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Moles and Molar Mass

Understand the mole concept, Avogadro's number, and how to calculate molar mass and convert between mass, moles, and particles.

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Moles and Molar Mass

The Mole Concept

The mole (mol) is the SI unit for amount of substance. It's one of the seven fundamental SI units.

Definition: One mole contains exactly 6.022×10236.022 \times 10^{23} particles.

This number is called Avogadro's number (NAN_A).

NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23} \text{ mol}^{-1}

Why We Need Moles

Atoms and molecules are incredibly small. A mole provides a convenient way to:

  • Count particles (atoms, molecules, ions, etc.)
  • Relate macroscopic measurements to atomic-scale quantities
  • Perform stoichiometric calculations

Analogy: Just like "dozen" means 12, "mole" means 6.022×10236.022 \times 10^{23}.

Converting Between Particles and Moles

Number of particles=moles×NA\text{Number of particles} = \text{moles} \times N_A

moles=Number of particlesNA\text{moles} = \frac{\text{Number of particles}}{N_A}

Example: How many atoms are in 2.5 moles of carbon?

atoms=2.5 mol×6.022×1023 atoms/mol=1.51×1024 atoms\text{atoms} = 2.5 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.51 \times 10^{24} \text{ atoms}

Molar Mass

Molar mass (MM) is the mass of one mole of a substance, expressed in grams per mole (g/mol).

For Elements

The molar mass of an element equals its atomic mass from the periodic table.

Examples:

  • Carbon (C): 12.01 g/mol
  • Oxygen (O): 16.00 g/mol
  • Hydrogen (H): 1.008 g/mol

For Compounds

Add up the molar masses of all atoms in the molecular formula.

Example: Water (H₂O)

MH2O=2(1.008)+1(16.00)=2.016+16.00=18.02 g/molM_{H_2O} = 2(1.008) + 1(16.00) = 2.016 + 16.00 = 18.02 \text{ g/mol}

Example: Glucose (C₆H₁₂O₆)

MC6H12O6=6(12.01)+12(1.008)+6(16.00)=72.06+12.10+96.00=180.16 g/molM_{C_6H_{12}O_6} = 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.10 + 96.00 = 180.16 \text{ g/mol}

Converting Between Mass and Moles

moles=mass (g)molar mass (g/mol)\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}

mass (g)=moles×molar mass (g/mol)\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}

Example: How many moles are in 25.0 g of NaCl?

First find molar mass: MNaCl=22.99+35.45=58.44M_{NaCl} = 22.99 + 35.45 = 58.44 g/mol

moles=25.0 g58.44 g/mol=0.428 mol\text{moles} = \frac{25.0 \text{ g}}{58.44 \text{ g/mol}} = 0.428 \text{ mol}

The Mole Road Map

To convert between different quantities, use this road map:

Mass (g)↔divide by Mmultiply by MMoles↔divide by NAmultiply by NAParticles\text{Mass (g)} \xleftrightarrow[\text{divide by } M]{\text{multiply by } M} \text{Moles} \xleftrightarrow[\text{divide by } N_A]{\text{multiply by } N_A} \text{Particles}

Percent Composition

Percent composition is the percent by mass of each element in a compound.

Percent of element=mass of element in 1 molmolar mass of compound×100%\text{Percent of element} = \frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100\%

Example: Find the percent composition of H₂O

%H=2(1.008)18.02×100%=2.01618.02×100%=11.19%\%H = \frac{2(1.008)}{18.02} \times 100\% = \frac{2.016}{18.02} \times 100\% = 11.19\%

%O=16.0018.02×100%=88.81%\%O = \frac{16.00}{18.02} \times 100\% = 88.81\%

Check: 11.19%+88.81%=100%11.19\% + 88.81\% = 100\% ✓

Empirical vs. Molecular Formulas

Empirical formula: Simplest whole-number ratio of atoms Molecular formula: Actual number of atoms in one molecule

Example:

  • Empirical formula: CH₂O
  • Molecular formula: C₆H₁₂O₆ (glucose)

The molecular formula is always a whole-number multiple of the empirical formula.

