NESA Chemistry Problem Solving

2 sample questions with marking guides and sample answers · Avg. score: 71.4%

Q39
2025
SCSA
14 marks
Q39
14 marks

Freon-11 is a colourless chlorofluorocarbon that boils at 23.77 °C. Prior to the knowledge of the ozone-depleting potential of chlorofluorocarbons (CFCs) and other possible harmful effects on the environment, it was used as a refrigerant.

The following data was used to determine that Freon-11 is trichlorofluoromethane, with a molecular formula of CCl3FCCl_3F.

A Freon-11 sample of 4.121 g was combusted in excess oxygen. All the carbon in the compound was converted to carbon dioxide and in a separate process, all its chlorine was converted into hydrochloric acid. The carbon dioxide produced had a mass of 1.320 g and the hydrochloric acid formed, required 85.70 mL of 1.050 mol L1^{-1} of ammonia solution for complete neutralisation.
Another sample of the Freon-11 with a mass of 3.721 g occupied a volume of 0.6068 L at a pressure of 120.00 kPa and temperature of 50.6 °C.

Using the same data, use calculations and reasoning to demonstrate that this is the correct molecular formula.

Reveal Answer

Carbon
n(C)=n(CO2)=1.320/44.01=0.02999n(C) = n(CO_2) = 1.320/44.01 = 0.02999 mol
m(C)=0.02299×12.01=0.3602m(C) = 0.02299 \times 12.01 = 0.3602 g

Chlorine
n(Cl)=n(HCl)=n(NH3)=1.050×0.08570=0.08999n(Cl) = n(HCl) = n(NH_3) = 1.050 \times 0.08570 = 0.08999 mol
m(Cl)=0.08999×35.45=3.1900m(Cl) = 0.08999 \times 35.45 = 3.1900 g

Fluorine
m(F)=4.121(0.3602+3.1900)=0.5708m(F) = 4.121 - (0.3602 + 3.1900) = 0.5708 g
n(F)=0.5708/19.00=0.03004n(F) = 0.5708 / 19.00 = 0.03004 mol

Mole Ratio
C: 0.02999
Cl: 0.08999
F: 0.03004

Simplify
Divide by 0.02999
C: 1
Cl: 3.00
F: 1.002

Empirical Formula
CCl3FCCl_3F

Molecular Formula
Empirical formula mass (EFM) = 137.36 amu/g mol1^{-1}
n=PV/RT=(120.0×0.6068)/(8.314×323.75)=0.027052476n = PV/RT = (120.0 \times 0.6068)/(8.314 \times 323.75) = 0.027052476 mol
Molecular formula mass (MFM) = m/n=3.721/0.0276503204=137.54m/n = 3.721/0.0276503204 = 137.54 g mol1^{-1}
Molecular formula = MFM/EFM ×\times Empirical formula = 137.54/137.36×CCl3F=1.0013×CCl3F137.54/137.36 \times CCl_3F = 1.0013 \times CCl_3F

Molecular formula = Empirical formula = CCl3FCCl_3F

Marking Criteria
DescriptorMarks

n(C) calculation

1

m(C) calculation

1

n(Cl) calculation

1

m(Cl) calculation

1

m(F) calculation

1

n(F) calculation

1

Mole Ratio setup

1

Simplify ratio

1

Empirical Formula

1

Empirical formula mass (EFM)

1

n = PV/RT calculation

1

Molecular formula mass (MFM) calculation

1

Molecular formula calculation

1

Molecular formula = Empirical formula statement

1
Q6
2020
VCAA
8 marks
Q6

Methane gas, CH4\text{CH}_4, can be captured from the breakdown of waste in landfills. CH4\text{CH}_4 is also a primary component of natural gas. CH4\text{CH}_4 can be used to produce energy through combustion.

Q6a
1 mark

Write the equation for the incomplete combustion of CH4\text{CH}_4 to produce carbon monoxide, CO.

Reveal Answer

2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) or
CH4(g) + 1.5O2(g) → CO(g) + 2H2O(l)

Marking Criteria
DescriptorMarks

Correct response.

