Chapter 4 of 8

All 8 chapters
  1. 1Chemical formulae, readunderstood
  2. 2Balancing equations, readunderstood
  3. 3The mole, readunderstood
  4. 4Reacting masses and yield, readunderstood
  5. 5Acids, bases and salts, readunderstood
  6. 6The reactivity series, readunderstood
  7. 7Electrolysis, readunderstood
  8. 8Energy changes and rates, readunderstood

S4.7.4 Chapter 4

Reacting masses and yield

If I burn 12 g of magnesium, how much magnesium oxide can I make?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Calculate the mass of product from a given mass of reactant
  • Calculate percentage yield and explain why it is below 100

A balanced equation is a recipe. In 2Mg + O₂ → 2MgO, two magnesium atoms and one oxygen molecule make two units of magnesium oxide. A chemist wants to know how much of each to weigh.

The numbers in front of the formulae count particles. So they also count moles. 2 mol of Mg and 1 mol of O₂ make 2 mol of MgO. They are not masses. A mole of each substance has its own mass.

Draw it first

From the equation to the masses

  1. The balanced equation drawn as particles: two Mg circles, a plus sign and an O₂ molecule, then an arrow to two pairs of touching circles, each Mg²⁺ with O²⁻, which are magnesium oxide. Under it, in step 2, the line 2 mol : 1 mol : 2 mol. In step 3, under that, the line 48 g : 32 g : 80 g.MgMgOOMg²⁺O²⁻Mg²⁺O²⁻

    1 Write the balanced equation: 2Mg + O₂ → 2MgO.

  2. The balanced equation drawn as particles: two Mg circles, a plus sign and an O₂ molecule, then an arrow to two pairs of touching circles, each Mg²⁺ with O²⁻, which are magnesium oxide. Under it, in step 2, the line 2 mol : 1 mol : 2 mol. In step 3, under that, the line 48 g : 32 g : 80 g.MgMgOOMg²⁺O²⁻Mg²⁺O²⁻2 mol : 1 mol : 2 mol

    2 The numbers in front are moles: 2 mol of Mg and 1 mol of O₂ make 2 mol of MgO.

  3. The balanced equation drawn as particles: two Mg circles, a plus sign and an O₂ molecule, then an arrow to two pairs of touching circles, each Mg²⁺ with O²⁻, which are magnesium oxide. Under it, in step 2, the line 2 mol : 1 mol : 2 mol. In step 3, under that, the line 48 g : 32 g : 80 g.MgMgOOMg²⁺O²⁻Mg²⁺O²⁻2 mol : 1 mol : 2 mol48 g : 32 g : 80 g

    3 Change moles into grams with Mr. 2 × 24 = 48 g, 1 × 32 = 32 g and 2 × 40 = 80 g.

To find the mass of a product, work in three steps. Change the mass you know into moles, with mass ÷ Mr. Use the equation to find the moles of the substance you want. Change those moles into mass, with moles × Mr.

Four boxes joined by arrows, from top to bottom. Mass of the substance you know. Moles of it: mass ÷ Mr. Moles of the other substance: use the equation ratio. Mass of the other substance: moles × Mr.Mass of the substance youknowMoles of it: mass ÷ MrMoles of the other substance:use the equation ratioMass of the other substance:moles × Mr
The method for every reacting mass question.

Worked example

What mass of magnesium oxide forms when 12 g of magnesium burns? Use 2Mg + O₂ → 2MgO.

  1. The ratio of Mg to MgO in the equation is 2 to 2, which is 1 to 1.
  2. Moles of Mg = 12 ÷ 24 = 0.5 mol.
  3. Moles of MgO = 0.5 mol, from the 1 to 1 ratio.
  4. Mr of MgO = 24 + 16 = 40.
  5. Mass of MgO = 0.5 × 40 = 20 g.

Answer: 20 g

Worked example

What mass of hydrogen chloride, HCl, reacts with 12 g of magnesium? Use Mg + 2HCl → MgCl₂ + H₂.

