0.28 g of a gaseous hydrocarbon was turned in excess oxygen. 0.88 g of carbon dioxide and 0.36 g of water were formed. The volume of 0.28 g of the hydrocarbon at 1.01 x10^5 Pa and 298 k is 123 cm^3. Work out the molecular formula of the hydrocarbon.

For a problem such as this, it is important to break down the question into the individual aspects and use some principles of chemistry that we have learnt previously to tackle the problem one step at a time.
Throughout this example I will be using my Whiteboard to demonstrate how any steps involving maths should be done and why they are important.
As with most numerical problems the information we need has been presented to us in the question, from the mass of the products we can calculate the number of moles of the products and reactants. (Prompting the student to recall how to do this and guiding them through one of the three required calculations). We can then use one of our chemistry principles to create an equation for the reaction based on our previous knowledge of combustion. (A short re cap can be included here if the student struggles). Using the number of moles of the reactant and product we can then find a ratio between the two and therefore use this to balance the symbol equation and correctly identify that the hydrocarbon in question is butene (C4H8). This can be confirmed by working in reverse and prompting the student to do this would help to know that they have fully engaged with the methodology in the question.