⚗️ Introduction to the Mole

Part 1 of 7 — Avogadro's Number & Counting by Weighing

Chemistry deals with incredibly tiny particles — atoms, molecules, and ions. A single drop of water contains roughly $10^{21}$ molecules. How do chemists keep track of such enormous numbers?

The answer is the mole — one of the most important concepts in all of chemistry.

What Is a Mole?

A mole (abbreviated mol) is a counting unit, just like a "dozen" means 12 items. But instead of 12, a mole is a very large number:

$1 \text{ mol} = 6.022 \times 10^{23} \text{ particles}$

This number is called Avogadro's number ($N_A$), named after Italian scientist Amedeo Avogadro.

Putting It in Perspective

• A dozen eggs = 12 eggs
• A gross of pencils = 144 pencils
• A ream of paper = 500 sheets
• A mole of atoms = $6.022 \times 10^{23}$ atoms

Why Such a Specific Number?

Avogadro's number is defined so that one mole of carbon-12 atoms has a mass of exactly 12 grams. This connects the atomic mass scale (in amu) to the laboratory mass scale (in grams).

Counting by Weighing

It would be impossible to count individual atoms one by one. Instead, chemists count by weighing.

The Analogy

Imagine you work at a hardware store and need to sell 500 nails. You could count each one, or you could:

1. Weigh one nail → say it is 2.0 g
2. Calculate: $500 \times 2.0 \text{ g} = 1000 \text{ g} = 1.0 \text{ kg}$
3. Simply weigh out 1.0 kg of nails!

Chemistry works the same way. We know the mass of one mole of any element or compound, so we weigh out the right mass to get the number of particles we need.

The Core Conversion

$\text{moles} = \frac{\text{number of particles}}{6.022 \times 10^{23}}$

$\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$

Mole ↔ Particle Conversions

Example 1: Moles → Particles

How many atoms are in 2.50 mol of iron (Fe)?

$\text{atoms} = 2.50 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.506 \times 10^{24} \text{ atoms}$

Example 2: Particles → Moles

How many moles is $3.011 \times 10^{23}$ molecules of CO₂?

$\text{moles} = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} = 0.500 \text{ mol}$

Watch Out for Polyatomic Molecules!

If you have 1 mol of H₂O, you have:

• $6.022 \times 10^{23}$ molecules of H₂O
• $2 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$ atoms of H
• $6.022 \times 10^{23}$ atoms of O
• Total: $3 \times 6.022 \times 10^{23} = 1.807 \times 10^{24}$ atoms overall

Mole Concept Quiz 🎯

Mole-Particle Conversion Drill 🧮

1. How many atoms are in 0.750 mol of aluminum (Al)? Express your answer in scientific notation as $a \times 10^{23}$ — enter only the value of $a$ (to 3 significant figures).

2. A sample contains $1.806 \times 10^{24}$ molecules of CO₂. How many moles is this? (to 3 significant figures)

3. How many individual oxygen atoms are in 2.00 mol of O₂? Express as $a \times 10^{24}$ — enter only $a$ (to 3 significant figures).

Mole Concept — Fill in the Blanks 🔽

Exit Quiz — The Mole ✅

⚖️ Molar Mass

Part 2 of 7 — The Mass of One Mole

Now that we know what a mole is, we need a way to connect moles to grams — something we can actually measure on a balance. That connection is the molar mass.

What Is Molar Mass?

The molar mass ($M$) of a substance is the mass in grams of one mole of that substance.

$M = \frac{\text{mass (g)}}{\text{moles (mol)}}$

Units: g/mol

For Elements

The molar mass of an element equals its atomic mass from the periodic table, but in grams per mole:

Key Insight

The number on the periodic table does double duty: it tells you the mass of one atom in amu and the mass of one mole of atoms in grams.

Molar Mass of Compounds

For a compound, add up the molar masses of all atoms in the formula.

