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CHEMISTRY, M7 2023 HSC 27

A student has been asked to produce 185 mL of ethanol (MM = 46.068 g mol\(^{-1} \)) by fermenting glucose using yeast, as shown in the equation.

\( \ce{C6H12O6(aq)} \rightarrow \ce{2C2H5OH(aq)} + \ce{2CO2(g)} \)

Given that the density of ethanol is 0.789 g mL\(^{-1} \), calculate the volume of carbon dioxide gas produced at 310 K and 100 kPa.  (4 marks)

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\(81.7\ \text{L}\)

Show Worked Solution

\(\text{Density}(\rho)\ = \dfrac{\text{m}}{\text{V}}\ \ \Rightarrow\ \ \text{m} = \rho \times\ \text{V} \)

\(\text{m(ethanol)}\ = 0.789 \times 185 = 146\ \text{g} \)

\(\text{n(ethanol)}\ = \dfrac{\text{m}}{\text{MM}} = \dfrac{146}{46.068} = 3.17\ \text{mol} \)

\(\text{V}\) \(=\dfrac{\text{n} RT}{P} \)  
  \(=\dfrac{3.17 \times 8.314 \times 310}{100}\)  
  \(=81.7\ \text{L}\)  

CHEMISTRY, M7 2018 HSC 24

The following apparatus was set up to test the reaction rate of fermentation of glucose at different temperatures.
 

 
 

  1. Write a balanced equation for the fermentation of glucose.  (1 mark)

  2. After 24 hours, 5.5 mL of gas was collected at 25°C and 100 kPa.
  3. Calculate the mass of glucose that would have been reacted.  (3 marks)

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a.    \(\ce{C6H12O6(aq) -> 2C2H5OH(l) + 2CO2(g)}\)
 

b.    \(\ce{Vol(CO2) = 5.5\ ÷\ 1000 = 0.0055 L}\)

\[\ce{n(CO2) = \frac{V(CO2)}{24.79} = 2.2186 \times 10^{-4} mol}\]

\[\ce{n(C6H12O6) = \frac{1}{2} \times n(CO2) = 1.1093 \times 10^{-4} mol}\]

\(\ce{MM(C6H12O6) = (6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00) = 180.156}\)

\(\ce{m(C6H12O6) = 1.1093 \times 10^{-4} \times 180.156 = 0.020 g}\)

Show Worked Solution
a.    \(\ce{C6H12O6(aq) -> 2C2H5OH(l) + 2CO2(g)}\)
 

Mean mark (a) 57%.
 

b.    \(\ce{Vol(CO2) = 5.5\ ÷\ 1000 = 0.0055 L}\)

\[\ce{n(CO2) = \frac{V(CO2)}{24.79} = 2.2186 \times 10^{-4} mol}\]

\[\ce{n(C6H12O6) = \frac{1}{2} \times n(CO2) = 1.1093 \times 10^{-4} mol}\]

\(\ce{MM(C6H12O6) = (6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00) = 180.156}\)

\(\ce{m(C6H12O6) = 1.1093 \times 10^{-4} \times 180.156 = 0.020 g}\)

CHEMISTRY, M7 2015 HSC 17 MC

What volume of carbon dioxide will be produced if 10.3 g of glucose is fermented at 25°C and 100 kPa?

  1. 1.30 L
  2. 1.42 L
  3. 2.57 L
  4. 2.83 L
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`D`

Show Worked Solution

\(\ce{C6H12O6 -> 2CO2 + 2C2H5OH}\)

\[\ce{n(C6H12O6) = \frac{10.3}{6 \times 12.01 + 12 \times 1.008 + 6 \times 16} = \frac{10.3}{180.156} = 0.057 mol}\]

\(\ce{n(CO2) = 2 \times n(C6H12O6) = 0.114 mol}\)

\(\ce{\text{Volume} (CO2) = 0.114 \times 24.79 = 2.83 L}\)

`=>D`

CHEMISTRY, M7 2019 HSC 34

The following reaction scheme can be used to synthesise ethyl ethanoate.
 


 

Outline the reagents and conditions required for each step and how the product of each step could be identified.   (7 marks) 

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Step 1:

→ To synthesise chloroethane (A) into ethanol (B), \(\ce{NaOH}\) is added and heated. \(\ce{KMnO4 / H+}\) is then added and heated.

→ The mixture is then treated with concentrated sulfuric acid and refluxed.

→ Ethanol (B) can be identified using infrared spectroscopy by looking for a broad absorption between 3230 cm ¯1 and 3550 cm ¯1, which indicates the presence of an \(\ce{O-H}\) bond. This absorption would not be present in chloroethane (A).

