LESSON PLAN | CHEMISTRY KS4

KS4

CHEMISTRY ABOUT THE AUTHOR:Dr Joanna L. Rhodes M.Chem, D.Phil, MRSC is a teacher of science at Shelley College, Huddersfield. The BLT proposal Teaching mole calculations and interpretations of chemical equations to Y10 students can be a challenge – but Dr Joanna Rhodes tasty ideas will help...

I have just finished teaching the new quantitative chemistry unit at Key Stage 4 to my year 10 students. The concepts and mathematical expectations are Why more advanced, and have moved closer to A-level chemistry in all respects. However, we are now teaching these concepts to youngsters two years junior in teach understanding and chemistry experience; the approach I would take with year 12 is very difficult to apply successfully to year 10. This has led me to completely this? rethink the strategies I use to teach moles, equations, concentration, volume, The new GCSE reacting masses, limiting reagents and yield. The unit is definitely one of the most chemistry specifications require challenging we will teach at key stage 4 – but with some new ideas and strategies mole calculations it can still engage and interest all students while preparing the most able with a and interpretation of superb grounding to progress onto A-level chemistry. chemical equations at a similar standard to those we usually This technical lesson provides strategies, hooks and ideas to introduce students to begin teaching at AS the main areas of quantitative chemistry, which are addressed in the new program of level. In my experience study. The main aim is to support different learning styles that are more appropriate key stage 4 students need a more visual to year 10 students by drawing on other curriculum areas including food technology, approach that helps computer science and maths, and on students’ visual and spatial awareness skills. them to see the bigger picture of what the different calculations involving moles are 3Ba trying to achieve. STARTER Using strategies from ACTIVITY 2Br 3L Br2Ba3L3T2 elsewhere in the curriculum including If I make a bacon lettuce computational and tomato sandwich then 2T analogies, food and the mass of the sandwich recipes and writing is equal to the mass of all the ingredients that went conservation of mass in chemical equations. Now let’s try frames, this lesson into it. The sandwich cannot giving each ingredient a ‘chemical symbol’ so we can see how aims to give you weigh more or less than a chemical equation is put together. In the illustration above, all ideas to crack the combined mass of the the ingredients are labeled. This allows us to write the ‘chemical’ the challenge of ingredients separately. So if equation for making a bacon sandwich as quantitative chemistry the total mass of 2 slices of 2Br + 3Ba + 3L + 2T → Br₂Ba₃L3T₂. in your classroom. bread, 3 rashers of bacon, 3 lettuce leaves and 2 slices Why not try to create more chemical ‘menus’ of your own that of tomato is 250g, then the students can use to create their own food equations, or ask sandwich will also weigh 250g. students to come up with the food equation for their favourite This is the principal of the sandwich or meal?

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What if I want to make more than one sandwich? Well the Examples of how the moles keyhole could be used equation tells me the ratio of each ingredient that I need. So in in calculations include: the case of the bacon sandwich it is 2 bread: 3 bacon: 3 lettuce: 2 tomato, 2:3:3:2. Ask students to use this ratio to determine 1. Calculate the maximum mass of iron that can the limiting reactant (the one which runs out first) in each of be made from 500g of iron oxide according to the equation: the following cases: 2Fe₂O₃ + 3C → 4Fe + 3CO₂ Step 1 (input): convert 500g of iron oxide into moles using the 1. I have 6 slices of bread, 9 rashers of bacon, 6 pieces of lettuce input equation moles = mass/Mr(Fe₂O₃) and 6 slices of tomato Step 2 (process): apply the mole ratio of 2:4 to work out the 2. I have 4 slices of bread, 4 slices of bacon, 6 pieces of lettuce moles of iron formed and 4 slices of tomato Step 3 (output): convert the moles of iron into mass using the 3. I have 2 slices of bread, 6 slices of bacon, 6 pieces of lettuce output equation mass = moles x Mr(Fe) and 4 slices of tomato. 2. Calculate the maximum volume of hydrogen gas that Of course I could put different amounts of bacon, lettuce or is produced when 15g of magnesium ribbon reacts with an tomato into my sandwich, but if I did the sandwich would have a excess of hydrochloric acid according to the equation: different ‘chemical formula’, it would not be the same. Students Mg + 2HCl → MgCl₂ + H₂ should now complete the calculations of Mr (the mass of one Step 1 (input): convert 15g of magnesium into moles using the mole of a substance) on the GCSE bitesize website [AR1] to input equation moles = mass/Mr(Mg) prepare them for the next part of the lesson. Step 2 (process): apply the mole ratio of 1:1 to work out the moles of hydrogen gas formed Step 3 (output): convert the moles of hydrogen into volume MAIN ACTIVITIES using the output equation vol = moles x 24.