  1. The ratio of Mg to HCl in the equation is 1 to 2.
  2. Moles of Mg = 12 ÷ 24 = 0.5 mol.
  3. Moles of HCl = 2 × 0.5 = 1 mol.
  4. Mr of HCl = 1 + 35.5 = 36.5.
  5. Mass of HCl = 1 × 36.5 = 36.5 g.

Answer: 36.5 g

A graph. The bottom is labelled magnesium (g) and runs from 0 to 24. The side is labelled magnesium oxide (g) and runs from 0 to 40. A straight line starts at 0 and 0 and rises to 24 and 40.036912151821240510152025303540magnesium (g)magnesium oxide (g)
More magnesium makes more oxide, in a fixed ratio. Double the magnesium and the oxide doubles. The line is straight and starts at 0.

The amount a perfect reaction would make is the theoretical yield. The amount we really collect is the actual yield. The actual yield is nearly always smaller.

The percentage yield compares them. percentage yield = actual yield ÷ theoretical yield × 100.

A bar chart labelled magnesium oxide (g), with two bars. The bar labelled theory reaches 20. The bar labelled actual reaches 18.0510152025theoryactualmagnesium oxide (g)
The theory says 20 g. A student collects 18 g.

Worked example

The theoretical yield of magnesium oxide is 20 g. A student collects 18 g. What is the percentage yield?

  1. percentage yield = actual yield ÷ theoretical yield × 100.
  2. Actual yield = 18 g. Theoretical yield = 20 g.
  3. 18 ÷ 20 = 0.9.
  4. 0.9 × 100 = 90%.

Answer: 90%

Why is the yield below 100%? Atoms are not lost. The product is somewhere, but not all of it reaches the scale. Some stays on the glass or the filter paper when it is moved. Another reaction may make a different product. The reaction may stop before all the reactants are used. Some product may stay dissolved.

A yield over 100% would be a mistake. The product may be wet or mixed with another substance, so it weighs too much.

See it three ways

As a picture

A straight line graph. The bottom is labelled magnesium (g) and runs from 0 to 24. The side is labelled magnesium oxide (g) and runs from 0 to 40. The line rises from 0 and 0.036912151821240510152025303540magnesium (g)magnesium oxide (g)

In words

The oxide is always 40 for every 24 of magnesium, which is 5 for every 3. A straight line through zero shows that the two masses are in a fixed ratio.

In symbols

mass of MgO = mass of Mg × 40 ÷ 24

Try it

Where does the yield go?

You need: four cups, or two that you wash and dry, a spoon, water, cooking oil, paper and a pencil.

Be safe: Taste nothing. Wipe up any oil that spills, because it makes the floor slippery.

  1. Count out 10 level spoons of water into one cup.
  2. Pour all of it into the second, dry cup. Pour slowly, and wait until the last drop falls.
  3. Use the same spoon to lift the water out of the second cup. Count the spoons, until you can lift no more.
  4. Do the same with 10 spoons of cooking oil, using two clean, dry cups.

What you will see: You get back a little less than you put in, and the oil loses more than the water. The liquid clings to the cup, so it never all arrives. Real reactions lose product the same way, so the actual yield is below the theoretical yield. Your own counts may differ from the picture.

If you cannot try it, this is what happens.
A table with the columns Liquid, In and Out. Water: 10 spoons in, 9 spoons out. Oil: 10 spoons in, 8 spoons out. These are an example, not a fixed result.LiquidInOutwater10 spoons9 spoonsoil10 spoons8 spoons

A wrong idea many people have

In 2Mg + O₂ → 2MgO the numbers say that 2 g of Mg reacts with 1 g of O₂.

Why it feels right: The numbers are in the equation, and they look like quantities to weigh out.

The test that breaks it: Work out the masses of 2 mol of Mg and 1 mol of O₂. They are 48 g and 32 g, which is not a ratio of 2 to 1.

What is true: The numbers in front are moles, or numbers of particles. To get masses, multiply the moles by Mr.

Key words

theoretical yield
the mass of product a perfect reaction would make.
actual yield
the mass of product really collected.
percentage yield
the actual yield divided by the theoretical yield, times 100.

