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CHEMISTRY, M8 2025 HSC 23

A student attempted to determine the % w/w of sulfate in a sample of solid fertiliser. They used the procedure described below.

  1. Weigh a clean, dry beaker.
  2. Add fertiliser to the beaker and weigh again.
  3. Add 250 mL of distilled water and stir thoroughly.
  4. Add 20 mL of 0.1 mol L\(^{-1} \ \ce{BaCl2}\)  solution.
  5. Filter out the \(\ce{BaSO4}\) precipitate, using distilled water to ensure all of the solid is transferred from the beaker to the filter paper.
  6. Put the filter paper and precipitate onto a weighed watch glass and leave them to dry for 20 minutes in the sun.
  7. Weigh the watch glass, the filter paper and the precipitate.
  8. Calculate the % w/w.

Justify TWO changes that can be made to the procedure to ensure more accurate results.   (3 marks)

Show Answers Only

Change 1: Dry the precipitate in an oven or desiccator instead of in the sun.

  • Drying in the sun is inconsistent and may leave moisture in the precipitate, causing an overestimation of mass.
  • Using an oven at a controlled temperature ensures all water is removed, giving a more accurate mass of precipitate.
  • This would also prevent impurities from the air such as dust or particles mixing with the sample.

Change 2: Measure the mass of the filter paper initially without the precipitate.

  • Weighing the filter paper, precipitate and watch glass together without knowing the mass of the precipitate (the watch glass has been weighed) inflates the mass of the \(\ce{BaSO4}\) causing inaccuracy in the results. 

Other answers could include:

  • Continue to dry and reweigh the sample for over 20 minutes until a constant mass is obtained.
  • Increase the concentration of \(\ce{BaCl2}\) to ensure all sulfate ions react.
Show Worked Solution

Change 1: Dry the precipitate in an oven or desiccator instead of in the sun.

  • Drying in the sun is inconsistent and may leave moisture in the precipitate, causing an overestimation of mass.
  • Using an oven at a controlled temperature ensures all water is removed, giving a more accurate mass of precipitate.
  • This would also prevent impurities from the air such as dust or particles mixing with the sample.

Change 2: Measure the mass of the filter paper initially without the precipitate.

  • Weighing the filter paper, precipitate and watch glass together without knowing the mass of the precipitate (the watch glass has been weighed) inflates the mass of the \(\ce{BaSO4}\) causing inaccuracy in the results. 

Other answers could include:

  • Continue to dry and reweigh the sample for over 20 minutes until a constant mass is obtained.
  • Increase the concentration of \(\ce{BaCl2}\) to ensure all sulfate ions react.

CHEMISTRY, M8 2013 VCE 2

The strength of the eggshell of birds is determined by the calcium carbonate, \(\ce{CaCO3}\), content of the eggshell.

The percentage of calcium carbonate in the eggshell can be determined by gravimetric analysis.

0.412 g of clean, dry eggshell was completely dissolved in a minimum volume of dilute hydrochloric acid.

\(\ce{CaCO3(s) + 2H+(aq)\rightarrow Ca^2+(aq) + CO2(g) + H2O(l)}\)

An excess of a basic solution of ammonium oxalate, \(\ce{(NH4)2C2O4}\), was then added to form crystals of calcium oxalate monohydrate, \(\ce{CaC2O4.H2O}\).

The suspension was filtered and the crystals were then dried to constant mass.

0.523 g of \(\ce{CaC2O4.H2O}\) was collected.

  1. Write a balanced equation for the formation of the calcium oxalate monohydrate precipitate.   (1 mark)

  2. Determine the percentage, by mass, of calcium carbonate in the eggshell.   (3 marks)

Show Answers Only

a.    \(\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l) \rightarrow CaC2O4.H2O(s) + 2NH4^+(aq)}\)

b.    \(86.9\)%

Show Worked Solution

a.    \(\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l)} \rightarrow \)

\(\ce{CaC2O4.H2O(s) + 2NH4^+(aq)}\)
 

♦♦♦ Mean mark (a) 30%.

b.    \(\ce{M(CaC2O4.H2O)= 40.08 + 2(12.01) + 4(16) + 2(1.008) + 16 = 146.116\ \text{g mol}^{-1}}\)