1. The Moles Keyhole 3. 25cm3 of H₂SO₄ is completely neutralised by 14.5cm3 of At A-level, I teach reacting masses, concentrations and titrations, 0.5mol/dm3 NaOH. Calculate the concentration of H₂SO₄. The gas volumes, mole ratios from chemical equations and theoretical equation for the reaction is H₂SO₄ + 2NaOH→ Na2SO₄ + 2H₂O. yield as separate concepts and expect students to put each of Step 1 (input): convert 14.5cm3 (14.5 x 10-3 dm3) of 0.5mol/dm3 these together to perform moles based calculations. This was NaOH into moles using the equation moles = conc x volume the first strategy I used to teach year 10; after all the examination Step 2 (process): apply the mole ratio of 2:1 to work out the specification is also structured in this way with each concept and moles of H2SO4 that have reacted. its associated equations contained within a different specification Step 3 (output): convert the moles of sulfuric acid into statement. I discovered however, that students had no idea how to concentration using the output equation conc = moles/ choose which equation to use; they couldn’t put them together to vol(H2SO4) work out reacting masses, volumes and concentrations and they struggled to link the calculations back to the chemical equation 4. 650g of Ti is formed from the reaction of 2000g of TiCl₄ or the mole ratio. It was this challenge that led me to employ the according to the equation TiCl₄ + 2Mg → Ti + 2MgCl₂ . What is ‘moles keyhole’ for two substances linked by a chemical equation. the percentage yield of the reaction? Step 1 (input): convert 200g of TiCl₄ into moles using the

Figure 3 – The moles keyhole equation moles = mass/Mr(TiCl4) The aim of the moles keyhole is to use a computer programming Step 2 (process): apply the mole ratio of 1:1 to work out the based approach with inputs and outputs and the mole ratio theoretical moles of Ti that could be formed process in the centre. To tackle any question, you need to Step 3 (output): convert the moles of Ti into mass using the convert a quantitative measurement into an input in moles, output equation mass = moles x Mr (Ti) you then apply the mole ratio and convert back out of moles to Step 4 (yield): calculate the percentage of the actual mass the appropriate quantitative measurement. The system can be formed (650g) compared to the theoretical maximum that you summarised as input→process→output. have just calculated in step 3.

Substance 1 Substance 2

Fig. 3 moles = mass / Mr (substance 1)

moles = mass x Mr (substance 2)

Solid MOLES Solid moles = conc / vol

moles = conc x vol moles = vol x 24

RATIO Solution Solution moles = vol/24 aA+bB → cC+dD To apply the ratio from the chemical equation divie by the coefficient (a,b,c,d) of substance 1 and multiply by the coefficient (a,b,c,d) of substance 2. Gas Gas

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2. Balanced equations from moles of substance The moles keyhole approach requires the mole ratio from the chemical equation to be known, or a balanced chemical equation to be given. Students may be required to produce the balanced chemical equation using reacting masses. Using a writing frame for this type of calculation may help to make the problem more visual for learners.

aA+bB → cC+dD

Coefficent = mass/Mr

Divide through by smallest value

Multiply up to whole numbers (if appropriate)

_A+_B → _C+_D

Figure 4 – calculation ‘writing frame’ for balanced equations Examples of how the calculation writing-frame can be used to produce balanced equations.