Back to your guess

The answer is B: More than 12 g, and the equation shows how much more.

The product is 20 g. Oxygen from the air joins the magnesium, so the oxide weighs more than the metal. The equation, with Mr, tells us how much.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The graph shows magnesium oxide made from magnesium. Read it. How many grams of magnesium oxide form from 15 g of magnesium?

    A graph. The bottom is labelled magnesium (g) and runs from 0 to 24. The side is labelled magnesium oxide (g) and runs from 0 to 40. A straight line starts at 0 and 0 and rises to 24 and 40.036912151821240510152025303540magnesium (g)magnesium oxide (g)
    Tap your answer.

    1. Not this one. That is the mass of magnesium. The oxide weighs more, because oxygen joins the magnesium. Go to S4.7.2. Try another.

    2. Yes, that is right. Go up from 15 g of magnesium to the line, then across to 25 on the side. The calculation agrees: 15 ÷ 24 × 40 = 25 g.

    3. Not this one. That is the top of the line, for 24 g of magnesium. Go up from 15 on the bottom scale to the line, and then across. Try another.

    See the answer

    25 g

    Go up from 15 g of magnesium to the line, then across to 25 on the side. The calculation agrees: 15 ÷ 24 × 40 = 25 g.

    • If you chose A: That is the mass of magnesium. The oxide weighs more, because oxygen joins the magnesium. Go to S4.7.2.
    • If you chose C: That is the top of the line, for 24 g of magnesium. Go up from 15 on the bottom scale to the line, and then across.

  2. In N₂ + 3H₂ → 2NH₃, how many moles of NH₃ form from 3 mol of H₂?

    Tap your answer.

    1. Not this one. The numbers in front are 3 for H₂ and 2 for NH₃. They are not equal. Go to S4.7.2. Try another.

    2. Not this one. That multiplies by 2, but the ratio is 3 H₂ to 2 NH₃. 3 mol of H₂ is one lot of the recipe. Try another.

    3. Yes, that is right. The ratio of H₂ to NH₃ is 3 to 2. So 3 mol of H₂ make 2 mol of NH₃.

    See the answer

    2 mol

    The ratio of H₂ to NH₃ is 3 to 2. So 3 mol of H₂ make 2 mol of NH₃.

    • If you chose A: The numbers in front are 3 for H₂ and 2 for NH₃. They are not equal. Go to S4.7.2.
    • If you chose B: That multiplies by 2, but the ratio is 3 H₂ to 2 NH₃. 3 mol of H₂ is one lot of the recipe.

  3. Calcium carbonate breaks down: CaCO₃ → CaO + CO₂. What mass of calcium oxide, CaO, forms from 10 g of calcium carbonate?

    Tap your answer.

    1. Yes, that is right. Mr of CaCO₃ is 100, so 10 g is 0.1 mol. The ratio is 1 to 1, so 0.1 mol of CaO forms. Mr of CaO is 56, so the mass is 0.1 × 56 = 5.6 g.

    2. Not this one. That is the mass of the carbon dioxide, 0.1 × 44. The question asks for the calcium oxide. Try another.

    3. Not this one. The product has less mass, because carbon dioxide gas leaves. Use moles: 10 g is 0.1 mol. Try another.

    See the answer

    5.6 g

    Mr of CaCO₃ is 100, so 10 g is 0.1 mol. The ratio is 1 to 1, so 0.1 mol of CaO forms. Mr of CaO is 56, so the mass is 0.1 × 56 = 5.6 g.

    • If you chose B: That is the mass of the carbon dioxide, 0.1 × 44. The question asks for the calcium oxide.
    • If you chose C: The product has less mass, because carbon dioxide gas leaves. Use moles: 10 g is 0.1 mol.

  4. A reaction should make 40 g of product. It makes 30 g. What is the percentage yield?

    Tap your answer.

