Much of the content for this topic has been covered in O Levels. There are some new concepts in this topic as well! One of the commonly asked question in this topic deals with the combustion of hydrocarbons, and is covered in our JC Chemistry tuition class.

Example:

10.0cm3 of a hydrocarbon, CaHb, is exploded with an excess of oxygen. A contraction of 35.0cm3 occurs, all volumes being measured at r.t.p. On treatment of the products with NaOH solution, a further contraction of 40.0 cm3 occurs. Deduce the molecular formula of the hydrocarbon.

 Some of the points to note for solving such questions include the following:

  1. Knowing the general useful equations used:
  2. Water is a liquid at room temperature and pressure (r.t.p.) and hence does not take up any volume.
  3. Since vol is proportional to amount in mol, direct use of volume is sufficient to solve for the hydrocarbon formula, there is no need to convert the volume back to amount in mol.
  4. Aim to solve for the vol of CO2 evolved after combustion and vol of O2 used for combustion.
  5. NaOH is used to absorb acidic gases like CO2 and SO2. Hence the decrease in volume after passing the gas mixture through an alkaline solution is due to the loss of the acidic gases.
  6. When dealing with contraction or expansion of volume after combustion, it is more tricky as the contraction is not due to just the loss of the hydrocarbons and oxygen and the expansion is not just due to the CO2 evolved.

In the above question, a contraction of 35.0cm3 occurs does not mean that the sum of vol of hydrocarbon and oxygen used for combustion is 35 cm3. However, there is a very simple and quick method to solve for such questions.

This concept came out in 2010 A level paper and many students were caught unprepared for it. During classes, content like this will be covered so that students will be able to avoid these pitfalls so as not to tread on the same path as others who have made the same mistakes.

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