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The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown - HSC - SSCE Chemistry - Question 29 - 2015 - Paper 1

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Question 29

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The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown. 2.00 g of a sampl... show full transcript

Worked Solution & Example Answer:The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown - HSC - SSCE Chemistry - Question 29 - 2015 - Paper 1

Step 1

Suggest modifications that could be made to the procedure to improve the results of this investigation. Justify your suggestions.

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Answer

  1. Heating the Mixture: After adding barium chloride, the mixture could be heated for some time. This would allow the fine barium sulfate particles to collide and form larger aggregates, which are more easily filtered.

  2. Using a Sintered Glass Filter: Instead of regular filter paper, using a sintered glass filter can prevent fine particles from passing through, ensuring a more complete collection of the precipitate.

  3. Taking Multiple Samples: To achieve more reliable results, multiple samples of the fertiliser could be tested and averaged. This would account for any discrepancies in concentration across the sample.

These modifications would help ensure that the precipitate is maximally collected and that the measurement of sulfate concentration is accurate.

Step 2

Calculate the percentage of sulfate in the original fertiliser sample.

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Answer

To calculate the percentage of sulfate in the fertiliser, we first note the reaction for the formation of barium sulfate:

Ba2+(aq)+SO42(aq)BaSO4(s)\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)

From the problem, the mass of precipitate (barium sulfate) collected is 2.23 g. Now, we calculate the moles of barium sulfate:

  • Molar mass of ( \text{BaSO}_4 ): 233.37 g/mol.

Calculating moles:

moles of BaSO4=2.23 g233.37 g/mol=0.00954 mol\text{moles of BaSO}_4 = \frac{2.23 \text{ g}}{233.37 \text{ g/mol}} = 0.00954\text{ mol}

Now let's find the mass of sulfate:

  • Molar mass of ( \text{SO}_4^{2-} ): 96.07 g/mol.

Calculating mass of sulfate:

mass of SO42=0.00954 mol×96.07 g/mol=0.9175 g\text{mass of SO}_4^{2-} = 0.00954 \text{ mol} \times 96.07 \text{ g/mol} = 0.9175 \text{ g}

Finally, we can calculate the percentage of sulfate in the original fertiliser sample:

Percentage of SO42=0.9175 g2.00 g×100=45.88%\text{Percentage of SO}_4^{2-} = \frac{0.9175 \text{ g}}{2.00 \text{ g}} \times 100 = 45.88\%

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