Example 1: Water (H₂O)

$M_{\text{H}_2\text{O}} = 2(1.008) + 1(16.00) = 2.016 + 16.00 = 18.02 \text{ g/mol}$

Example 2: Sodium Chloride (NaCl)

$M_{\text{NaCl}} = 22.99 + 35.45 = 58.44 \text{ g/mol}$

Example 3: Calcium Carbonate (CaCO₃)

$M_{\text{CaCO}_3} = 40.08 + 12.01 + 3(16.00) = 40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}$

Example 4: Glucose (C₆H₁₂O₆)

$M_{\text{C}_6\text{H}_{12}\text{O}_6} = 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.10 + 96.00 = 180.16 \text{ g/mol}$

Tip: Watch for Parentheses!

For $\text{Ca(OH)}_2$:

$M = 40.08 + 2(16.00 + 1.008) = 40.08 + 2(17.008) = 40.08 + 34.02 = 74.10 \text{ g/mol}$

Molar Mass Concept Quiz 🎯

Molar Mass Calculation Drill 🧮

Use these atomic masses: H = 1.008, C = 12.01, N = 14.01, O = 16.00, Na = 22.99, S = 32.07, Cl = 35.45, K = 39.10, Ca = 40.08, Fe = 55.85

1. Calculate the molar mass of ammonia (NH₃) in g/mol. (to 3 significant figures)

2. Calculate the molar mass of potassium permanganate (KMnO₄) in g/mol. (Mn = 54.94; to 3 significant figures)

3. Calculate the molar mass of iron(III) oxide (Fe₂O₃) in g/mol. (to 3 significant figures)

Molar Mass Concepts — Fill in the Blanks 🔽

Exit Quiz — Molar Mass ✅

🔄 Mole-Mass Conversions

Part 3 of 7 — Grams, Moles, and Particles

You now know what a mole is and what molar mass means. The next step is the most practical skill in chemistry: converting between grams, moles, and particles. This is the foundation of all stoichiometry.

Grams → Moles

To convert from grams to moles, divide by the molar mass:

$n = \frac{m}{M}$

where $n$ = moles, $m$ = mass in grams, $M$ = molar mass in g/mol.

Example 1

How many moles are in 36.04 g of water (H₂O)?

$n = \frac{36.04 \text{ g}}{18.02 \text{ g/mol}} = 2.000 \text{ mol}$

Example 2

How many moles are in 100.0 g of NaCl? ($M = 58.44$ g/mol)

$n = \frac{100.0}{58.44} = 1.711 \text{ mol}$

Moles → Grams

To convert from moles to grams, multiply by the molar mass:

$m = n \times M$

Example 3

What is the mass of 0.250 mol of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$, $M = 180.16$ g/mol)?

$m = 0.250 \times 180.16 = 45.04 \text{ g}$

Multi-Step Conversions: Grams ↔ Moles ↔ Particles

Often you need to go from grams to particles (or vice versa). This requires two steps:

$\text{grams} \xrightarrow{\div M} \text{moles} \xrightarrow{\times N_A} \text{particles}$

$\text{particles} \xrightarrow{\div N_A} \text{moles} \xrightarrow{\times M} \text{grams}$

Example: Grams → Particles

How many molecules are in 9.01 g of H₂O?

Step 1: Convert to moles: $n = \frac{9.01}{18.02} = 0.500 \text{ mol}$

Step 2: Convert to molecules: $\text{molecules} = 0.500 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$

Example: Particles → Grams

What is the mass of $1.505 \times 10^{24}$ atoms of iron (Fe, $M = 55.85$ g/mol)?

Step 1: Convert to moles: $n = \frac{1.505 \times 10^{24}}{6.022 \times 10^{23}} = 2.499 \approx 2.50 \text{ mol}$

Step 2: Convert to grams: $m = 2.50 \times 55.85 = 139.6 \text{ g}$

Conversion Concept Quiz 🎯

Mole-Mass Conversion Drill 🧮

Use: H = 1.008, C = 12.01, O = 16.00, Na = 22.99, Cl = 35.45, Ca = 40.08

1. How many moles are in 25.0 g of CaCO₃ ($M = 100.09$ g/mol)? (to 3 significant figures)

2. What is the mass (in grams) of 0.400 mol of NaCl ($M = 58.44$ g/mol)? (to 3 significant figures)

3. How many molecules are in 5.00 g of CO₂ ($M = 44.01$ g/mol)? Express as $a \times 10^{22}$ — enter $a$ (to 3 significant figures).