1. 8.1g of zinc oxide (ZnO) reacts completely with 0.60g of carbon to form 2.2 g of carbon dioxide and 6.5g of zinc. Use these masses to construct the balanced symbol equation.

_ZnO _C → _CO₂ _Zn

8.1/81 0.6/12 =6.5/65 Coefficent = mass/Mr 2.2/44 =0.1 =0.05 =0.05 =0.1

Divide through by 0.1/0.05 0.05/0.05 0.05/0.05 0.1/0.05 smallest value =2 =1 =1 =2

2ZnO C → CO₂ 2Zn

So the equation is 2ZnO + C→ CO₂ + 2Zn

2. In a chemical reaction, 36g of magnesium (Mg) reacts with 98g of phosphoric acid (H₃PO₄) to make 131g of magnesium phosphate (Mg₃(PO₄)₂) and 3g of hydrogen (H₂). Use these masses to construct the balanced symbol equation.

_H₃PO₄ →_Mg₃(PO₄)₂ _Mg H₂ SUMMARY Coefficent = mass/Mr 36/24 98/98 131/262 3/2 =1.5 =1.0 =0.5 =1.5 The website www.gcsescience.com [AR2] contains examples of moles calculations between mass, volume, Multiply up to whole molecules and concentration as well as conversions and numbers (if appropriate) 3 2 1 3 example calculations. Students should log on at home or in school to complete the revision questions [AR3] as 2H₃PO₄ → Mg₃(PO₄)₂ directed by the teacher. 3Mg 3H₂

So the equation is 3Mg + 2H3PO4 → Mg3(PO4)2 + 3H2

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EXTENSION ACTIVITY connects the balloon to a flask containing marble chips (calcium carbonate) and hydrochloric acid, again 15dm3 is collected and the balloon is sealed”. Later the students measure the mass of their Gas Volumes balloons and are surprised to discover it is different. Explain why The equations for volume given in the moles keyhole example are using calculations to support your answer. based on a temperature of 25°C and atmospheric pressure. As The solution to the problem is achieved by calculating n for an extension for your more able, or gifted and talented students, constant pressure and volume but with a temperature of -78.5°C introduce the ideal gas equation: pV=nRT where p is the pressure (194.5K) for the carbon dioxide collected from cold dry ice and in Pa, V is the volume in m , n is the number of moles, R is the gas 3 a temperature of 25°C (298K) for the carbon dioxide collected -1 -1 constant 8.314 J K mol and T is the temperature in K. Also give from the reaction at room temperature (which may even be students a number of exercises to practice converting between a little bit higher as the reaction is exothermic). The masses cm3 and dm3 and m3 and between temperature in °C and K. calculated using these values and atmospheric pressure (101325 Then set them the following problem: “Two students have been Pa) will be approximately 41g for the cold balloon and 27g for the 3 given the task of filling a balloon with a volume of 15dm of carbon warm balloon. For further opportunities to practice their moles dioxide. One of them decides to connect the balloon to a flask calculations students could be directed to complete the online quiz containing dry ice and the balloon inflates as the dry ice evaporates from the Royal Society of Chemistry [AR4] until 15dm3 is collected and the balloon is sealed. The other student

ADDITIONAL RESOURCES

[AR1] Calculating Mr http://www.bbc.co.uk/education/guides/zysk7ty/revision [AR2] Moles definitions and examples http://www.gcsescience.com/imoles.htm [AR3] Moles revision questions http://www.gcsescience.com/m31.htm [AR4]Moles calculations involving gases http://www.rsc.org/learn-chemistry/wiki/Quiz:GeneralA003:_Mole_calculations

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