1
Q6b
2 marks

If 20.0 g of CH4\text{CH}_4 is kept in a 5.0 L sealed container at 25C25 ^\circ\text{C}, what would be the pressure in the container?

Reveal Answer

n(CH4)=20.0/16.0=1.25 moln(\text{CH}_4) = 20.0 / 16.0 = 1.25 \text{ mol} *
Pressure(CH4)=nRT/V=1.25×8.31×298/5.0=6.2×102 kPa\text{Pressure}(\text{CH}_4) = nRT / V = 1.25 \times 8.31 \times 298 / 5.0 = 6.2 \times 10^2 \text{ kPa} *

Marking Criteria
DescriptorMarks

Calculates correct amount of CH4 (1.25 mol).

1

Calculates correct pressure of CH4 (6.2 x 10^2 kPa).

1
Q6c
3 marks

A Bunsen burner is used to heat a beaker containing 350.0 g of water. Complete combustion of 0.485 g of CH4\text{CH}_4 raises the temperature of the water from 20C20 ^\circ\text{C} to 32.3C32.3 ^\circ\text{C}.

Calculate the percentage of the Bunsen burner's energy that is lost to the environment.

Reveal Answer

Energy from CH4=0.485 g×55.6 kJ g1=27.0 kJ\text{Energy from CH}_4 = 0.485 \text{ g} \times 55.6 \text{ kJ g}^{-1} = 27.0 \text{ kJ}

Energy absorbed by water=4.18 J g1 K1×350.0×(32.320.0)\text{Energy absorbed by water} = 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 350.0 \times (32.3 - 20.0)
=1.80×104 J= 1.80 \times 10^4 \text{ J}
=18.0 kJ= 18.0 \text{ kJ}

Energy lost to environment=27.018.0=9.0 kJ\text{Energy lost to environment} = 27.0 - 18.0 = 9.0 \text{ kJ}

% energy lost=(9.0/27.0)×100=33.3%\% \text{ energy lost} = (9.0 / 27.0) \times 100 = 33.3 \%

Marking Criteria
DescriptorMarks

Calculating the energy from CH4.

1

Calculating the energy absorbed by water.

1

Working out the percentage of energy loss.

1
Q6d
2 marks

Compare the environmental impact of CH4\text{CH}_4 obtained from landfill to the environmental impact of CH4\text{CH}_4 obtained from natural gas.

Reveal Answer

Similarity – methane from both sources

  • Both produce atmospheric carbon dioxide through combustion.
  • Methane from both sources contains small amounts of nitrogen and sulfur; combustion of natural gas leads to the formation of acidic oxides such as SOx and NOx.

Difference – landfill versus natural gas

  • Methane from landfill can be produced renewably, whereas methane from natural gas releases stored carbon.
  • Methane from landfill is more carbon neutral, methane from natural gas increases atmospheric CO2 levels.
  • Obtaining methane from natural gas via fracking causes additional significant environmental damage, whereas when obtaining methane from a landfill the damage has already been done in the formation of the landfill.
  • Landfill gases contain less methane and release more CO2 (for the same amount of energy generated), natural gas contains more methane and releases comparatively less CO2.
  • Methane captured from landfill and used as a source on energy may have a positive impact as it is a more potent greenhouse gas than CO2.
  • CH4 from landfill is more easily collected compared to fracking/sourcing methane from fossil fuels.
Marking Criteria
DescriptorMarks

1 mark for each valid comparison point (any 2 of): Both produce atmospheric carbon dioxide through combustion; Methane from both sources contains small amounts of nitrogen and sulfur; Methane from landfill can be produced renewably, whereas methane from natural gas releases stored carbon; Methane from landfill is more carbon neutral; Obtaining methane from natural gas via fracking causes additional significant environmental damage; Landfill gases contain less methane and release more CO2; Methane captured from landfill may have a positive impact as it is a more potent greenhouse gas than CO2; CH4 from landfill is more easily collected.

2

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