Conversion Roadmap — Fill in the Blanks 🔽

Exit Quiz — Mole-Mass Conversions ✅

📊 Percent Composition

Part 4 of 7 — What's in Your Compound?

When chemists analyze a substance, one of the first questions they ask is: "What percentage of the mass comes from each element?" This is called percent composition, and it is a powerful tool for identifying compounds and determining formulas.

Mass Percent Formula

The mass percent of an element in a compound is:

$\% \text{ by mass} = \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \times 100\%$

Example 1: Water (H₂O)

$M_{\text{H}_2\text{O}} = 18.02$ g/mol

• Mass of H in 1 mol: $2(1.008) = 2.016$ g
• Mass of O in 1 mol: $1(16.00) = 16.00$ g

$\%\text{H} = \frac{2.016}{18.02} \times 100\% = 11.19\%$

$\%\text{O} = \frac{16.00}{18.02} \times 100\% = 88.79\%$

Check: $11.19 + 88.79 = 99.98\% \approx 100\%$ ✓ (small rounding difference is fine)

Example 2: Glucose (C₆H₁₂O₆)

$M = 180.16$ g/mol

$\%\text{C} = \frac{6(12.01)}{180.16} \times 100\% = \frac{72.06}{180.16} \times 100\% = 40.00\%$

$\%\text{H} = \frac{12(1.008)}{180.16} \times 100\% = \frac{12.10}{180.16} \times 100\% = 6.71\%$

$\%\text{O} = \frac{6(16.00)}{180.16} \times 100\% = \frac{96.00}{180.16} \times 100\% = 53.28\%$

Finding Empirical Formula from Percent Composition

If you know the percent composition, you can determine the empirical formula (simplest whole-number ratio of atoms).

Method

1. Assume 100 g of the compound (so percentages become grams directly)
2. Convert grams to moles for each element: $n = m/M$
3. Divide all by the smallest mole value to get the ratio
4. Round to the nearest whole number (or multiply if you get values like 1.5, 2.5, etc.)

Example

A compound is 40.0% C, 6.7% H, and 53.3% O by mass. Find the empirical formula.

Step 1: Assume 100 g → 40.0 g C, 6.7 g H, 53.3 g O

Step 2: Convert to moles:

• C: $40.0 / 12.01 = 3.33$ mol
• H: $6.7 / 1.008 = 6.65$ mol
• O: $53.3 / 16.00 = 3.33$ mol

Step 3: Divide by smallest (3.33):

• C: $3.33/3.33 = 1.00$
• H: $6.65/3.33 = 2.00$
• O: $3.33/3.33 = 1.00$

Step 4: Ratio = 1 : 2 : 1 → Empirical formula: CH₂O

Percent Composition Concept Quiz 🎯

Percent Composition Calculation Drill 🧮

Use: H = 1.008, C = 12.01, N = 14.01, O = 16.00, S = 32.07

1. What is the percent by mass of nitrogen in ammonia (NH₃, $M = 17.03$ g/mol)? (to 3 significant figures)

2. What is the percent by mass of sulfur in SO₃ ($M = 80.07$ g/mol)? (to 3 significant figures)

3. A compound is 85.7% C and 14.3% H by mass. What is the ratio of C to H in the empirical formula? (Enter as a single number: if the ratio is C₁H₂, enter 2)

Percent Composition — Fill in the Blanks 🔽

Exit Quiz — Percent Composition ✅

🔬 Empirical and Molecular Formulas

Part 5 of 7 — From Ratios to Real Formulas

In Part 4, we learned how to find the empirical formula — the simplest whole-number ratio of atoms. But many compounds share the same empirical formula. To find the molecular formula (the actual number of atoms in each molecule), we need one more piece of information: the molar mass.

