Yield and Atom Economy

Measure how efficient a reaction is.

  • Define and explain Yield and Atom Economy in your own words
  • Use key terms such as relative formula mass (Mr) accurately
  • Apply what you have learned to new examples and questions
  • Avoid the common mistakes learners make with this topic

This lesson focuses on Yield and Atom Economy: measure how efficient a reaction is.

Definition: Yield and Atom Economy

Measure how efficient a reaction is.

Key ideas

Real reactions are never perfect

Percentage yield compares what you actually make with the theoretical maximum: (actual ÷ theoretical) × 100. Yields fall short because of reversible reactions, side reactions and losses in filtering or transfer. Atom economy goes further, asking what fraction of the reactant atoms end up in the desired product — important for sustainable industry.

The mole connects mass and particles

One mole of any substance contains 6.02 × 10²³ particles and has a mass in grams equal to its relative formula mass — so 18 g of water is one mole. The relationship moles = mass ÷ Mr lets you convert between the two in seconds. This is the key that unlocks every reacting-mass calculation.

Key term — relative formula mass (Mr): The sum of the relative atomic masses in a formula, e.g. Mr of H₂O = 18.

Worked example: Yield and Atom Economy

Explain why a chemist might prefer a reaction with 100% atom economy even if its percentage yield is lower.

High atom economy means most reactant atoms end up in the desired product rather than waste, saving raw materials and reducing waste disposal — even if some product is lost in practice.

Answer: High atom economy means most reactant atoms end up in the desired product rather than waste, saving raw materials and reducing waste disposal — even if some product is lost in practice.

Common mistakes
  • Giving percentage yield above 100% without questioning it A yield over 100% means something is wrong: the product is probably wet or impure, or a mass was mismeasured. Check your results rather than reporting an impossible yield.
  • Forgetting to double the oxygen in O₂ Mr of O₂ is 32, not 16 — the molecule has two atoms. The same trap applies to H₂, N₂ and the halogens: read the formula, not just the element.

Practice

A reaction should theoretically make 8.40 g of product but only 6.72 g is obtained. Calculate the percentage yield.
(actual ÷ theoretical) × 100.

(6.72 ÷ 8.40) × 100 = 80.0%.

Calculate the number of moles in 10.0 g of sodium hydroxide, NaOH. (Ar: Na = 23.0, H = 1.0, O = 16.0)
First find Mr of NaOH.

Mr = 23.0 + 16.0 + 1.0 = 40.0. Moles = 10.0 ÷ 40.0 = 0.250 mol.

In the worked example, what mass of oxygen reacts with the 12.0 g of magnesium?
Use the 2:1 ratio of Mg to O₂.

Moles of Mg = 0.494; moles of O₂ = 0.494 ÷ 2 = 0.247 mol; mass = 0.247 × 32.0 = 7.90 g. (Check: 12.0 + 7.90 = 19.9 g of product.).

Calculate the concentration of a solution containing 0.20 mol of solute in 0.50 dm³ of solution.
Concentration = moles ÷ volume.

0.20 ÷ 0.50 = 0.40 mol/dm³.

Quick check

Yield and Atom Economy — quick check

Which of these best defines "relative formula mass (Mr)"?

The sum of the relative atomic masses in a formula, e.g. Mr of H₂O = 18.

5.84 g of NaCl is dissolved to make 1.00 dm³ of solution. What is its concentration? (Mr of NaCl = 58.4)

Moles = 5.84 ÷ 58.4 = 0.100 mol; concentration = 0.100 ÷ 1.00 = 0.100 mol/dm³.
Key takeaways
  • Yield and Atom Economy: measure how efficient a reaction is.
  • Real reactions are never perfect: Percentage yield compares what you actually make with the theoretical maximum: (actual ÷ theoretical) × 100.
  • mole: The amount of substance containing 6.02 × 10²³ particles (Avogadro's constant).
  • Watch out for: giving percentage yield above 100% without questioning it