Question

Vanadyl sulfate (VOSO4, FM 163.00), as supplied commercially, is contaminated with H2SO4 (FM 98.08) and H2O. A solution was prepared by dissolving 0.244 7 g of impure VOSO4 in 50.0 mL of water. Spectrophotometric analysis indicated that the concentration of the blue VO2+ ion was 0.024 3 M. A 5.00-mL sample was passed through a cation-exchange column loaded with H+. When VO2+from the 5.00-mL sample became bound to the column, the H+ released required 13.03 mL of 0.022 74 M NaOH for titration.Find the weight percent of each component (VOSO4, H2SO4, and H2O) in the vanadyl sulfate.

Answer 1

The weight percent of each component in the impure
`\( VOSO_4 \)` sample is as follows:

-
`\( VOSO_4 \)`: 80.93%

-
`\( H_2SO_4 \)`: 5.94%

-
`\( H_2O \)`: 13.13%

To find the weight percent of each component in the impure vanadyl sulfate sample, we will need to follow these steps:

1. Calculate the moles of
`\( \text{VO}^(2+) \)` ion in the original 50.0 mL solution using the concentration obtained from spectrophotometric analysis.

2. Calculate the mass of pure
`\( \text{VOSO}_4 \)` that corresponds to the moles of
`\( \text{VO}^(2+) \)` ion.

3. Calculate the moles of
`\( \text{H}^+ \)` released from the cation-exchange column, which is equal to the moles of
`\( \text{NaOH} \)` used in the titration.

4. Calculate the mass of
`\( \text{H}_2\text{SO}_4 \)` that corresponds to the moles of
`\( \text{H}^+ \)` ions.

5. Subtract the mass of
`\( \text{VOSO}_4 \)` and
`\( \text{H}_2\text{SO}_4 \)` from the initial mass of the impure sample to find the mass of
`\( \text{H}_2\text{O} \)`.

6. Calculate the weight percent of each component.

Let's start with step 1:

1. Calculate the moles of
`\( \text{VO}^(2+) \)` ion in the original 50.0 mL solution:

The concentration of
`\( \text{VO}^(2+) \)` is given as 0.0243 M, which means there are 0.0243 moles of
`\( \text{VO}^(2+) \)` per liter of solution. Since we have 50.0 mL, we need to adjust for the volume:

`\[ \text{Moles of } \text{VO}^(2+) = 0.0243 \, \text{M} * \frac{50.0 \, \text{mL}}{1000 \, \text{mL/L}} \]`

Let's do this calculation.

The moles of
`\( VO^(2+) \)` ion in the original 50.0 mL solution are 0.001215 moles.

2. Calculate the mass of pure
`\( VOSO_4 \)` that corresponds to the moles of
`\( VO^(2+) \)` ion:

To find the mass of
`\( VOSO_4 \)`, we multiply the moles of
`\( VO^(2+) \)` by the formula weight (FM) of
`\( VOSO_4 \)`.

`\[ \text{Mass of } VOSO_4 = \text{moles of } VO^(2+) * \text{FM of } VOSO_4 \]`

The formula weight (FM) of
`\( VOSO_4 \)` is given as 163.00 g/mol.

`\[ \text{Mass of } VOSO_4 = 0.001215 \text{ mol} * 163.00 \text{ g/mol} \]`

Let's calculate this.

The mass of pure
`\( VOSO_4 \)` that corresponds to the moles of
`\( VO^(2+) \)` ion is 0.1980 g.

3. Calculate the moles of
`\( H^+ \)` released from the cation-exchange column, which is equal to the moles of
`\( NaOH \)` used in the titration:

`\[ \text{Moles of } NaOH = \text{Volume of } NaOH * \text{Molarity of } NaOH \]`

`\[ \text{Moles of } NaOH = 13.03 \text{ mL} * \frac{0.02274 \text{ mol}}{1000 \text{ mL}} \]`

Let's calculate the moles of
`\( NaOH \).`

The moles of
`\( NaOH \)` used in the titration (and thus the moles of
`\( H^+ \)` released) are 0.0002963 moles.

4. Calculate the mass of
`\( H_2SO_4 \)` that corresponds to the moles of \( H^+ \) ions. Each mole of
`\( H_2SO_4 \)` releases 2 moles of \( H^+ \), so we need to halve the moles of
`\( NaOH \)` to find the moles of
`\( H_2SO_4 \):`

`\[ \text{Moles of } H_2SO_4 = \frac{\text{Moles of } NaOH}{2} \]`

Then, we calculate the mass:

`\[ \text{Mass of } H_2SO_4 = \text{Moles of } H_2SO_4 * \text{FM of } H_2SO_4 \]`

The formula weight (FM) of
`\( H_2SO_4 \)` is given as 98.08 g/mol.

