Question

A sample of 0.3220 g of an ionic compound containing the iodide ion (I-) is dissolved in water and treated with an excess of AgHCO3. If the mass of the AgI precipitate that forms is 0.235 g, what is the percent by mass of I in the original compound? The molar mass of AgI is 234.8 g.

Answer 1

Answer:

39.44 %

Explanation:

Step 1: Write the balanced equation

I⁻(aq) + AgHCO₃(aq) = AgI(s) + HCO₃⁻(aq)

Step 2: Calculate the moles corresponding to 0.235 g of AgI

The molar mass of AgI is 234.8 g/mol.

$0.235g \times \frac{1mol}{234.8g} = 1.001 \times 10^{-3} mol$

Step 3: Calculate the moles of I⁻ that form 1.001 × 10⁻³ mol of AgI

The molar ratio of I⁻ to AgI is 1:1. The reacting moles of I⁻ are 1/1 × 1.001 × 10⁻³ mol = 1.001 × 10⁻³ mol

Step 4: Calculate the mass corresponding to 1.001 × 10⁻³ mol of I⁻

The molar mass of I⁻ is 126.90 g/mol.

$1.001 \times 10^{-3} mol \times \frac{126.90g}{mol} = 0.1270 g$

Step 5: Calculate the percent by mass of I⁻ in the original compound

$\frac{0.1270g}{0.3220g} \times 100 \% = 39.44 \%$