Nine tips for scaffolding multi-step calculations

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Nine tips for scaffolding multi-step calculations Principles to help students sequence longer maths tasks From Education in Chemistry rsc.li/2paxfig

Build up the basics Ensure you have secured the following maths skills before applying them to a multi-step problem

1. Units Include units throughout a calculation. This makes it easier to see if you need to convert units. For example, if a concentration is given in g dm -3 but a volume is given in cm 3, the volume should be converted to dm 3 before substituting into the formula. It s also a useful way of checking for errors. If you end up with the units for concentration as dm 3 g -1, you know you ve made a mistake. Additionally, you will become more familiar with which units are used for each type of measurement (eg dm 3 for volume, g for mass), and recognise that something measured in dm 3 relates to a volume.

2. Making sense of formulas Understanding that concentration is how much solute is present in a certain volume of liquid helps to make sense of the formula, because the formula represents, in words, the amount per volume. For example, something concentrated has a large amount in a relatively small volume. concentration = amount volume Also understand that, for example, the molar mass is the mass per mole of a substance. So, a very large atom or molecule will have a large molar mass because each atom or molecule is bigger.

3. Speaking the language Talk through what you re doing in a calculation this will help you learn the appropriate language. For example, it helps to make the link between per and divide.

4. Using ratio and proportion Clearly set out the answers to questions involving ratio and proportion. For example, if we know that 1 mole of CO 2 gas occupies 24 dm 3, what volume would 0.162 mole of CO 2 gas occupy? Think through this calculation carefully. We have a ratio of 1 mole CO 2 to 24 dm 3 volume. To get from 1 to 24 we multiply by 24. So to work out the volume for 0.162 mole CO 2, we multiply 0.162 by 24 to get 3.9 dm 3.

Scaffold the process Once you re comfortable with calculation basics, build up the process for solving a multistep calculation.

5. Narrative Describe what the question tells you and what it asks you to do. Make 20.0 g of copper chloride from copper carbonate and dilute acid. The equation for the reaction is: CuCO 3 + 2HCl CuCl 2 + H 2 O + CO 2 Relative atomic masses, Ar: H = 1; C = 12; O = 16; Cl = 35.5; Cu = 63.5 Calculate the mass of copper carbonate you should react with dilute hydrochloric acid to make 20.0 g of copper chloride. For this example, you might say, The balanced chemical equation tells me one mole of copper carbonate gives me one mole of copper chloride. Moles are our currency in chemistry. I know the mass of copper chloride. So, I first have to work out the number of moles that corresponds to. Once I know this, I can say how many moles of copper carbonate would be required, and then I can convert that into grams to get the answer.

6. Signposts Write signposts that describe what is done at each step.

7. Colour code Colour helps you make connections between the different parts of the chemical equation and the calculations.

8. Put equations before numbers Write the equation needed before substituting numbers to do the calculation.

9. Format carefully Setting out working in plenty of white space. Set fractions out visually, for example: not inline as a / b. a b

Make 20.0 g of copper chloride from copper carbonate and dilute acid. The equation for the reaction is: CuCO 3 + 2HCl CuCl 2 + H 2 O + CO 2 Relative atomic masses, A r : H = 1; C = 12; O = 16; Cl = 35.5; Cu = 63.5 Calculate the mass of copper carbonate you should react with dilute hydrochloric acid to make 20.0 g of copper chloride. The balanced chemical equation tells us how many moles of each reactant and product are involved in the reaction. We are given the mass of CuCl 2 so we need to convert from g to moles. CuCO 3 + 2HCl CuCl 2 + H 2 O + CO 2 20 g formula mass CuCl 2 = 63.5 + 2 x 35.5 = 134.5 Convert g to moles. number of moles = mass formula mass 20 g = 0.149 mol number of moles = 134.5 g mol 1 From the balanced chemical equation we know that 0.149 mole of CuCl 2 is made from 0.149 mole of CuCO 3. 1 mole CuCO 3 1 mole CuCl 2 0.149 mole CuCO 3 0.149 mole CuCl 2 We are asked for the mass of copper carbonate so convert moles to g. mass = number of moles x formula mass formula mass CuCO 3 = 63.5 + 12 + 3 x 16 = 123.5 mass CuCO 3 = 0.149 mol x 123.5 g mol 1 = 18.4 g You need 18.4 g copper carbonate to react with dilute hydrochloric acid to make 20.0 g copper chloride.

A student used a pipette to add 25.0 cm 3 of sodium hydroxide of unknown concentration to a conical flask. The student carried out a titration to find out the volume of 0.100 mol/dm 3 sulfuric acid needed to neutralise the sodium hydroxide. The student carried out five titrations with the results shown in the table. Concordant results are within 0.10 cm 3 of each other. Use the student s concordant results to work out the mean volume of 0.100 mol/dm 3 sulfuric acid added. Calculate the concentration of the sodium hydroxide. Give your answer to three significant figures Titration 1 Titration 2 Titration 3 Titration 4 Titration 5 Volume of 0.100 mol dm -3 sulfuric acid in cm 3 27.40 28.15 27.05 27.15 27.15 calculate moles of sulfuric acid from concentration and volume sulfuric acid conc = 0.100 mol dm 3 volume = 27.1 cm 3 sodium hydroxide volume = 25.0 cm 3 concentration =? amount in moles = 0.100 mol dm 3 x 27.1 cm 3 = 0.100 mol dm 3 x 27.1 dm 3 amount in moles = 0.00271 mol 1000 write balanced equation 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O use equation to calculate moles in flask 2 moles NaOH reacts with 1 mole H 2 SO 4 2 x 0.00271 mol NaOH reacts with 0.00271 mol H 2 SO 4 0.00542 mol NaOH reacts with 0.00271 mol H 2 SO 4 use moles and volume to get concentration sodium hydroxide amount = 0.00542 mol, volume = 25.0 cm 3 concentration = 0.00542 mol = 0.217 mol dm 3 0.025 dm 3 The concentration of the sodium hydroxide is 0.217 mol dm 3