Problem 53 - Chemistry: Principles and Reactions

Question

A certain hydrate of potassium aluminum sulfate (alum) has the formula \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot x \mathrm{H}_{2} \mathrm{O}\). When a hydrate sample weighing \(5.459 \mathrm{~g}\) is heated to remove all the water, \(2.583 \mathrm{~g}\) of \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2}\) remains. What is the mass percent of water in the hydrate? What is \(x ?\)

Step-by-Step Solution

Answer

Answer: The mass percent of water in the hydrate is 52.70%, and the value of x is 29, giving the formula KAl(SO_4)_2∙29H_2O.

Step 1: Calculate the mass of water in the hydrate

To find the mass of water in the hydrate, subtract the mass of KAl(SO_4)_2 from the initial mass of the hydrate.#tag_math#m_{\text{water}} = m_{\text{hydrate}} - m_{\text{KAl(SO}_4\text{)}_2} = 5.459 - 2.583 = 2.876 \ \text{g}

Step 2: Calculate the mass percent of water in the hydrate

To find the mass percent of water, divide the mass of water by the initial mass of the hydrate, then multiply by 100.#tag_math#\text{Mass percent of water} = \frac{m_{\text{water}}}{m_{\text{hydrate}}} \times 100 = \frac{2.876}{5.459} \times 100 \approx 52.70 \%

Step 3: Determine the molecular weight of KAl(SO_4)_2 and H_2O

To find the value of x, we need the molecular weight of KAl(SO_4)_2 and H_2O. The molecular weights for potassium (K), aluminum (Al), sulfur (S), and oxygen (O) can be found in the periodic table.#tag_math#MW_{\text{KAl(SO}_4\text{)}_2} = 1\times \text{(K)} + 1\times \text{(Al)} + 2\times \text{(S)} + 8\times \text{(O)} = 1\times 39.1 +1\times 26.98 + 2\times 32.07 + 8\times 16.00 \approx 474.21 \ \text{g/mol}\\ MW_{\text{H}_2 \text{O}} = 2\times \text{(H)} + 1\times \text{(O)} = 2\times 1.008 + 1\times 16.00 \approx 18.02 \ \text{g/mol}

Step 4: Calculate the moles of KAl(SO_4)_2 and H_2O

Now, we'll find the moles of KAl(SO_4)_2 and H_2O by dividing their respective masses by their molecular weights.#tag_math#n_{\text{KAl(SO}_4\text{)}_2} = \frac{m_{\text{KAl(SO}_4\text{)}_2}}{MW_{\text{KAl(SO}_4\text{)}_2}} = \frac{2.583}{474.21} \approx 0.005442 \ \text{mol} \\ n_{\text{H}_2 \text{O}} = \frac{m_{\text{H}_2 \text{O}}}{MW_{\text{H}_2 \text{O}}} = \frac{2.876}{18.02} \approx 0.1596 \ \text{mol}

Step 5: Find the value of x

To find the value of x, divide the moles of H_2O by the moles of KAl(SO_4)_2. This will give the ratio of moles of water to moles of KAl(SO_4)_2 in the hydrate. Round the value of x to the nearest whole number.#tag_math#x = \frac{n_{\text{H}_2 \text{O}}}{n_{\text{KAl(SO}_4\text{)}_2}} = \frac{0.1596}{0.005442} = 29.32 \approx 29

Step 6: Final Answer

The mass percent of water in the hydrate is 52.70%, and the value of x is 29. Therefore, the formula for the hydrate is KAl(SO_4)_2∙29H_2O.

Key Concepts

Mass Percent Composition

Mass percent composition is crucial for chemists to understand the makeup of a compound. It denotes the fraction of a particular element or component within a substance, expressed as a percentage of the total mass. To calculate the mass percent of water in a hydrated salt, like the potassium aluminum sulfate hydrate in our exercise, you need to measure both the initial mass of the hydrate and the mass remaining after heating, which typically causes water to evaporate.

The formula to calculate mass percent is straightforward:
\[\begin{equation}\text{Mass percent of water} = \frac{m_{\text{water}}}{m_{\text{hydrate}}} \times 100 \end{equation}\]
where \(m_{\text{water}}\) is the mass of water that was in the hydrate, and \(m_{\text{hydrate}}\) is the initial total mass of the hydrate. In the given exercise, heating the sample resulted in water loss, which allowed us to compute that the mass percent of water in the alum hydrate was approximately 52.70%.

Molar Mass Calculation

Understanding Molar Mass

Molar mass is defined as the mass of one mole of any substance (elements or compounds). To find the molar mass of any chemical compound, one needs to sum the molar masses of the individual elements present in it, multiplied by their respective numbers in the formula unit. The unit for molar mass is grams per mole (g/mol). For hydrated salts, it's essential to include the water molecules in the molar mass calculation.

The periodic table provides atomic weights required for the computation which, when multiplied by the respective number of atoms in the molecule, gives the molar mass. For example:\[\begin{equation}MW_{\text{KAl(SO}_4\text{)}_2} = 1\times 39.1 +1\times 26.98 + 2\times 32.07 + 8\times 16.00 \end{equation}\]

• K = Potassium, with a molar mass of 39.1 g/mol,
• Al = Aluminum, 26.98 g/mol,
• S = Sulfur, 32.07 g/mol,
• O = Oxygen, 16.00 g/mol.

Adding these values together as shown in the exercise solution gives the molar mass of the anhydrous part of potassium aluminum sulfate hydrate (alum).

Empirical Formula Determination

The empirical formula is a simple whole number ratio defining proportion of elements in a compound. Hydrates have distinct empirical formulas that include water molecules as a ratio to the anhydrous salt. In order to compute the empirical formula, you first establish the number of moles of each element or component present. This involves using the molar masses of those components.

As applied in our exercise, to determine the exact formula of the hydrate, \(x\), which represents the number of water molecules associated with each formula unit of the salt, the ratio of moles of water to moles of salt must be calculated:\[\begin{equation}x = \frac{n_{\text{H}_2 \text{O}}}{n_{\text{KAl(SO}_4\text{)}_2}}\end{equation}\]
In this case, dividing the moles of water by the moles of KAl(SO4)2 as determined from their masses and molar masses, gives us a numerical value that corresponds to \(x\). Often, this value is rounded to the nearest whole number for empirical formula, because whole numbers represent the actual number of water molecules per formula unit. In this example, \(x\) is approximately 29, indicating the empirical formula KAl(SO4)2∙29H2O.