Question

\(5.0 \mathrm{~g}\) of a certain metal, \(\mathrm{X}\) (atomic mass \(=27\) ) is converted into \(61.7 \%\) crystalline sulphate containing \(48.6 \%\) by mass of water of crystallization. The simplest formula of the compound is: (a) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} 18 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} 14 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} 16 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} 12 \mathrm{H}_{2} \mathrm{O}\)

Short answer

The compound's simplest formula is (d) \(\mathrm{X}_2(\mathrm{SO}_4)_3 12 \mathrm{H}_2\mathrm{O}\).

Step-by-step solution

Determine the mass of anhydrous sulfate

Given that 61.7% of the compound is crystalline sulphate, and 5.0 g of metal X is used to make the compound, we calculate the mass of crystalline sulphate produced.Let the total mass of the compound be \( m \). Thus, \( m \times 0.617 \) is the mass of the crystalline sulphate, and the remaining is water of crystallization. The mass of sulphate: \[ m \times 0.617 = 5.0 \text{ g} \]Solving for \( m \), we get:\[ m = \frac{5.0}{0.617} \approx 8.1 \text{ g} \]

Calculate the mass of water in the compound

We already need water's percentage by mass in the compound. Given: water is 48.6% by mass. So, \[ m \times 0.486 \] is the mass of water.Plugging in the values: \[ 8.1 \text{ g} \times 0.486 \approx 3.93 \text{ g} \]

Find the mass of sulfate and the composition with Element X

At first, find the mass of sulfate (as the metal in it can be deduced by calculating the part left after excluding water).Mass of anhydrous sulfate: \[ 8.1 \text{ g} - 3.93 \text{ g} = 4.17 \text{ g} \]Now, as 5.0 g of metal \( X \) has formed this compound, it intends to be totally combined with metal sulfate \( SO_4^2- \).

Calculate Moles of Sulphate and concluding metal linkage

Given: Atomic mass of \( X \) (where X = Metal) = 27, and its mass = 5.0 gNow calculate the moles of \( X \):\[ \therefore \text{Moles of } X = \frac{5}{27} \approx 0.185 \text{ moles} \]Assuming the general formula, \( \text{X}_a(\text{SO}_4)_b\cdot n\text{H}_2\text{O} \):As both X-metal and sulfate mass suggest, - \( a \) being based on moles of element- \( b \) occurs balanced general sulfate such as in sulphate anion being -8 mass For resulting water via hydration:\[ \left( \frac{4.17}{80+64} \right) \approx 0.058 \text{ moles sulfate}\]Balance to prepare:\( [\mathrm{X}]_2\left(\mathrm{SO}_4\right)_3 \cdot n\mathrm{H}_2\mathrm{O} \) which makes .\( n = 12 \) approximated

Key concepts

Crystalline compounds

Crystalline compounds are a fascinating concept in chemistry, referring to solids whose atoms are arranged in a highly ordered, repeating pattern. This structural order translates into a defined crystal lattice, which gives these substances unique properties such as determined melting points, specific electrical conductivity, and predictable fracture patterns.
They are not just lab phenomena—they play a significant role in everyday materials. Several minerals and gemstones, for instance, are crystalline, resulting in their durable, sometimes dazzling properties. These compounds can also be synthetic, forming the basis for materials like semiconductors. In the context of stoichiometry, understanding crystalline compounds is vital, as it allows chemists to predict how these substances interact in reactions, and stoichiometry helps in determining the proportions of elements in these compounds.

Hydration in compounds

Hydration refers to the inclusion of water molecules within the structure of a compound. In crystalline compounds, this usually appears as water of crystallization, where water molecules are a part of the crystal lattice but not bonded by strong chemical bonds as in typical hydrogen bonds.
The presence of water significantly affects the appearance and properties of a compound. For example, cobalt(II) chloride is blue when anhydrous but transforms to a pinkish color when hydrated. Recognizing hydration is key when determining the empirical formula of hydrated compounds in stoichiometry. The proportion of water can influence the mass and balance of a chemical reaction, and thus it's crucial to account for it accurately when performing calculations to ensure precision in determining chemical formulas.

Chemical formula determination

Determining the chemical formula of a compound involves understanding the proportion of elements within it. This process often involves steps such as:

• Calculating masses of elements present in the compound.
• Determining the moles of each element based on their atomic masses.
• Finding the simplest ratio of these elements which forms the empirical formula.

Stoichiometry provides the mathematical framework to perform these calculations. For example, in hydrate compounds, it’s essential to calculate both the anhydrous part and the water content separately before combining them to get the complete formula.
In educational setups, exercises such as determining the simplest formula of hydrated crystalline compounds solidify students' understanding of how stoichiometric coefficients reflect quantities on a molecular scale. Accurate formula determination enables chemists to write balanced chemical equations, which are fundamental in predicting the outcomes of reactions.