{Na}_2\text{SO}_4\)) is composed of:

• 2 sodium atoms: \(23 \text{ g/mol} \times 2 = 46 \text{ g/mol}\)
• 1 sulfur atom: \(32 \text{ g/mol}\)
• 4 oxygen atoms: \(16 \text{ g/mol} \times 4 = 64 \text{ g/mol}\)

Adding these together gives the molecular weight of sodium sulfate:

\[46 + 32 + 64 = 142 \text{ g/mol}\]

Similarly, a water molecule (\(\text{H}_2\text{O}\)) has a molecular weight of:

• 2 hydrogen atoms: \(1 \text{ g/mol} \times 2 = 2 \text{ g/mol}\)
• 1 oxygen atom: \(16 \text{ g/mol}\)

Thus, the molecular weight of water is:
\[2 + 16 = 18 \text{ g/mol}\]

With these calculations, one can confidently determine the formulas of hydrates and other compounds by comparing the calculated weights with experimental data.

Hydrate Water Content

When dealing with hydrated salts, one must account for the water molecules bonded to the compound.

These water molecules, known as "water of hydration," can be quantitatively significant. Their presence in a crystalline salt affects the overall molecular weight and influences chemical properties and reactions. In our earlier example regarding sodium sulfate hydrate, realizing how many water molecules are present is essential for determining its complete formula.

To find the water content, weigh the hydrate before and after heating to remove the water. The change in weight reflects how much water was present. This is quantified by the formula:
\[\text{Percentage Weight Loss} = \frac{\text{Weight of Water Lost}}{\text{Initial Weight of Hydrate}} \times 100\]
Applying this concept allows for finding how many water molecules were initially present. For example, if a compound shows a 55.9% weight loss when heated, we'll compute how many units of\(\text{H}_2\text{O}\)are associated with the crystalline salt. Solving this weight percentage loss gives us the formula for the hydrate, as seen when we established that sodium sulfate has ten water molecules bonded.

Sodium Sulfate Hydration

Sodium sulfate (\(\text{Na}_2\text{SO}_4\)) is a versatile and widely used chemical, known for forming hydrates—compounds in which sodium sulfate is bonded to water molecules.

The most recognized hydrate of sodium sulfate is sodium sulfate decahydrate, also known as Glauber's salt, where ten water molecules are associated with each formula unit of sodium sulfate.
This compound's formula is \(\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}\), emphasizing the importance of correctly determining the number of water molecules via methods like weight loss calculation during heating.

• Low-cost and readily available, sodium sulfate hydrates play an essential role in various industries, such as glass manufacturing and as a filler in powdered detergents.
• Glauber's salt is also used for its easy water dissolution and cooling properties, important in thermal energy storage.

The formation and dehydration of these hydrates are crucial for both industrial usage and educational purposes. Understanding sodium sulfate's ability to form hydrates highlights the complex interactions between ionic salts and water, providing deeper insights into chemical bonding and physical properties.