Let's perform these calculations.

The moles of
`\( H_2SO_4 \)` are 0.0001481 moles, and the corresponding mass of
`\( H_2SO_4 \)` is approximately 0.01453 g.

5. Subtract the mass of
`\( VOSO_4 \)` and
`\( H_2SO_4 \)` from the initial mass of the impure sample to find the mass of
`\( H_2O \):`

`\[ \text{Mass of impure } VOSO_4 \text{ sample} - (\text{Mass of } VOSO_4 + \text{Mass of } H_2SO_4) = \text{Mass of } H_2O \]`

The initial mass of the impure
`\( VOSO_4 \)` sample is given as 0.2447 g.

`\[ \text{Mass of } H_2O = 0.2447 \text{ g} - (0.1980 \text{ g} + 0.01453 \text{ g}) \]`

Let's calculate the mass of
`\( H_2O \)`.

The mass of
`\( H_2O \)` in the impure
`\( VOSO_4 \)` sample is approximately 0.03212 g.

6. Calculate the weight percent of each component:

`\[ \text{Weight percent of } VOSO_4 = \left( \frac{\text{Mass of } VOSO_4}{\text{Mass of impure sample}} \right) * 100 \]`

`\[ \text{Weight percent of } H_2SO_4 = \left( \frac{\text{Mass of } H_2SO_4}{\text{Mass of impure sample}} \right) * 100 \]`

`\[ \text{Weight percent of } H_2O = \left( \frac{\text{Mass of } H_2O}{\text{Mass of impure sample}} \right) * 100 \]`

The weight percent of each component in the impure
`\( VOSO_4 \)` sample is as follows:

-
`\( VOSO_4 \)`: 80.93%

-
`\( H_2SO_4 \)`: 5.94%

-
`\( H_2O \)`: 13.13%

Answer 2

The weight percent of VOSO₄ is 8.07%, the weight percent of H₂SO₄ is 4.87%, and the weight percent of H₂O is 6.03% in the vanadyl sulfate solution.

Let's perform the calculations step by step:

Step 1: moles of VO₂⁺ ions in the sample:

Mole = molarity x volume

`\[ \text{Moles of VO}_2^+ = 0.0243 \, \text{M} * 0.005 \, \text{L} = 0.0001215 \, \text{mol} \]`

Step 2: moles of VOSO₄:

`\[ \text{Moles of VOSO}_4 = 0.0001215 \, \text{mol} \]`

Step 3: moles of H₂SO₄ released during titration:

`\[ \text{Moles of H}_2\text{SO}_4 = 0.0001215 \, \text{mol} \]`

Step 4: moles of H₂O in the solution:

`\[ \text{Moles of H}_2\text{O} = \frac{0.2447 \, \text{g}}{163.00 \, \text{g/mol}} - 0.0001215 \, \text{mol} - 0.0001215 \, \text{mol} = 0.0008193 \, \text{mol} \]`

Step 5: Mass of each component:

`\[ \text{Mass of VOSO}_4 = 0.0001215 \, \text{mol} * 163.00 \, \text{g/mol} = 0.01975 \, \text{g} \]`

`\[ \text{Mass of H}_2\text{SO}_4 = 0.0001215 \, \text{mol} * 98.08 \, \text{g/mol} = 0.01192 \, \text{g} \]`

`\[ \text{Mass of H}_2\text{O} = 0.0008193 \, \text{mol} * 18.02 \, \text{g/mol} = 0.01476 \, \text{g} \]`

Finally, the weight percent of each component can be calculated as:

`\[ \text{Weight percent of VOSO}_4 = \left( \frac{0.01975 \, \text{g}}{0.2447 \, \text{g}} \right) * 100 = 8.07\% \]`

`\[ \text{Weight percent of H}_2\text{SO}_4 = \left( \frac{0.01192 \, \text{g}}{0.2447 \, \text{g}} \right) * 100 = 4.87\% \]`

`\[ \text{Weight percent of H}_2\text{O} = \left( \frac{0.01476 \, \text{g}}{0.2447 \, \text{g}} \right) * 100 = 6.03\% \]`

So, the weight percent of VOSO₄ is 8.07%, the weight percent of H₂SO₄ is 4.87%, and the weight percent of H₂O is 6.03% in the vanadyl sulfate solution.