Common Calculations

1. Mass to Moles

Given mass, find moles: n=mMn = \frac{m}{M}

2. Moles to Particles

Given moles, find particles: N=n×NAN = n \times N_A

3. Mass to Particles

Combine: N=mM×NAN = \frac{m}{M} \times N_A

4. Molar Mass from Formula

Add atomic masses from periodic table

📚 Practice Problems

1Problem 1easy

❓ Question:

How many molecules are in 8.50 g of CO₂?

💡 Show Solution

Solution:

Given: 8.50 g of CO₂ Find: Number of molecules

Step 1: Find molar mass of CO₂

MCO2=12.01+2(16.00)=12.01+32.00=44.01 g/molM_{CO_2} = 12.01 + 2(16.00) = 12.01 + 32.00 = 44.01 \text{ g/mol}

Step 2: Convert mass to moles

n=mM=8.50 g44.01 g/mol=0.193 moln = \frac{m}{M} = \frac{8.50 \text{ g}}{44.01 \text{ g/mol}} = 0.193 \text{ mol}

Step 3: Convert moles to molecules

N=n×NA=0.193 mol×6.022×1023 molecules/molN = n \times N_A = 0.193 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}

N=1.16×1023 moleculesN = 1.16 \times 10^{23} \text{ molecules}

Answer: 1.16×10231.16 \times 10^{23} molecules of CO₂

Verification:

  • Used correct molar mass for CO₂ ✓
  • Converted g → mol → molecules ✓
  • Answer has correct units ✓

2Problem 2medium

❓ Question:

A sample contains 2.50 moles of calcium carbonate (CaCO₃). (a) Calculate the mass of this sample in grams. (b) How many formula units of CaCO₃ are present? (c) How many total atoms are in this sample?

💡 Show Solution

Solution:

(a) Mass calculation:

  • Molar mass of CaCO₃ = Ca + C + 3(O) = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol
  • Mass = moles × molar mass = 2.50 mol × 100.09 g/mol = 250 g

(b) Formula units:

  • Use Avogadro's number: 6.022 × 10²³ formula units/mol
  • Formula units = 2.50 mol × 6.022 × 10²³ = 1.51 × 10²⁴ formula units

(c) Total atoms:

  • Each CaCO₃ has 5 atoms (1 Ca + 1 C + 3 O)
  • Total atoms = 1.51 × 10²⁴ × 5 = 7.53 × 10²⁴ atoms

3Problem 3medium

❓ Question:

Calculate the molar mass and percent composition of calcium nitrate, Ca(NO₃)₂.

💡 Show Solution

Solution:

Given: Ca(NO₃)₂ Find: Molar mass and percent composition

Step 1: Identify atoms in the formula

Ca(NO₃)₂ contains:

  • 1 Ca atom
  • 2 N atoms (from 2 × NO₃)
  • 6 O atoms (from 2 × 3O)

Step 2: Calculate molar mass

From periodic table:

  • Ca: 40.08 g/mol
  • N: 14.01 g/mol
  • O: 16.00 g/mol

M=1(40.08)+2(14.01)+6(16.00)M = 1(40.08) + 2(14.01) + 6(16.00) M=40.08+28.02+96.00=164.10 g/molM = 40.08 + 28.02 + 96.00 = 164.10 \text{ g/mol}

Step 3: Calculate percent composition

%Ca=40.08164.10×100%=24.42%\%Ca = \frac{40.08}{164.10} \times 100\% = 24.42\%

%N=28.02164.10×100%=17.07%\%N = \frac{28.02}{164.10} \times 100\% = 17.07\%

%O=96.00164.10×100%=58.50%\%O = \frac{96.00}{164.10} \times 100\% = 58.50\%

Answer:

  • Molar mass: 164.10 g/mol
  • Percent composition: 24.42% Ca, 17.07% N, 58.50% O

Verification:

  • 24.42+17.07+58.50=99.99≈100%24.42 + 17.07 + 58.50 = 99.99 \approx 100\% ✓ (rounding)

4Problem 4medium

❓ Question:

A chemist has 45.0 g of hydrated copper(II) sulfate (CuSO₄·5H₂O). (a) Calculate the number of moles of the hydrated compound. (b) How many moles of water molecules are in the sample? (c) What mass of anhydrous CuSO₄ would remain if all the water were removed?