    1. Not this one. That divides the wrong way round. The actual yield goes on top: 30 ÷ 40. Try another.

    2. Yes, that is right. percentage yield = actual ÷ theoretical × 100 = 30 ÷ 40 × 100 = 75%.

    3. Not this one. That is the percentage lost, 10 ÷ 40. The yield is the percentage collected. Try another.

    See the answer

    75%

    percentage yield = actual ÷ theoretical × 100 = 30 ÷ 40 × 100 = 75%.

    • If you chose A: That divides the wrong way round. The actual yield goes on top: 30 ÷ 40.
    • If you chose C: That is the percentage lost, 10 ÷ 40. The yield is the percentage collected.

  5. The bars show a theoretical yield and an actual yield. What is the percentage yield?

    A bar chart labelled product (g) with two bars. The bar labelled theory reaches 50. The bar labelled actual reaches 40.0102030405060theoryactualproduct (g)
    Tap your answer.

    1. Yes, that is right. percentage yield = actual ÷ theoretical × 100 = 40 ÷ 50 × 100 = 80%.

    2. Not this one. That divides the theoretical yield by the actual yield. The actual yield goes on top. Try another.

    3. Not this one. That is the difference in grams, 10 g, not a percentage of the theoretical yield. Try another.

    See the answer

    80%

    percentage yield = actual ÷ theoretical × 100 = 40 ÷ 50 × 100 = 80%.

    • If you chose B: That divides the theoretical yield by the actual yield. The actual yield goes on top.
    • If you chose C: That is the difference in grams, 10 g, not a percentage of the theoretical yield.

  6. A student works out a yield of more than 100%. What is the most likely reason?

    Tap your answer.

    1. Not this one. Atoms are never made in a reaction. The product must have weighed more for another reason. Go to S4.7.2. Try another.

    2. Not this one. Losing product makes the yield smaller, not larger. Try another.

    3. Yes, that is right. Water or an impurity adds mass, so the product weighs more than it should. The mistake is in the weighing, not in the chemistry.

    See the answer

    The product was still wet or mixed with another substance.

    Water or an impurity adds mass, so the product weighs more than it should. The mistake is in the weighing, not in the chemistry.

    • If you chose A: Atoms are never made in a reaction. The product must have weighed more for another reason. Go to S4.7.2.
    • If you chose B: Losing product makes the yield smaller, not larger.

  7. From an earlier chapter

    How many moles are in 11 g of carbon dioxide, CO₂?

    Tap your answer.

    1. Not this one. That is 44 ÷ 11, upside down. Moles are mass divided by Mr. Go to S4.7.3. Try another.

    2. Yes, that is right. Mr of CO₂ is 44. moles = mass ÷ Mr = 11 ÷ 44 = 0.25 mol.

    3. Not this one. Mr of CO₂ is 12 + 2 × 16 = 44, so 11 g is a quarter of a mole, not a half. Go to S4.7.3. Try another.

    See the answer

    0.25 mol

    Mr of CO₂ is 44. moles = mass ÷ Mr = 11 ÷ 44 = 0.25 mol.

    • If you chose A: That is 44 ÷ 11, upside down. Moles are mass divided by Mr. Go to S4.7.3.
    • If you chose C: Mr of CO₂ is 12 + 2 × 16 = 44, so 11 g is a quarter of a mole, not a half. Go to S4.7.3.

  8. Magnesium burns: 2Mg + O₂ → 2MgO. What mass of magnesium oxide forms from 6 g of magnesium?

    See the answer

    10 g

    6 ÷ 24 = 0.25 mol of Mg. The ratio is 1 to 1, so 0.25 mol of MgO forms. Its mass is 0.25 × 40 = 10 g.

  9. The theoretical yield is 40 g. The actual yield is 34 g. What is the percentage yield?

    See the answer

    85%

    percentage yield = actual ÷ theoretical × 100 = 34 ÷ 40 × 100 = 85%.

Keep it

Copy this into your exercise book.

Numbers in an equation are moles. Mass to moles with Mr, moles to moles with the ratio, moles to mass with Mr. Percentage yield = actual ÷ theoretical × 100.

Draw: The flow of four boxes: mass, moles, other moles, other mass, with the three operations on the arrows.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (907 KB). It opens on any phone.