Empirical vs. Molecular Formulas

Key Relationship

$\text{Molecular formula} = n \times (\text{Empirical formula})$

where:

$n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}}$

$n$ must be a positive integer (1, 2, 3, ...).

Examples of Empirical ↔ Molecular

Step-by-Step Method

Finding the Empirical Formula from Experimental Data

1. Start with mass or percent of each element
2. Convert to moles: $n_i = m_i / M_i$
3. Divide all by the smallest mole value
4. If needed, multiply to get whole numbers

Finding the Molecular Formula

1. Calculate the molar mass of the empirical formula ($M_{\text{emp}}$)
2. Divide the given molecular molar mass by $M_{\text{emp}}$: $n = M_{\text{mol}} / M_{\text{emp}}$
3. Multiply all subscripts in the empirical formula by $n$

Worked Example

A compound is 40.0% C, 6.7% H, and 53.3% O by mass. Its molar mass is 180.2 g/mol. Find the molecular formula.

Step 1–4 (from Part 4): Empirical formula = CH₂O

Step 5: $M_{\text{emp}} = 12.01 + 2(1.008) + 16.00 = 30.03$ g/mol

Step 6: $n = 180.2 / 30.03 = 6.00$

Step 7: Molecular formula = $\text{C}_{1 \times 6}\text{H}_{2 \times 6}\text{O}_{1 \times 6} = \text{C}_6\text{H}_{12}\text{O}_6$ (glucose!)

Empirical & Molecular Formula Quiz 🎯

Empirical & Molecular Formula Drill 🧮

Use: H = 1.008, C = 12.01, N = 14.01, O = 16.00, P = 30.97

1. A compound is 43.6% P and 56.4% O by mass. What is the mole ratio of P to O? Give as the number of O per 1 P (to 3 significant figures).

2. The empirical formula from question 1 is P₂O₅. Calculate $M_{\text{emp}}$ in g/mol. (to 3 significant figures)

3. If the molar mass of the compound is 283.88 g/mol, what is $n$ (the multiplier to get the molecular formula)? (whole number)

Formulas — Fill in the Blanks 🔽

Exit Quiz — Empirical & Molecular Formulas ✅

🧪 Problem-Solving Workshop

Part 6 of 7 — Multi-Step Conversions & Real-World Applications

You now have all the core tools: moles, molar mass, percent composition, and formulas. This part puts them all together with multi-step problems, real lab scenarios, and mixed practice.

The Mole Conversion Map

Here is the complete conversion roadmap you should have memorized:

$\text{Particles} \xleftrightarrow{\times \text{ or } \div \; N_A} \text{Moles} \xleftrightarrow{\times \text{ or } \div \; M} \text{Grams}$

Key Formulas Summary

Strategy for Multi-Step Problems

1. Identify what you are given and what you need to find
2. Plan the conversion path (grams → moles → particles, etc.)
3. Set up conversion factors so units cancel
4. Calculate and check significant figures
5. Verify — does the answer make sense?

Lab Scenario: Analyzing an Unknown

A student in the lab weighs out 11.0 g of an unknown white solid. Analysis shows it is pure calcium carbonate (CaCO₃, $M = 100.09$ g/mol).

Questions to Answer

a) How many moles? $n = \frac{11.0}{100.09} = 0.1099 \text{ mol}$

b) How many formula units? $N = 0.1099 \times 6.022 \times 10^{23} = 6.62 \times 10^{22}$

c) How many total atoms? (CaCO₃ has 5 atoms per formula unit: 1 Ca + 1 C + 3 O) $\text{atoms} = 5 \times 6.62 \times 10^{22} = 3.31 \times 10^{23}$

d) How many grams of calcium are in the sample? $m_{\text{Ca}} = 0.1099 \text{ mol} \times 40.08 \text{ g/mol} = 4.41 \text{ g}$

Or using percent composition: $\%\text{Ca} = 40.08/100.09 \times 100 = 40.04\%$; $m_{\text{Ca}} = 0.4004 \times 11.0 = 4.40$ g.

