Quantitative chemistry: masses, moles and yield
Quantitative chemistry turns equations into predictions: how much product you should get, and how much you actually did.
Lesson overview
What you'll learn in this lesson
Carry out quantitative chemistry calculations and explain chemical changes.
Key learning points
- • Relative formula mass and moles
- • Using equations
- • Yield and atom economy
This lesson at a glance
- 30 minutes
- 19 parts to scroll through
- 4 quick checks
- Marked quiz at the end
- Gentle pace: short sittings with pauses
Words to know
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Part 1 of 19
Visual introductionPicture this
Quantitative chemistry: masses, moles and yield
Quantitative chemistry turns equations into predictions: how much product you should get, and how much you actually did.
In a nutshell
Carry out quantitative chemistry calculations and explain chemical changes.
Learning cycle
Part 2 of 19
Learning cycle 1 of 2
Part 1 · Relative formula mass and moles
A short piece of teaching, then a check to make sure it has landed.
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Relative formula mass and moles
Relative formula mass, Mr, is the sum of the relative atomic masses in a formula. Moles = mass ÷ Mr, so 44 g of CO₂ (Mr = 44) is one mole. Setting out mass, Mr and moles in a small table before calculating prevents most errors.
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Learn
Using equations
The balancing numbers give the ratio of moles reacting. Convert the known mass to moles, apply the ratio, then convert back to a mass. A limiting reactant is the one that runs out first and therefore determines how much product forms; the other is in excess.
Quick check
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Quick check
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Pause
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Learning cycle 2 of 2
Part 2 · Yield and atom economy
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Yield and atom economy
Percentage yield = actual ÷ theoretical × 100, and is below 100% because of incomplete reactions, losses during transfer and side reactions. Atom economy = Mr of the desired product ÷ total Mr of reactants × 100, and measures how much of the starting material becomes something useful.
Quick check
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Quick check
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Worked example
Worked answer: 10 g of calcium carbonate decomposes to calcium oxide and carbon dioxide. Find the theoretical mass of CaO (4 marks)
Mr of CaCO₃ = 40 + 12 + 48 = 100, and Mr of CaO = 40 + 16 = 56 (1). Moles of CaCO₃ = 10 ÷ 100 = 0.1 mol (1). The equation CaCO₃ → CaO + CO₂ is a 1 : 1 ratio, so 0.1 mol of CaO forms (1). Mass = 0.1 × 56 = 5.6 g (1). If only 4.8 g were collected, the percentage yield would be 4.8 ÷ 5.6 × 100 = 86%.
Challenge round
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Game · Sort it
Which of these are true?
Drag each card into the right column. Tap a card first if dragging is fiddly.
True
Not true
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Part 16 of 19
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Challenge round
Part 17 of 19
Game · Recall cards
Moles equals
Card 1 of 4
Mastery quiz
Part 18 of 19
Marked quiz
End of lesson quiz: Quantitative chemistry: masses, moles and yield
4 questions, marked with the reasoning shown. No timer.
1. Moles equals
2. The limiting reactant is the one that
3. Percentage yield is
4. Atom economy measures
Lesson round-up
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Lesson round-up
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