Problem 68 - Chemistry and Chemical Reactivity

Question

Which of the following samples has the largest number of ions? (a) \(1.0 \mathrm{g}\) of \(\mathrm{BeCl}_{2}\) (d) \(1.0 \mathrm{g}\) of \(\mathrm{SrCO}_{3}\) (b) \(1.0 \mathrm{g}\) of \(\mathrm{MgCl}_{2}\) (e) \(1.0 \mathrm{g}\) of \(\mathrm{BaSO}_{4}\) (c) \(1.0 \mathrm{g}\) of \(\mathrm{CaS}\)

Step-by-Step Solution

Answer

BeCl₂ has the largest number of ions.

Step 1: Calculate Molar Mass for Each Compound

First, find the molar mass of each compound.- For \( \text{BeCl}_2 \): \( \text{Be} = 9.01 \; \text{g/mol}, \text{Cl} = 35.45 \; \text{g/mol} \times 2 \). Molar mass = \( 9.01 + 2 \times 35.45 = 79.91 \; \text{g/mol} \).- For \( \text{SrCO}_3 \): \( \text{Sr} = 87.62 \; \text{g/mol}, \text{C} = 12.01 \; \text{g/mol}, \text{O} = 16.00 \; \text{g/mol} \times 3 \). Molar mass = \( 87.62 + 12.01 + 3 \times 16.00 = 147.63 \; \text{g/mol} \).- For \( \text{MgCl}_2 \): \( \text{Mg} = 24.31 \; \text{g/mol}, \text{Cl} = 35.45 \; \text{g/mol} \times 2 \). Molar mass = \( 24.31 + 2 \times 35.45 = 95.21 \; \text{g/mol} \).- For \( \text{BaSO}_4 \): \( \text{Ba} = 137.33 \; \text{g/mol}, \text{S} = 32.07 \; \text{g/mol}, \text{O} = 16.00 \; \text{g/mol} \times 4 \). Molar mass = \( 137.33 + 32.07 + 4 \times 16.00 = 233.39 \; \text{g/mol} \).- For \( \text{CaS} \): \( \text{Ca} = 40.08 \; \text{g/mol}, \text{S} = 32.07 \; \text{g/mol} \). Molar mass = \( 40.08 + 32.07 = 72.15 \; \text{g/mol} \).

Step 2: Calculate Moles of Each Compound

Use the formula \( n = \frac{m}{M} \) where \( n \) is the number of moles, \( m \) is the mass (1.0 g), and \( M \) is the molar mass.- \( \text{BeCl}_2: n = \frac{1.0}{79.91} = 0.0125 \; \text{mol} \).- \( \text{SrCO}_3: n = \frac{1.0}{147.63} = 0.0068 \; \text{mol} \).- \( \text{MgCl}_2: n = \frac{1.0}{95.21} = 0.0105 \; \text{mol} \).- \( \text{BaSO}_4: n = \frac{1.0}{233.39} = 0.0043 \; \text{mol} \).- \( \text{CaS}: n = \frac{1.0}{72.15} = 0.0139 \; \text{mol} \).

Step 3: Determine Number of Ions per Mole

Count the ions produced from the dissociation of one formula unit.- \( \text{BeCl}_2 \) dissociates into 1 \( \text{Be}^{2+} \) and 2 \( \text{Cl}^- \) ions (3 ions total).- \( \text{SrCO}_3 \) dissociates into 1 \( \text{Sr}^{2+} \) and 1 \( \text{CO}_3^{2-} \) ions (2 ions total).- \( \text{MgCl}_2 \) dissociates into 1 \( \text{Mg}^{2+} \) and 2 \( \text{Cl}^- \) ions (3 ions total).- \( \text{BaSO}_4 \) is largely insoluble, dissociating into minimal ions.- \( \text{CaS} \) dissociates into 1 \( \text{Ca}^{2+} \) and 1 \( \text{S}^{2-} \) ions (2 ions total).

Step 4: Calculate Total Number of Ions

Multiply the moles of each compound by the number of ions per mole.- \( \text{BeCl}_2: 0.0125 \times 3 = 0.0375 \; \text{mol ions} \).- \( \text{SrCO}_3: 0.0068 \times 2 = 0.0136 \; \text{mol ions} \).- \( \text{MgCl}_2: 0.0105 \times 3 = 0.0315 \; \text{mol ions} \).- \( \text{BaSO}_4: ≈ 0 \; \text{mol ions} \) (insoluble).- \( \text{CaS}: 0.0139 \times 2 = 0.0278 \; \text{mol ions} \).

Step 5: Conclusion

The compound with the largest number of ions is \( \text{BeCl}_2 \) with 0.0375 mol ions in 1.0 g.

Key Concepts

Molar Mass Calculation

Understanding molar mass is essential for solving chemistry problems. Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It’s calculated by adding up the atomic masses of all the atoms in a molecule. For instance, to find the molar mass of \( \text{BeCl}_2 \):

• Beryllium (Be) has an atomic mass of 9.01 g/mol.
• Chlorine (Cl) is 35.45 g/mol, and since there are two chlorine atoms, it becomes \(2 \times 35.45 = 70.90\) g/mol.

Add these together: \(9.01 + 70.90 = 79.91\) g/mol. You repeat this process for each compound, ensuring that every element's mass is correctly included. This calculation is vital for determining how much of a substance you have in moles.

Ion Dissociation

Ion dissociation is the process by which a compound separates into its constituent ions when dissolved in water. This concept is crucial for understanding how to calculate the number of ions present in a solution. Each ionic compound will dissociate differently based on its chemical formula. For example:

• \( \text{BeCl}_2 \) dissociates into 1 beryllium ion \( \text{Be}^{2+} \) and 2 chloride ions \( \text{Cl}^- \), producing a total of 3 ions.
• \( \text{SrCO}_3 \) splits into 1 strontium ion \( \text{Sr}^{2+} \) and 1 carbonate ion \( \text{CO}_3^{2-} \), totaling 2 ions.
• Note that some compounds, like \( \text{BaSO}_4 \), are mostly insoluble and do not dissociate significantly in water.

Understanding this helps in calculating the total ions, as only dissociated ions affect the ion count.

Mole Calculation

Mole calculations allow chemists to quantify substances and assess reactions in terms of moles rather than grams. The basic equation for finding moles is \( n = \frac{m}{M} \), where:

• \( n \) is the number of moles.
• \( m \) is the mass of the substance (in grams).
• \( M \) is the molar mass (in g/mol).

Suppose you have 1.0 g of \( \text{MgCl}_2 \) with a molar mass of 95.21 g/mol. To find the moles:

• Calculate: \( n = \frac{1.0}{95.21} \approx 0.0105 \text{ mol} \).

Once you know the number of moles, you can further determine the number of ions by multiplying by the ions per mole resulting from dissociation. This way, you can find out which sample contains the largest number of ions.