→ Alternative ways to identify ethanol include: mass spectrum analysis (single ion peak at m/z = 46), reactivity tests, and \( \ce{^1H NMR}\) spectrum analysis (3 signals vs 2 for chloroethane).
 

Step 2:

→ Ethanol (B) can be converted into ethanoic acid (C) by combining it with a strong oxidant like sodium carbonate, which produces carbon dioxide bubbles, confirming the presence of a carboxylic acid.

→ Ethanol will not react as above and the compounds can be distinguished.

→ Alternative ways to identify ethanoic acid include: IR or \( \ce{^13C NMR}\) spectrum analysis, litmus indicators, mass spectrum analysis (ion peak at m/z = 60 vs m/z = 46)
 

Step 3

→ Ethyl ethanoate (D) can be synthesised by heating a mixture of ethanol, ethanoic acid and concentrated sulfuric acid under reflux.

→ A \( \ce{^1H NMR}\) spectrum can be used to identify ethyl ethanoate as it will have 3 signals versus ethanol and ethanoic acid that will only have 2 each.

→ Alternative ways to identify ethyl ethanoate include: a distinct smell, no \(\ce{O-H}\) peaks in the IR spectrum or mass spectrum analysis (ion peak at m/z = 102).

Show Worked Solution

Step 1:

→ To synthesise chloroethane (A) into ethanol (B), \(\ce{NaOH}\) is added and heated. \(\ce{KMnO4 / H+}\) is then added and heated.

→ The mixture is then treated with concentrated sulfuric acid and refluxed.

→ Ethanol (B) can be identified using infrared spectroscopy by looking for a broad absorption between 3230 cm ¯1 and 3550 cm ¯1, which indicates the presence of an \(\ce{O-H}\) bond. This absorption would not be present in chloroethane (A).

→ Alternative ways to identify ethanol include: mass spectrum analysis (single ion peak at m/z = 46), reactivity tests, and \( \ce{^1H NMR}\) spectrum analysis (3 signals vs 2 for chloroethane).
 

Step 2:

→ Ethanol (B) can be converted into ethanoic acid (C) by combining it with a strong oxidant like sodium carbonate, which produces carbon dioxide bubbles, confirming the presence of a carboxylic acid.

→ Ethanol will not react as above and the compounds can be distinguished.

→ Alternative ways to identify ethanoic acid include: IR or \( \ce{^13C NMR}\) spectrum analysis, litmus indicators, mass spectrum analysis (ion peak at m/z = 60 vs m/z = 46)
 

Step 3

→ Ethyl ethanoate (D) can be synthesised by heating a mixture of ethanol, ethanoic acid and concentrated sulfuric acid under reflux.

→ A \( \ce{^1H NMR}\) spectrum can be used to identify ethyl ethanoate as it will have 3 signals versus ethanol and ethanoic acid that will only have 2 each.

→ Alternative ways to identify ethyl ethanoate include: a distinct smell, no \(\ce{O-H}\) peaks in the IR spectrum or mass spectrum analysis (ion peak at m/z = 102).


♦♦ Mean mark 38%.

CHEMISTRY, M7 2021 HSC 25

A student conducted an experiment in the school laboratory under standard laboratory conditions (25°C, 100 kPa) to determine the volume of carbon dioxide gas produced during the fermentation of glucose. The following apparatus was set up.
 


 

The following data were collected.
 

Assume the total volume of gas produced was due to the production of carbon dioxide.

Calculate the mass of ethanol produced by the fermentation reaction. Include a relevant chemical equation in your answer.   (4 marks)

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`text{C}_6 text{H}_12 text{O}_6(aq) → 2 text{C}_2 text{H}_5 text{OH}(aq) + 2 text{CO}_2(g)`

`1.869  text{g}`

Show Worked Solution

The fermentation reaction:

`text{C}_6 text{H}_12 text{O}_6(aq) → 2 text{C}_2 text{H}_5 text{OH}(aq) + 2 text{CO}_2(g)`

`text{v}_(text{(CO}_2 text{)})= 1006  text{mL}= 1.006  text{L}`

`text{n}_(text{(CO}_2 text{)})= text{v} / text{V}_m= 1.006 / 24.79= 0.04058087939  text{mol}`

`text{n}_(text{(C}_2 text{H}_5 text{OH)})=text{n}_(text{(CO}_2 text{)})= 0.04058087939  text{mol}`

`text{m}_(text{(C}_2 text{H}_5 text{OH)})` `= text{n}  xx  text{MM}`
  `= 0.04058087939 xx 46.068 `
  `= 1.869  text{g}`

♦ Mean mark 55%.