This quiz will assess your knowledge from the previous lesson and previous lessons

H₂ (g) + Cl₂ (g) → 2HCl (g). Starting with 4 g of hydrogen, what mass of hydrogen chloride would be produced? (Relative atomic masses: H = 1; Cl = 35.5)

Which of these is NOT a reason why the percentage yield is less than 100%?

Which of these is a reason for wanting high atom economy?

Calculate the number of moles in 20 cm³ of a 1.25 M solution of NaOH.

Calculate the volume of O₂ needed to completely burn 18 g of methane (CH₄). Relative atomic mass of C = 12; H = 1.

Review: Quantitative Chemistry - Triple

﻿Hello, I'm Mrs. Begum and welcome to another quantitative chemistry lesson. This lesson, we'll be focusing on looking at the practical application of some of the calculations that you've learnt in this topic. For this lesson, you will need a pen, paper, calculator and a copy of the periodic table. Please make sure that you are distraction free so all the tabs are closed, make sure your mobile phone is on silent, and also make sure you're working in a quiet environment. This is so that you could focus on the lesson and get the most out of this lesson. As always, don't give up, make sure you complete the lesson right to the end. Okay, so when you're ready, let's begin. Okay, so in this lesson, you're going to apply a lot of the calculations that you've learnt to a practical example to make a specific mass of salt in a lab. So we're looking at calculating moles. Then looking at calculating volumes from moles, and looking at writing a method for making a salt. Then finally, looking at calculating yield and atom economy. Okay, so let's start off with a quick warmup. Pause the lesson here to answer these questions. You will need a copy of the periodic table to help you answer these questions, resume the lesson once you have completed it. Okay, let's check the answers. How did you do? Okay, so, question number one, the answer is 0.05 moles. Question number two is 0.002 moles. Question number three, four grammes of O two will react with six grammes of magnesium. Question number four, you will need 2.4 grammes of magnesium to react with 7.3 grammes of HCl and the added 4.8 grammes of mg. So the MG is an excess. So the HCl is the limiting factor. Number five, to calculate yield use actual yield divided by the theoretical yield times it by 100. Number six, the answer is 77.7%. Well done. Fantastic effort. Okay, so the task for this lesson is to make copper sulphate salts. So first of all, let's think about what chemicals do you need to make copper sulphate salts. Can we do it with copper and sulfuric acid? Well, no we can't because copper is an unreactive metal and you add it to acid it won't work. So we can't use that. Copper Oxide or sulfuric acid? Yes, we can use that. Copper oxide and sulfuric acid produced copper sulphate and water. What about copper carbonate and sulfuric acid? Well, yes, we can use that as well and that would produce copper sulphate, carbon dioxide and water. But for this lesson, we're going to use copper oxide and sulfuric acid. Okay, so this is your look on the screen. This is the equipment that you would need to produce the salt. So you have, first of all you have the sulfuric acid and copper oxide, labelled on the screen there. What are these two pieces of equipment called? That's right, filter paper and funnel and then this piece of equipment is called an evaporating basin or evaporating dish. And you will need a Bunsen burner, a Tripod and a Gauze and a Heat Proof Mat. Okay, so to produce the copper sulphate salt the first step you would do is you would heat the acid up very gently. Then you take the acid gently heated up acid and the other one spatula of copper oxide at a time. Okay once you've added a copper oxide it's really important that you stir to make sure that it's fully reacted okay this shows that all the copper oxide has dissolved is reacted. So we need to add more keep adding more copper oxide until no more will dissolve the copper oxide is in excess. Okay, so that shows that it is reacted and the copper oxide isn't dissolving anymore. It's not reacting anymore. Okay, so then you take you need to obviously you need to get rid of the insoluble copper oxide, excess copper oxide. So use the conical flask, a funnel, and a piece of filter paper. Okay and then you filter the copper sulphate solution with the excess copper oxide. and because the copper oxide is insoluble it will stay on the filter paper, the copper sulphate solution will easily pass through the paper because it's small the particles are small enough to pass through and it will collect in the conical flask. Okay, so once you have your copper sulphate solution, it's important that you crystallise it or evaporate the water to produce the copper sulphate salts. Okay, time for a quick quiz. Why was the copper oxide stirred after each addition? Stirring is fun, to cool the reaction to make sure it is fully reacted, Or to give the particles less energy. That's right. It's make sure if fully reacted, What observation was made when the copper oxide could no longer react? The bubbling stopped, the solution cooled to below room temperature. The solution turned from blue to colourless, or the insoluble copper oxide remained in the beaker. Fantastic it is insoluble copper oxide remained in the beaker. What word means adding more of a reactant than is needed to fully react? Is it excess, is it limiting reactant, equilibrium or exothermic? Super, its excess. And what phrase means a substance that will run out first in the reaction? Is it excess, is it limiting reactant, equilibrium, or exothermic? Fantastic, it's limiting reactant. Now that you know how to make copper sulphate salts, your task for this lesson is to make five grammes of copper sulphate salts. So that's the task so how do we do it? So, looking at the reaction that we're going to use, the copper oxide plus sulfuric acid to produce copper sulphate plus water. That's the reaction that we're going to use. So, first step is to work out how to make five grammes of copper sulphate is to find the MR to the copper oxide has an MR of 78 79.5 H2SO4 is 98 copper sulphate is 159.5 and water is 18. So quick, quick check here, the mass in equal the mass out. Okay, so then let's use a value from the question. We want to make five grammes of copper sulphate not hundred 159.5. So work out the scale factor. So