💡 Show Solution

Solution:

(a) Moles of hydrated compound:

  • Molar mass of CuSO₄·5H₂O = 63.55 + 32.07 + 4(16.00) + 5(18.02) = 249.72 g/mol
  • Moles = 45.0 g ÷ 249.72 g/mol = 0.180 mol

(b) Moles of water:

  • Each formula unit has 5 water molecules
  • Moles of H₂O = 0.180 mol × 5 = 0.900 mol

(c) Mass of anhydrous CuSO₄:

  • Molar mass of CuSO₄ = 63.55 + 32.07 + 4(16.00) = 159.62 g/mol
  • Mass = 0.180 mol × 159.62 g/mol = 28.7 g

5Problem 5hard

❓ Question:

A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. If its molar mass is 180 g/mol, what is its molecular formula?

💡 Show Solution

Solution:

Given:

  • 40.0% C, 6.7% H, 53.3% O
  • Molar mass = 180 g/mol Find: Molecular formula

Step 1: Assume 100 g sample (percents become grams)

  • 40.0 g C
  • 6.7 g H
  • 53.3 g O

Step 2: Convert to moles

mol C=40.0 g12.01 g/mol=3.33 mol\text{mol C} = \frac{40.0 \text{ g}}{12.01 \text{ g/mol}} = 3.33 \text{ mol}

mol H=6.7 g1.008 g/mol=6.65 mol\text{mol H} = \frac{6.7 \text{ g}}{1.008 \text{ g/mol}} = 6.65 \text{ mol}

mol O=53.3 g16.00 g/mol=3.33 mol\text{mol O} = \frac{53.3 \text{ g}}{16.00 \text{ g/mol}} = 3.33 \text{ mol}

Step 3: Find simplest ratio (divide by smallest)

Smallest = 3.33

C:3.333.33=1C: \frac{3.33}{3.33} = 1 H:6.653.33=2H: \frac{6.65}{3.33} = 2 O:3.333.33=1O: \frac{3.33}{3.33} = 1

Empirical formula: CH₂O

Step 4: Find empirical formula mass

MCH2O=12.01+2(1.008)+16.00=30.03 g/molM_{CH_2O} = 12.01 + 2(1.008) + 16.00 = 30.03 \text{ g/mol}

Step 5: Find multiplier

n=Molecular massEmpirical mass=18030.03=6n = \frac{\text{Molecular mass}}{\text{Empirical mass}} = \frac{180}{30.03} = 6

Step 6: Multiply empirical formula by n

Molecular formula=(CH2O)×6=C6H12O6\text{Molecular formula} = (CH_2O) \times 6 = C_6H_{12}O_6

Answer: C₆H₁₂O₆ (glucose)

Verification:

  • Molar mass of C₆H₁₂O₆ = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol ✓
  • Matches given molar mass ✓
Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple ChoiceMCQ6090 min50%✅
Free Response (Long)FRQ369 min30%✅
Free Response (Short)FRQ436 min20%✅

📊 Scoring: 1-5

5
Extremely Qualified
~12%
4
Well Qualified
~16%
3
Qualified
~24%
2
Possibly Qualified
~24%
1
No Recommendation
~24%

💡 Key Test-Day Tips

  • ✓Memorize common polyatomic ions
  • ✓Practice dimensional analysis
  • ✓Know your gas laws

⚠️ Common Mistakes: Moles and Molar Mass

Avoid these 3 frequent errors

🌍 Real-World Applications: Moles and Molar Mass

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

22 mol of H2H_2 reacts with 11 mol of O2O_2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O2H_2 + O_2 \to 2H_2O)

2Determine the limiting reagent
3Calculate moles of product
4Convert moles to grams

📌 Related Topics in Atomic Structure and Properties

❓ Frequently Asked Questions

What is Moles and Molar Mass?▾
Understand the mole concept, Avogadro's number, and how to calculate molar mass and convert between mass, moles, and particles.
How can I study Moles and Molar Mass effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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