A soluble fertiliser contains phosphorus in the form of phosphate ions (PO₄³⁻). To determine the PO₄³⁻ content by gravimetric analysis, 5.97 g of the fertiliser powder was completely dissolved in water to make a volume of 250.0 mL. A 20.00 mL volume of this solution was pipetted into a conical flask and the PO₄³⁻ ions in the solution were precipitated as MgNH₄PO₄. The precipitate was filtered, washed with water and then converted by heating into Mg₂P₂O₇. The mass of Mg₂P₂O₇ was 0.352 g. b. Calculate the amount, in mole, of phosphorus in the 20.00 mL volume of solution. c. Calculate the amount, in mole, of phosphorus in 5.97 g of fertiliser. d. Calculate the percentage of phosphate ions (PO₄³⁻) by mass in the fertiliser. Ensure you express your answer to an appropriate number of significant figures. e. i. Several actions which could occur during this analytical procedure are listed below (A–D). For each action, indicate the likely effect on the calculated percentage of phosphate ions in the fertiliser by placing a tick in the appropriate box. Action A. The MgNH₄PO₄ precipitate was not washed with water. B. The conical flask had been previously washed with water but not dried. C. A 25.00 mL pipette was unknowingly used instead of a 20.00 mL pipette. D. The mass of the fertiliser was recorded incorrectly. The recorded mass was 0.2 g. ii. In the case of action B above, explain your reasoning for the answer that you have given.

SSCE VCE Chemistry Measuring Solubility and Concentration: A soluble fertiliser contains ph

A soluble fertiliser contains phosphorus in the form of phosphate ions (PO₄³⁻) - VCE - SSCE Chemistry - Question 3 - 2006 - Paper 1

Question 3

A soluble fertiliser contains phosphorus in the form of phosphate ions (PO₄³⁻). To determine the PO₄³⁻ content by gravimetric analysis, 5.97 g of the fertiliser powd... show full transcript

Worked Solution & Example Answer:A soluble fertiliser contains phosphorus in the form of phosphate ions (PO₄³⁻) - VCE - SSCE Chemistry - Question 3 - 2006 - Paper 1

Step 1

a. Calculate the amount, in mole, of Mg₂P₂O₇.

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Answer

To find the amount in moles of Mg₂P₂O₇, use the formula:

$n = \frac{m}{M}$

Where:

$n$ is the number of moles
$m$ is the mass (0.352 g)
$M$ is the molar mass of Mg₂P₂O₇.

Calculating the molar mass:

Mg: 24.31 g/mol (2 atoms) = 48.62 g/mol
P: 30.97 g/mol (2 atoms) = 61.94 g/mol
O: 16.00 g/mol (7 atoms) = 112.00 g/mol

The total molar mass of Mg₂P₂O₇ is:

$M = 48.62 + 61.94 + 112.00 = 222.56 \, \text{g/mol}$

Now substitute values:

$n = \frac{0.352}{222.56} = 0.00158 \, \text{moles}$ .

Thus, the amount in moles of Mg₂P₂O₇ is approximately 0.00158 moles.

Step 2

b. Calculate the amount, in mole, of phosphorus in the 20.00 mL volume of solution.

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Answer

The amount of phosphorus can be calculated using the stoichiometry of the reaction. Each mole of MgNH₄PO₄ contains one mole of phosphorus.

From part a, the moles of MgNH₄PO₄ precipitated is equal to the moles of Mg₂P₂O₇ formed:

$n_{P} = 0.00158 \, \text{moles}.$

Therefore, there are also 0.00158 moles of phosphorus in the 20.00 mL volume of solution.

Step 3

c. Calculate the amount, in mole, of phosphorus in 5.97 g of fertiliser.

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Answer

From the initial solubility, the moles of phosphorus in the total sample can be calculated.

The 20.00 mL solution represents 1/12.5 (since 250 mL is the total), so:

$n_{fertiliser} = 0.00158 imes 12.5 = 0.01975 \, \text{moles}.$

Thus, the amount in moles of phosphorus in 5.97 g of fertiliser is approximately 0.01975 moles.

Step 4

d. Calculate the percentage of phosphate ions (PO₄³⁻) by mass in the fertiliser.

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Answer

The mass of phosphorus can be derived from the number of moles:

$m_{P} = n_{P} \times M_{P} = 0.01975 \, \text{moles} \times 30.97 \, \text{g/mol} = 0.612 \, \text{g}.$

The mass percentage of phosphate ions in the fertiliser:

$\text{Percentage} = \left( \frac{m_{P}}{m_{fertiliser}} \right) \times 100 = \left( \frac{0.612}{5.97} \right) \times 100 \approx 10.24\%.$

Step 5

e. i. A. The MgNH₄PO₄ precipitate was not washed with water.

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Answer

If the precipitate is not washed, impurities may be included in the mass measurement of Mg₂P₂O₇, leading to an inaccurately low calculation of the phosphate percentage.

Step 6

e. i. B. The conical flask had been previously washed with water but not dried.

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Water left in the flask will dilute the precipitate, resulting in a calculated percentage that is too low since the mass of the dry precipitate would appear to be less.

Step 7

e. i. C. A 25.00 mL pipette was unknowingly used instead of a 20.00 mL pipette.

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Using a larger volume will dilute the phosphates in the analysis. Therefore, the calculated percentage will be too low compared to using the correct volume.

Step 8

e. i. D. The mass of the fertiliser was recorded incorrectly. The recorded mass was 0.2 g.

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If the fertiliser mass is recorded too low, the percentage of phosphates will be calculated as higher than it truly is.

Step 9

e. ii. In the case of action B above, explain your reasoning for the answer that you have given.

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Answer

The residual water in the conical flask increases the overall mass without contributing any phosphate content, which distorts the actual mass of the Mg₂P₂O₇ precipitate, thereby leading to a lower calculated percentage of phosphate ions.

A soluble fertiliser contains phosphorus in the form of phosphate ions (PO₄³⁻) - VCE - SSCE Chemistry - Question 3 - 2006 - Paper 1

Worked Solution & Example Answer:A soluble fertiliser contains phosphorus in the form of phosphate ions (PO₄³⁻) - VCE - SSCE Chemistry - Question 3 - 2006 - Paper 1

a. Calculate the amount, in mole, of Mg₂P₂O₇.

b. Calculate the amount, in mole, of phosphorus in the 20.00 mL volume of solution.

c. Calculate the amount, in mole, of phosphorus in 5.97 g of fertiliser.

d. Calculate the percentage of phosphate ions (PO₄³⁻) by mass in the fertiliser.

e. i. A. The MgNH₄PO₄ precipitate was not washed with water.

e. i. B. The conical flask had been previously washed with water but not dried.

e. i. C. A 25.00 mL pipette was unknowingly used instead of a 20.00 mL pipette.

e. i. D. The mass of the fertiliser was recorded incorrectly. The recorded mass was 0.2 g.

e. ii. In the case of action B above, explain your reasoning for the answer that you have given.

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