\(\ce{n(CaC2O4.H2O)=\dfrac{0.523}{146.116}=0.003579\ \text{mol}}\)

\(\ce{n(CaC2O4.H2O) = n(Ca^2+)= n(CaCO3) = 0.003579\ \text{mol}}\)

\(\ce{m(CaCO3)=0.003579 \times (40.08 +12.01 + 3(16))= 0.358\ \text{g}}\)

\(\text{% Mass}\ =\dfrac{0.358}{0.412} \times 100 = 86.9\%\)

CHEMISTRY, M8 2016 VCE 2*

A common iron ore, fool’s gold, contains the mineral iron pyrite, \(\ce{FeS2}\).

Typically, the percentage by mass of \(\ce{FeS2}\) in a sample of fool’s gold is between 90% and 95%. The actual percentage in a sample can be determined by gravimetric analysis.

The sulfur in \(\ce{FeS2}\) is converted to sulfate ions, \(\ce{SO4^2–}\) as seen below:

\(\ce{4FeS2 + 11O2 \rightarrow 2Fe2O3 + 8SO4^2-}\)

This is then mixed with an excess of barium chloride, \(\ce{BaCl2}\), to form barium sulfate, \(\ce{BaSO4}\), according to the equation

\(\ce{Ba^2+(aq) + SO4^2–(aq)\rightarrow BaSO4(s)}\)

When the reaction has gone to completion, the \(\ce{BaSO4}\) precipitate is collected in a filter paper and carefully washed. The filter paper and its contents are then transferred to a crucible. The crucible and its contents are heated until constant mass is achieved.

The data for an analysis of a mineral sample is as follows.
 

\(\text{initial mass of mineral sample}\) \(\text{14.3 g}\)
\(\text{mass of crucible and filter paper}\) \(\text{123.40 g}\)
\(\text{mass of crucible, filter paper and dry}\ \ce{BaSO4}\) \(\text{174.99 g}\)
\(\ce{M(FeS2)}\) \(\text{120.0 g mol}^{-1}\)
\(\ce{M(BaCl2)}\) \(\text{208.3 g mol}^{-1}\)
\(\ce{M(BaSO4)}\) \(\text{233.4 g mol}^{-1}\)
     
  1. Calculate the percentage by mass of \(\ce{FeS2}\) in this mineral sample.  (4 marks)

  2. State one assumption that was made in completing the calculations for this analysis.  (1 mark)

Show Answers Only

a.  \(\ce{\% FeS2} =92.7\%\ \text{(3 sig.fig.)}\)

b.    Answers could have included one of the following:

  • All the sulfur is converted to \(\ce{SO4^2-}\).
  • Precipitate was pure.
  • No \(\ce{BaSO4}\) was lost in washing the sample.
  • \(\ce{BaSO4}\) was the only precipitate.
  • The sample was fully dehydrated.
Show Worked Solution

a.    \(\ce{m(BaSO4 \text{final})} = 174.99-123.40=51.59\ \text{g}\)

\(\ce{n(BaSO4(s))}=\dfrac{51.59}{233.4}=0.221\ \text{mol}\)

\(\Rightarrow \ce{n(SO4^2-(aq))}=0.221\ \text{mol}\)
 

Molar ratio  \(\ce{FeS2 : SO4^2-} = 1:2\)

\(\Rightarrow \ce{n(FeS2)}=0.221 \times \dfrac{1}{2}=0.1105\ \text{mol}\)

\(\ce{m(FeS2)}=0.1105 \times 120.0 =13.26\ \text{g}\)

\(\ce{\% FeS2} = \dfrac{13.26}{14.3} \times 100=92.7\%\ \text{(3 sig.fig.)}\)
 

b.    Answers could have included one of the following:

  • All the sulfur is converted to \(\ce{SO4^2-}\).
  • Precipitate was pure.
  • No \(\ce{BaSO4}\) was lost in washing the sample.
  • \(\ce{BaSO4}\) was the only precipitate.
  • The sample was fully dehydrated.
♦ Mean mark (b) 39%.

