Moles, Molar Mass, and Percent Composition in General Chemistry

Moles and Molar Mass

Understanding the Mole Concept

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. One mole contains exactly $6.022 \times 10^{23}$ entities (Avogadro's number), such as atoms, molecules, or ions.

• Mole (mol): The amount of substance containing as many entities as there are atoms in 12 g of carbon-12.

• Molar Mass: The mass of one mole of a substance, expressed in grams per mole (g/mol).

• Example: For CO2:

  • 1 molecule CO2: 1 C atom, 2 O atoms, mass = $7.31 \times 10^{-23}$ g

  • 1 mole CO2: $6.022 \times 10^{23}$ molecules, mass = 44.01 g

Atoms in Molecules and Moles

Calculating the number of atoms in a molecule or a mole is essential for stoichiometric calculations.

• Example: In one molecule of (NH4)2HPO4 (ammonium hydrogen phosphate):

  • 2 N atoms, 9 H atoms, 1 P atom, 4 O atoms

  • Masses of each atom can be calculated using atomic masses and Avogadro's number

• In one mole of (NH4)2HPO4:

  • $1.2044 \times 10^{24}$ N atoms (2 mol N atoms)

  • $5.420 \times 10^{24}$ H atoms (9 mol H atoms)

  • $6.022 \times 10^{23}$ P atoms (1 mol P atom)

  • $2.409 \times 10^{24}$ O atoms (4 mol O atoms)

Converting Between Grams, Moles, and Atoms

To convert between mass, moles, and number of atoms, use the molar mass and Avogadro's number.

• Formula:

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

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

• Example 1: How many moles of He atoms are in 6.46 g of He?

  • Molar mass of He = 4.003 g/mol

  • $\frac{6.46\,\text{g He}}{4.003\,\text{g/mol}} = 1.61\,\text{mol He}$

• Example 2: How many grams of Pb are in 12.4 moles?

  • Molar mass of Pb = 207.2 g/mol

  • $12.4\,\text{mol Pb} \times 207.2\,\text{g/mol} = 2.57 \times 10^{3}\,\text{g Pb}$

Counting Atoms in Molecules and Moles

Atoms in Molecules and Moles

To determine the number of atoms in a given number of molecules or moles, use the molecular formula and Avogadro's number.

• Example 1: H atoms in 50 molecules of urea [(NH2)2CO]:

  • Each molecule has 4 H atoms

  • $50 \times 4 = 2.0 \times 10^{2}$ H atoms

• Example 2: H atoms in 10.0 moles of urea:

  • Molar mass of urea = 60.06 g/mol

  • $10.0\,\text{mol} \times 6.022 \times 10^{23}\,\text{molecules/mol} \times 4\,\text{H atoms/molecule} = 2.41 \times 10^{25}$ H atoms

• Example 3: H atoms in 72.5 g of ethanol (C2H6O):

  • Molar mass = 46.0684 g/mol

  • $\frac{72.5\,\text{g}}{46.0684\,\text{g/mol}} = 1.574\,\text{mol}$

  • $1.574\,\text{mol} \times 6.022 \times 10^{23}\,\text{molecules/mol} \times 6\,\text{H atoms/molecule} = 5.69 \times 10^{24}$ H atoms

Conversions Involving Atoms and Grams

Atoms to Grams and Vice Versa

Conversions between atoms and grams require using both Avogadro's number and the molar mass.

• Example 1: Grams of potassium in 551 atoms:

  • Molar mass of K = 39.10 g/mol

  • $1\,\text{mol K} = 6.022 \times 10^{23}$ atoms

  • $551\,\text{atoms} \times \frac{1\,\text{mol K}}{6.022 \times 10^{23}\,\text{atoms}} \times 39.10\,\text{g/mol} = 3.58 \times 10^{-21}\,\text{g K}$

• Example 2: Potassium atoms in 0.551 mol:

  • $0.551\,\text{mol} \times 6.022 \times 10^{23}\,\text{atoms/mol} = 3.32 \times 10^{23}$ atoms

Percent Composition

Definition and Calculation

Percent composition expresses the mass percentage of each element in a compound. It is useful for determining empirical and molecular formulas.

• Formula:

  • $\%\,\text{element} = \frac{n \times \text{mass of element}}{\text{mass of compound}} \times 100$

  • Alternate: $\%\,\text{element} = \frac{\text{mass of element}}{\text{mass of compound}} \times 100$

• Example: For a 12.04 g sample containing 7.34 g C, 1.85 g H, and 2.85 g N:

  • $\%\,\text{C} = \frac{7.34}{12.04} \times 100 = 61.0\%$

  • $\%\,\text{H} = \frac{1.85}{12.04} \times 100 = 15.4\%$

  • $\%\,\text{N} = \frac{2.85}{12.04} \times 100 = 23.7\%$

Percent Composition from Molecular Formula

Percent composition can also be determined using the molecular formula and atomic masses.

• Example: Aspirin (C9H8O4):

  • Molar mass = $9 \times 12.011 + 8 \times 1.00794 + 4 \times 15.9994 = 180.159$ g/mol

  • $\%\,\text{C} = \frac{9 \times 12.011}{180.159} \times 100 = 60.00\%$

  • $\%\,\text{H} = \frac{8 \times 1.00794}{180.159} \times 100 = 4.48\%$

  • $\%\,\text{O} = \frac{4 \times 15.9994}{180.159} \times 100 = 35.52\%$

Summary Table: Key Conversions and Formulas

Conversion	Formula	Example
Grams to Moles	$\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$	$6.46\,\text{g He} \rightarrow 1.61\,\text{mol He}$
Moles to Grams	$\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$	$12.4\,\text{mol Pb} \rightarrow 2.57 \times 10^{3}\,\text{g Pb}$
Moles to Atoms	$\text{atoms} = \text{moles} \times 6.022 \times 10^{23}$	$0.551\,\text{mol K} \rightarrow 3.32 \times 10^{23}\,\text{atoms}$
Atoms to Grams	$\text{grams} = \text{atoms} \times \frac{1\,\text{mol}}{6.022 \times 10^{23}\,\text{atoms}} \times \text{molar mass}$	$551\,\text{atoms K} \rightarrow 3.58 \times 10^{-21}\,\text{g K}$
Percent Composition	$\%\,\text{element} = \frac{\text{mass of element}}{\text{mass of compound}} \times 100$	$\%\,\text{C} = 61.0\%$ (sample calculation)

Additional info: These notes cover foundational concepts in general chemistry, including the mole concept, molar mass, conversions between mass, moles, and atoms, and percent composition. Mastery of these topics is essential for quantitative chemical calculations and stoichiometry.