Moles and Molar Mass

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

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 $6.022 \times 10^{23}$ particles.

This number is called Avogadro's number ( $N_A$ ).

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

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 $6.022 \times 10^{23}$ .

Converting Between Particles and Moles

$\text{Number of particles} = \text{moles} \times N_A$

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

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

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

Molar Mass

Molar mass ( $M$ ) 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)

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

Example: Glucose (C₆H₁₂O₆)

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

Converting Between Mass and Moles

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

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

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

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

$\text{Mass (g)} \xleftrightarrow[\text{divide by } M]{\text{multiply by } M} \text{Moles} \xleftrightarrow[\text{divide by } N_A]{\text{multiply by } N_A} \text{Particles}$

Percent Composition

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

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

Example: Find the percent composition of H₂O

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

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

Check: $11.19\% + 88.81\% = 100\%$ ✓

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

2. Moles to Particles

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

3. Mass to Particles

Combine: $N = \frac{m}{M} \times N_A$

4. Molar Mass from Formula

Add atomic masses from periodic table

📚 Practice Problems

1Problem 1medium

❓ 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

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

2Problem 2easy

❓ 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

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

Step 2: Convert mass to moles

$n = \frac{m}{M} = \frac{8.50 \text{ g}}{44.01 \text{ g/mol}} = 0.193 \text{ mol}$

Step 3: Convert moles to molecules

$N = n \times N_A = 0.193 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}$

$N = 1.16 \times 10^{23} \text{ molecules}$

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

Verification:

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

3Problem 3medium

❓ Question:

💡 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

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

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

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 5medium

❓ Question:

💡 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

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

6Problem 6medium

❓ 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

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

Step 3: Calculate percent composition

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

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

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

Answer:

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

Verification:

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

7Problem 7hard

❓ 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

$\text{mol C} = \frac{40.0 \text{ g}}{12.01 \text{ g/mol}} = 3.33 \text{ mol}$

$\text{mol H} = \frac{6.7 \text{ g}}{1.008 \text{ g/mol}} = 6.65 \text{ mol}$

$\text{mol O} = \frac{53.3 \text{ g}}{16.00 \text{ g/mol}} = 3.33 \text{ mol}$

Step 3: Find simplest ratio (divide by smallest)

Smallest = 3.33

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

Empirical formula: CH₂O

Step 4: Find empirical formula mass

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

Step 5: Find multiplier

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

Step 6: Multiply empirical formula by n

$\text{Molecular formula} = (CH_2O) \times 6 = C_6H_{12}O_6$

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 ✓

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