CHEMISTRY, M8 2018 HSC 16 MC

An investigation was carried out to determine the calcium ion concentration of a 2.0 L sample of tap water. Excess `text{Na}_2 text{CO}_3` was added to the sample. The precipitate was filtered, dried and weighed. The mass of the dried precipitate was 400 mg.

What was the concentration of calcium ions in the sample of tap water?

  1. `80 \ text{mg L}^-1`
  2. `160 \ text{mg L}^-1`
  3. `200 \ text{mg L}^-1`
  4. `400 \ text{mg L}^-1`
Show Answers Only

`A`

Show Worked Solution

\(\ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s)}\)

\[\ce{n(CaCO3) = \frac{m}{MM} = \frac{0.400}{100.09} = 3.9964 \times 10^{-3} mol}\]

\(\ce{n(Ca^2+) = n(CaCO3) = 3.9964 \times 10^{-3} mol}\)

\(\ce{m(Ca^2+) = n \times MM = 3.9964 \times 10^{-3}  \times 40.08 = 160 mg}\)

\[\therefore \ce{[Ca^2+] = \frac{160}{2.0} = 80 mg L^{-1}}\]

`=>A`


♦ Mean mark 41%.

CHEMISTRY, M8 2016 HSC 19 MC

Excess barium nitrate solution is added to 200 mL of 0.200 mol L¯1 sodium sulfate.

What is the mass of the solid formed?

  1. 4.65 g
  2. 8.69 g
  3. 9.33 g
  4. 31.5 g
Show Answers Only

`C`

Show Worked Solution

\(\ce{n(Na_2SO_4) = c\ × V = 0.2\ ×\ 0.2 = 0.04 mol}\)

\(\ce{n(BaSO_4) = 0.04 mol (1 : 1 mole ratio)}\)

\(\ce{n(Ba_2SO_4) = n × MM = 0.04\ × (137.3 + 32.07 + 16\ × 4) =9.3348 g = 9.33 (3 sig fig)}\)

`=>C`

CHEMISTRY, M8 2015 HSC 29

The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown.
 

  1. Suggest modifications that could be made to the procedure to improve the results of this investigation. Justify your suggestions.   (4 marks)

  2. Calculate the percentage of sulfate in the original fertiliser sample.   (3 marks)

Show Answers Only

a.   Modification 1:

  • The \(\ce{BaCl2}\) should be added slowly. This would help more of the sulfate ions come into contact with barium and form a precipitate.

Modification 2:

  • Barium sulfate is a very fine precipitate. By heating the mixture gently for some time, the \(\ce{BaSO4}\) precipitate is able to flocculate into larger particles, which would be trapped by the filter.

Modification 3:

  • The pores of most filter paper are still too large to effectively capture the \(\ce{BaSO4}\) precipitate. The experiment should use a small pore filter, like a sintered glass crucible.

Modification 4:

  • Multiple samples of the fertiliser should be analysed and the average value calculated to achieve more valid results. This would mitigate inaccuracies caused by unevenly distributed sulfate in the mixture.

b.   45.9%

Show Worked Solution

a.   Modification 1:

  • The \(\ce{BaCl2}\) should be added slowly. This would help more of the sulfate ions come into contact with barium and form a precipitate.

Modification 2:

  • Barium sulfate is a very fine precipitate. By heating the mixture gently for some time, the \(\ce{BaSO4}\) precipitate is able to flocculate into larger particles, which would be trapped by the filter.

Modification 3:

  • The pores of most filter paper are still too large to effectively capture the \(\ce{BaSO4}\) precipitate. The experiment should use a small pore filter, like a sintered glass crucible.

Modification 4:

  • Multiple samples of the fertiliser should be analysed and the average value calculated to achieve more valid results. This would mitigate inaccuracies caused by unevenly distributed sulfate in the mixture. 

♦ Mean mark (a) 49%.

b.   \(\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}\)

\(\ce{m_{fertiliser} = 2.00 g, \ \ m_{precipitate} = 2.23 g}\)

\[\ce{m(SO4^2-) = \frac{MM (SO4^2-)}{MM (BaSO4)} \times 2.23 = \frac{96.07}{233.37} \times 2.23 = 0.918 mol}\]

\[\therefore \text{% of} \ce{(SO4^2-) = \frac{0.918}{2.00} = 45.9\text{%}}\]


Mean mark (b) 54%.