Multi-Step Problem Quiz 🎯

Multi-Step Calculation Drill 🧮

Use: H = 1.008, C = 12.01, N = 14.01, O = 16.00, Na = 22.99, Cl = 35.45, Fe = 55.85

1. How many grams of iron (Fe) contain the same number of atoms as 12.01 g of carbon (C)? (to 3 significant figures)

2. A chemist has 25.0 g of NaCl ($M = 58.44$). How many chloride ions ($\text{Cl}^-$) are present? Express as $a \times 10^{23}$ — enter $a$ (to 3 significant figures).

3. How many hydrogen atoms are in 36.04 g of water ($M_{\text{H}_2\text{O}} = 18.02$)? Express as $a \times 10^{24}$ — enter $a$ (to 3 significant figures).

Problem-Solving Strategy — Fill in the Blanks 🔽

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Connecting Concepts & Exam Preparation

This final part ties together everything from Parts 1–6. You will practice AP-style multiple-choice questions, tackle free-response-style calculations, and learn the most common mistakes to avoid on exam day.

Concept Connections

All of the mole concepts are interconnected:

$\text{Percent Composition} \rightarrow \text{Empirical Formula} \xrightarrow{+\,M} \text{Molecular Formula}$

$\text{Mass (g)} \xleftrightarrow{M} \text{Moles} \xleftrightarrow{N_A} \text{Particles}$

The Big Ideas

1. The mole bridges the atomic and macroscopic worlds
2. Molar mass connects mass (measurable) to amount (moles)
3. Percent composition reveals the elemental makeup of compounds
4. Empirical formulas come from mole ratios of elements
5. Molecular formulas require knowing the molar mass

Common Exam Topics

• Converting between grams, moles, and particles
• Calculating molar mass of compounds
• Determining empirical and molecular formulas
• Percent composition calculations
• Multi-step conversion problems

Common Mistakes to Avoid ⚠️

1. Confusing Atomic Mass and Molar Mass Units

• Atomic mass → amu (for single atoms)
• Molar mass → g/mol (for one mole)
• The numbers are the same; only the units differ!

2. Forgetting Subscripts in Molar Mass Calculations

• $M_{\text{Ca(OH)}_2} \neq 40.08 + 16.00 + 1.008$
• $M_{\text{Ca(OH)}_2} = 40.08 + 2(16.00) + 2(1.008) = 74.10$ g/mol ✓

3. Atoms vs. Molecules

• 1 mol H₂O = $6.022 \times 10^{23}$ molecules (not atoms!)
• Total atoms in 1 mol H₂O = $3 \times 6.022 \times 10^{23} = 1.807 \times 10^{24}$ atoms

4. Rounding Mole Ratios Too Soon

• A ratio of 1.33 is NOT 1 — it is $4/3$, so multiply all by 3
• A ratio of 1.50 is NOT 2 — it is $3/2$, so multiply all by 2
• A ratio of 1.25 is $5/4$, so multiply all by 4

5. Forgetting to Use the Mole Ratio (Not Mass Ratio)

• In stoichiometry, always convert to moles first, never work directly with grams

AP-Style Multiple Choice 🎯

AP-Style Free Response Calculations 🧮

A student analyzes a pure sample of hydrated copper(II) sulfate, CuSO₄·5H₂O.

Use: Cu = 63.55, S = 32.07, O = 16.00, H = 1.008

1. Calculate the molar mass of CuSO₄·5H₂O in g/mol. (to 3 significant figures)

2. What is the percent by mass of water in CuSO₄·5H₂O? (to 3 significant figures)

3. If the student has 50.0 g of CuSO₄·5H₂O, how many moles of water molecules are present? (to 3 significant figures)

Comprehensive Review — Fill in the Blanks 🔽

Final Exit Quiz — Moles & Molar Mass ✅