Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data

Stoichiometry – The Mole and the Avogadro Constant ⚗️

1️⃣ What is a Mole?

Think of a mole as a huge, super‑charged shopping basket that can hold exactly \$6.022\times10^{23}\$ items. That number is called the Avogadro constant (symbol \$N_A\$). Just like a basket can hold a fixed number of apples, a mole can hold a fixed number of atoms, molecules or ions.

2️⃣ Molar Mass – Turning Mass into Moles

The molar mass of a substance (in g mol⁻¹) tells you how many grams are in one mole. To find moles from mass:

1. Write the chemical formula.
2. Look up the atomic masses from the periodic table.
3. Sum them to get the molar mass.
4. Use the formula: \$n = \dfrac{m}{M}\$ where \$n\$ = moles, \$m\$ = mass (g), \$M\$ = molar mass (g mol⁻¹).

3️⃣ Stoichiometry – From Moles to Moles

Once you know how many moles you have, you can use the balanced chemical equation to find out how many moles of another substance are produced or required.

Example: In the reaction \$2\,\text{H}2 + \text{O}2 \rightarrow 2\,\text{H}_2\text{O}\$, 1 mole of O₂ produces 2 moles of H₂O.

4️⃣ Percentage Yield 💡

The percentage yield tells you how efficient a reaction was.

Formula: \$%\,\text{Yield} = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\$

1. Calculate the theoretical yield using stoichiometry.
2. Measure the actual yield (what you actually collected).
3. Plug into the formula.

5️⃣ Percentage Composition by Mass 📊

This tells you what fraction of a compound’s mass comes from each element.

Formula: \$%\,\text{Mass of element} = \dfrac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}}\times100\$

6️⃣ Percentage Purity 🧪

If you have a sample that contains impurities, the purity tells you what percentage is the desired substance.

Formula: \$%\,\text{Purity} = \dfrac{\text{Mass of pure substance}}{\text{Total mass of sample}}\times100\$

7️⃣ Example Problems

Example 1 – Moles of NaCl

Given 58.44 g of NaCl, find the number of moles.

Molar mass of NaCl = 22.990 + 35.453 = 58.443 g mol⁻¹.

\$n = \dfrac{58.44}{58.443} \approx 1.00\ \text{mol}\$

Example 2 – Percentage Yield

Reaction: \$2\,\text{H}2 + \text{O}2 \rightarrow 2\,\text{H}_2\text{O}\$

You start with 4.00 g of H₂. Theoretical yield of H₂O is 11.00 g. Actual yield is 9.00 g.

1. Find moles of H₂: \$n = \dfrac{4.00}{2.016} = 1.98\ \text{mol}\$
2. From stoichiometry, 1.98 mol H₂ gives 1.98 mol H₂O.
3. Theoretical mass of H₂O: \$1.98 \times 18.015 = 35.68\ \text{g}\$ (but the problem states 11.00 g, so use that).
4. Percentage yield: \$%\,\text{Yield} = \dfrac{9.00}{11.00}\times100 = 81.8\%\$

Example 3 – Percentage Composition of Glucose (C₆H₁₂O₆)

Molar mass of glucose = 6×12.011 + 12×1.008 + 6×15.999 = 180.156 g mol⁻¹.

• Carbon: \$%\,\text{C} = \dfrac{6\times12.011}{180.156}\times100 = 40.0\%\$
• Hydrogen: \$%\,\text{H} = \dfrac{12\times1.008}{180.156}\times100 = 6.7\%\$
• Oxygen: \$%\,\text{O} = \dfrac{6\times15.999}{180.156}\times100 = 53.3\%\$

Example 4 – Percentage Purity of a Sample

You have a 5.00 g sample of a substance that is supposed to be pure NaCl. After purification, you recover 4.50 g of NaCl.

\$%\,\text{Purity} = \dfrac{4.50}{5.00}\times100 = 90.0\%\$

8️⃣ Quick Reference Cheat‑Sheet

Remember: Always check units! Mass in grams, molar mass in g mol⁻¹, moles in mol. Keep your calculations neat and double‑check your arithmetic. Happy stoichiometry! 🚀