CHEMISTRY, M8 2015 HSC 18-19 MC

Question 18

How could the reliability of the analysis of the pond water be improved?

  1. Analyse more samples from the same pond
  2. Use 50 mL of distilled water as a control sample
  3. Analyse samples from different ponds on the site
  4. Remove other contaminants from the sample before the analysis

 
Question 19

What was the concentration of lead ions in the sample?

  1. `5.0 × 10^{-3} \ text{mol L}^(-1)`
  2. `5.8 × 10^{-3} \ text{mol L}^(-1)`
  3. `9.3 × 10^{-3} \ text{mol L}^(-1)`
  4. `10.7 × 10^{-3} \ text{mol L}^(-1)`
Show Answers Only

`18.  A`

`19.  C`

Show Worked Solution

Question 18

`=>A`
 

Question 19

\[\ce{n(PbCl2) = \frac{0.13}{207.2 + 2 \times 35.45} = \frac{0.13}{278.1} = 4.67 \times 10^{-4} mol}\]

\(\ce{n(Pb^2+) = n(PbCl2) = 4.67 \times 10^{-4} mol}\)

\[\ce{[Pb^2+] = \frac{4.67 \times 10^{-4}}{0.050} = 9.3 \times 10^{-3} mol L^{-1}}\]

`=>C`


♦ Mean mark (Q19) 52%.

CHEMISTRY, M8 2017 HSC 19 MC

The sulfate content of a fertiliser is 48% by mass. 1.20 g of this fertiliser is completely dissolved in water and an excess of \( \ce{Ba(NO3)2(aq)} \) is added.

What mass of precipitate would be formed?

  1. 0.006 g
  2. 0.58 g
  3. 1.40 g
  4. 1.57 g
Show Answers Only

`C`

Show Worked Solution

\(\ce{m(SO4^2-) = 0.48 \times 1.20 = 0.576\ g}\)

\[\ce{n(SO4^2-) = \frac{m}{MM} = \frac{0.576}{32.07 + 4 \times 16.00} = 5.996 \times 10^{-3} mol} \]

Chemical equation

\(\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \)

\(\ce{n(BaSO4) = n(SO4^2-) = 5.996 \times 10^{-3} mol}\)

\(\ce{m(BaSO4) = n \times MM = (5.996 \times 10^{-3}) \times 233.37 = 1.40 g} \)

\( \Rightarrow C \)


Mean mark 57%.

CHEMISTRY, M8 2022 HSC 28

The iron content of an impure sample (4.32 g) was determined by the process shown in the flow chart.
 

  1. Identify the brown precipitate formed at the end of step 3.   (1 mark)
  1. Calculate the percentage of iron in the original impure sample if 4.21 g of iron(`text{III}`) oxide `(\text{Fe}_2\text{O}_3)` was collected. Assume that all the iron was converted to iron(`text{III}`) oxide.   (4 marks)
Show Answers Only

a.   `text{Fe(OH)}_3`

b.   `text{68.2%}`

Show Worked Solution

a.  `text{Fe(OH)}_3`

b.  `text{n(Fe}_2 text{O}_3)= text{m} / text{MM}=4.21 / [2 xx55.85+3 xx 16.00]=0.0264  text{mol}`

`text{n(Fe)}= 2 xx text{n(Fe}_2 text{O}_3 )=0.0527  text{mol}`

`text{m(Fe)}= text{n} xx  text{MM}=0.0527238 xx 55.85=2.94\ text{g}`

`text{Fe(%)}= text{m(Fe)} / text{m(original sample)} xx 100%=2.944627 / 4.32 xx 100%=68.2%`

Therefore, the percentage of iron in the original impure sample is 68.2%.

CHEMISTRY, M8 2021 HSC 28

A 5.30 g sample of an alkali metal hydroxide was dissolved in water. After mixing with excess \(\ce{Cu(NO3)2}\), the precipitate was collected, dried, measured and found to have a mass of 4.61 g.

Identify the alkali metal hydroxide. Support your answer with calculations and a balanced equation.   (4 marks)

Show Answers Only

Since alkali metals are in group 1, they have a +1 charge.

