What is the molecular formula of pentanoic acid?
- \(\ce{C_5H_9O}\)
- \(\ce{C_5H_10O}\)
- \(\ce{C_5H_10O_2}\)
- \(\ce{C_5H_11O_2}\)
CHEMISTRY, M6 2015 HSC 26
A sodium hydroxide solution was titrated against citric acid \(\ce{(C6H8O7)}\) which is triprotic.
- Draw the structural formula of citric acid (2-hydroxypropane-1,2,3-tricarboxylic acid). (1 mark)
- How could a computer-based technology be used to identify the equivalence point of this titration? (2 marks)
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- The sodium hydroxide solution was titrated against 25.0 mL samples of 0.100 mol L ¯1 citric acid. The average volume of sodium hydroxide used was 41.50 mL.
- Calculate the concentration of the sodium hydroxide solution. (4 marks)
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a.
b. Technology solution
→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.
→ The equivalence point would be identified by a steep rise in the pH on the graph.
c. \(\ce{C6H8O7 + 3NaOH -> C6H5O7Na3 + 3H2O}\)
\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 mol}\)
\(\ce{n(NaOH) = 3 \times 0.00250 = 0.00750 mol}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.181 mol L^{-1} (3 sig fig)}\]
Show Worked Solution
a.
♦ Mean mark (a) 46%.
b. Technology solution
→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.
→ The equivalence point would be identified by a steep rise in the pH on the graph.
♦ Mean mark (b) 42%.
c. \(\ce{C6H8O7 + 3NaOH -> C6H5O7Na3 + 3H2O}\)
\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 mol}\)
\(\ce{n(NaOH) = 3 \times 0.00250 = 0.00750 mol}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.181 mol L^{-1} (3 sig fig)}\]
Mean mark (c) 56%.
CHEMISTRY, M7 2019 HSC 21
- The structural formula for 2-methylpropan-2-ol is shown in the table.
- Draw one structural isomer of this alcohol and state its name. (2 marks)
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- The structural formulae for two compounds are shown below.
- Why are these two compounds classed as functional group isomers? (2 marks)
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- A chemical test is required to distinguish between the isomers in part (b).
- Identify a suitable test and explain the expected observations. (3 marks)
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a. Successful answers should have one of the following:
b. Functional Group isomers
→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.
→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
c. Tollens’ Test:
→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).
→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.
→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.
Show Worked Solution
a. Successful answers should have one of the following:
b. Functional Group isomers
→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.
→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
c. Tollens’ Test:
→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).
→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.
→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.
♦♦ Mean mark (c) 39%.
CHEMISTRY, M7 2019 HSC 10 MC
Which class of organic compound must contain at least three carbon atoms?
- Aldehydes
- Alkenes
- Carboxylic acids
- Ketones
CHEMISTRY, M7 2020 HSC 3 MC
Which of the following compounds is the most basic?
- Ethane
- Ethanol
- Ethanamine
- Ethyl ethanoate