you do 159.5 divided by 159.5 times it by five. Okay, so that would mean that I would need to do the same to the copper oxide 2.49 grammes of copper oxide, but the copper oxide we added excess anyway. So that isn't the limiting reactant. We want to make sure that the H2SO4 the sulfuric acid isn't the limiting reactant. That's what we want to find out. So we always add the copper oxide in excess. So we need to make sure that the acid isn't the limiting reactant. So how much acid do we need? Okay, so how can we make sure we add enough H2SO4? So the acid sulfuric acid is one molar. So first step is to find out how many moles are there in five grammes of CuSO4. So we do five, that's the mass divided by the MR which we worked out doing 159.5 and answer is 0.031 moles. So if you look at the balanced equation, it shows a ratio of one to one for copper sulphate and hydrosulfuric acid. So if there's zero point if you need 0.031 moles of copper sulphate, then you would also need 0.0 0.031 moles of H2SO4. So what volume do we need? So we know that it's 0.031 moles. So volume equals moles divided by concentration. So we know the moles, we know the concentration. So that tells us in decimeters cubed that we need 0.031 decimeters cubed. Well obviously, we're in a lab we're going to work in centimetres cube. So therefore we need to convert that into centimetres cubed by multiplying a by 1000 and we need 31 centimetres cubed. Okay, so now that we know what mass and volume of each each one that we need, we can start writing the method. So the method for this reaction would be to measure 31 centimetres cubed of H2SO4 and pour it into the conical flask. You gently heat the acid add copper oxide, one spatula at a time until no more will dissolve. Filter to get rid of the excess insoluble CuO. And then evaporate the water from the copper sulphate solution and you should be left with the copper sulphate crystals. Okay, so I'll just give you the moment just to make sure that you have read the method fully. Okay, pause the lesson here to complete this task. Resume once you have finished. Okay, how do you do? Let's look at the answers. So what why do you gently heat the acid? Well, the acid is gently heated to make the reaction faster and to make sure all the copper oxide reacts. Why is excess copper oxide added? Excess copper oxide is added to make sure all the acid reacts. And the last question, why do you need to filter? Well, we filter to get rid of the in soluble copper oxide Time for you to have a go, the first task is I want you to I want you to produce 6.5 grammes of copper sulphate using 1.5 molar H2SO4. The first step is work out the volume of acid you would need to make sure that the acid isn't the limiting reactant. Pause the lesson here to complete this task, resume once you have finished. If you need an extra little bit of help, I've put some hints and steps for you to use to complete this task. Okay, so how did you do? So the number of moles in 6.5 grammes of copper sulphate is 0.041. So the balance equation shows a ratio of one to one. So therefore you would also need 0.041 moles of H2SO4. So then you calculate the volume, so it's the moles 0.041 divided by the concentration which is 1.5. And that gives 0.027 decimeters cubed but we're working in centimetres cubed so we need to convert that into centimetres cube. So 0.027 times it by 1000 which is 27 centimetres cubed. That's the volume of acid you would need. So the next step I would like you to do, is I would like you to write a method for your experiment. So thinking back to the method I described, we saw on the video, write a method for this experiment, for your experiment to make 6.5 grammes of copper sulphate. Okay, let's look at what your method should be like. Pause the lesson here to check your method. Okay, so yield. Okay, so we've looked at yield before. So what is the yield for your experiment? Obviously, you haven't carried this experiment, but I will give you some data in which you can calculate the yield. So, how do we calculate yield? The actual mass obtained from your experiment divided by the theoretical mass times it but 100% Okay, so let's look at how to calculate your yield. So your task was to produce 6.5 grammes of copper sulphate. You identified the volume of acid needed, which is fantastic. So, a student carried out your experiment and produced 4.9 grammes of copper sulphate. Okay, what is the percentage yield? And suggest reasons from the experiment that you've seen in the video why the yield might be less than 100% Pause the lesson here to complete those questions. Resume once you have finished. Okay, so let's look at the answers. So student produced 4.9 grammes and you were supposed to produce 6.5. So the yield would be 4.9 divided by 6.5 times it by 100. And the answer is 75.4% yield. So, suggest reasons why the yield is less than 100%. Some of the reasons from your from that experiment could be some copper sulphate remains on the filter paper. So if you don't wash it through some of the copper sulphate might have remained on the filter paper and some of the copper sulphate may have been spilled or lost during the transfer from the conical flask to the evaporating dish. Atom economy. So let's look at the atom economy for this reaction. So, atom economy is a relative formula mass of the desired product, in this case is a copper sulphate, divided by the total formula mass of all the reactants, so its the copper oxide and the sulfuric acid, times it by 100%. Calculate the atom of your reaction. Pause the lesson here to work out the atom economy. Resume once you've finished. Okay, so how did How did you do? Okay, so first step work out the MR And the atom economy would be 159.5, which is the MR for copper sulphate, the desired product, and then you divide by the mass of all the reactants and the answer is 89.9%. Well done. So in this lesson, you look at the practical applications of some of the calculations you've learnt in this topic. So, we looked at how to calculate moles, calculating volumes from moles, then looking at a method for making a salt, then finally, calculating yield and atom economy, just to put some of the calculation into context and where it is used. So that's the end of this lesson. Please make sure you complete the plenary quiz and well done for your fantastic efforts this lesson and I hope to see you all again in the next lesson.