`text{Cu(NO}_3 )_2 (aq) + 2 text{XOH} (aq)  →  text{Cu(OH)}_2 (s) + 2 text{XNO}_3 (aq)`

`text{m(Cu(OH)}_2 ) = 4.61\ text{g}`

`text{n(Cu(OH)}_2 )= text{m} / text{MM}= 4.61 / (97.566)= 0.04725\ text{mol}`

`2text{n}(text{OH}^(-))= text{n(Cu(OH)}_2)= 0.09450\ text{OH}`

`text{n}(text{XOH}^(-))=2text{n}(text{OH}^(-))= 0.09450\ text{mol}`
 

`text{n}=text{m} / text{MM}\ \ =>\ \ text{MM}= text{m} / text{n}`

`text{MM (XOH)}= 5.30 / 0.0945= 56.08\ text{g mol}^-1`

`text{MM (X}^+)` `= text{MM (XOH)}-text{MM (OH}^- )`  
  `= 56.08-(16 + 1.008)`  
  `= 39.077\ text{g mol}^-1`  

 

  • This is consistent with the molar mass of potassium (`text{K}`) from the periodic table.
  • Therefore, the unknown alkali metal hydroxide is `text{KOH}` (potassium hydroxide).
Show Worked Solution

Since alkali metals are in group 1, they have a +1 charge.

`text{Cu(NO}_3 )_2 (aq) + 2 text{XOH} (aq)  →  text{Cu(OH)}_2 (s) + 2 text{XNO}_3 (aq)`

`text{m(Cu(OH)}_2 ) = 4.61\ text{g}`

`text{n(Cu(OH)}_2 )= text{m} / text{MM}= 4.61 / (97.566)= 0.04725\ text{mol}`

`2text{n}(text{OH}^(-))= text{n(Cu(OH)}_2)= 0.09450\ text{OH}`

`text{n}(text{XOH}^(-))=2text{n}(text{OH}^(-))= 0.09450\ text{mol}`
 

`text{n}=text{m} / text{MM}\ \ =>\ \ text{MM}= text{m} / text{n}`

`text{MM (XOH)}= 5.30 / 0.0945= 56.08\ text{g mol}^-1`

`text{MM (X}^+)` `= text{MM (XOH)}-text{MM (OH}^- )`  
  `= 56.08-(16 + 1.008)`  
  `= 39.077\ text{g mol}^-1`  

 

  • This is consistent with the molar mass of potassium (`text{K}`) from the periodic table.
  • Therefore, the unknown alkali metal hydroxide is `text{KOH}` (potassium hydroxide).

♦ Mean mark 46%.

CHEMISTRY, M8 2021 HSC 17 MC

A sample was contaminated with sodium phosphate. The sample was dissolved in water and added to an excess of acidified \(\ce{(NH_4)_2MoO_4}\) to produce a precipitate of \(\ce{(NH_4)_3PO_4.12MoO_3\ \ \ \ \ \ ($MM$ = 1877 g mol^{-1})} \).

If 24.21 g of dry \(\ce{(NH_4)_3PO_4.12MoO_3}\) was obtained, what was the mass of sodium phosphate in the original sample?

  1. 1.225 g
  2. 1.521 g
  3. 1.818 g
  4. 2.115 g
Show Answers Only

`D`

Show Worked Solution

`text{n((NH4)}_3 text{PO}_4 cdot 12text{MoO}_3\text{)}= text{m} / text{MM}= 24.21/1877= 0.01289…\ text{moles}`

Each precipitate molecule has one molecule of `text{PO}_(4) ^ (\ 3-)`

Each `text{Na}_(3) text{PO}_(4)` molecule has one molecule of `text{PO}_(4) ^ (\ 3-)`

  • `text{n(Na}_3 text{PO}_4) =\ text{n((NH_4)}_3 text{PO}_4 cdot 12text{MoO}_3)` 
  • `text{n(Na}_3 text{PO}_4)` `=\ text{m} / text{MM}`
`text{m(Na}_3 text{PO}_4)` `=\ text{n × MM}`  
  `=0.01289… xx (3 xx 22.99 + 30.97 + 4 xx 16)`  
  `= 2.115\ text{g}`  

 
`=> D`


♦♦ Mean mark 36%.