This quiz will test your knowledge across the calculations unit

What phrase is given to the reactant that runs out first in a reaction?

Which of these is a valid reason for less than 100% yield in a reaction between magnesium and oxygen to produce magnesium oxide?

Which statement is a correct definition for atom economy?

One way to produce copper chloride salts is shown by the reaction: CuO + 2HCl -> CuCl₂   + H₂O. How many moles of acid are needed for every mole of product?

What volume of 2 M H₂SO₄ l is needed if 0.068 moles of product are required and the ratio of acid to product is 1:1?

Practical application of Quantitative Chemistry - Triple

Extension review on choosing the right concentration to make a named salt, covering limiting reactant, reacting mass, yield and atom economy

This lesson, students will learn how to conduct a practical to obtain copper sulfate salts using the calculations learnt in this unit.

Chemistry

Y10 Chemistry

Key Stage 4

Year 10

Practical application of quantitative chemistry

Quantitative Chemistry

UNIT

Relative formula mass (FT only)

Relative formula mass (HT only)

Moles and Avogadro's constant (HT only)

Balancing equations using moles (HT only)

Reacting masses (HT only)

Reacting masses and yield (GCSE Chemistry)

Atom economy

Concentration

Titration calculations

Limiting reactants

Gas Volumes

Review (HT only)

Review (GCSE Chemistry)

What is the theoretical yield of a product?

H₂ (g) + Cl₂ (g) → 2HCl (g). Starting with 2 g of hydrogen, what mass of hydrogen chloride would be produced? (Relative atomic masses: H = 1; Cl = 35.5)

Which of these is the number of atoms in a mole?

Which of these reactions for the production of ethanol has an atom economy of 100%?

How do you calculate atom economy?

How can you improve the atom economy of a reaction?

What is the atom economy for the production of Copper Oxide (CuO) using the reaction shown?       CuCO₃ --> CuO + CO₂     The Ar for Cu=63.5, C=12, O=16

The formula of magnesium sulfate is MgSO₄. Calculate the relative formula mass of magnesium sulfate. Relative atomic masses: Mg = 24; S= 32; O = 16.

Calculate the relative formula mass of iron oxide, Fe₂O₃.  Relative atomic masses: O = 16; Fe = 56.

A chemist was asked to identify a nitrogen compound. The chemist carried out an experiment to find the relative formula mass (Mr) of the compound. The Mr of the compound was 44. Relative atomic masses: N = 14; O = 16. What is the formula of the compound?

Calculate the percentage of O in Be(OH)₂. Be = 9; O = 16; H = 1.

What type of reaction happens when a compound breaks down due to heating?

A metal oxide has a relative formula mass (Mr) of 81. The formula of this metal oxide is XO.                            The relative atomic mass (Ar) of oxygen is 16.  Calculate the relative atomic mass (Ar) of metal X.

Which of the following equations is balanced?

Which number is missing to balance the equation : MgO + ___HCl → MgCl₂ + H₂O

What is the concentration of a solution with 0.15 g in a volume of 400 cm³?

What is the mass of copper chloride in 50 cm³ of a 300 g/dm³ solution?

How many atoms are in 1 mole of sodium?

What is the mass of 2 moles of Na₂O? Ar of Na = 23, O = 16

What volume does 1 mole of a gas occupy?

How many cm³ in 24 dm³?

Calculate the volume of gas in 12 g of CO₂? Relative atomic mass of C = 12; O = 16

Calculate the mass of 1.5 moles of MgSO₄. Relative atomic mass of Mg = 24; S = 32; O 16

Calculate the number of moles in 35 g NaCl. Relative formula mass Na = 23; Cl = 35.5

Calculate the number of moles in  14 cm³ of 2 M HCl

Lesson overview:Practical application of quantitative chemistry

Quantitative Chemistry: