Analyzing Compounds: Molar Mass, Percent Composition, and Chemical Formulas

Analyzing Compounds

Introduction

This unit introduces essential concepts for analyzing chemical compounds, including the mole, Avogadro's number, molar mass, percent composition, and empirical and molecular formulas. Mastery of these topics is fundamental for understanding chemical quantities and relationships in General Chemistry.

The Mole and Avogadro's Number

Definition and Significance

• Mole (mol): The SI base unit for measuring the amount of substance. It is a counting unit, similar to a dozen, but much larger.

• Avogadro's Number (NA): The number of particles (atoms, molecules, ions, etc.) in one mole of a substance.

Key Facts:

• 1 mole = $6.022 \times 10^{23}$ items

• This number is used to relate macroscopic amounts of material to the number of fundamental particles present.

Example: 1 mole of carbon atoms contains $6.022 \times 10^{23}$ carbon atoms.

Molar Mass

Definition and Calculation

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

• For elements, the molar mass is the atomic mass from the periodic table (in g/mol).

• For compounds, sum the molar masses of all atoms in the formula.

Examples:

• Carbon (C): 12.01 g/mol

• Water (H2O): $2 \times 1.008 + 16.00 = 18.02$ g/mol

• Diphosphorus pentoxide (P2O5): $2 \times 30.97 + 5 \times 16.00 = 141.94$ g/mol

Molar Conversions

Using Avogadro's Number and Molar Mass

• Avogadro's number allows conversion between moles and number of particles.

• Molar mass allows conversion between mass and moles.

Conversion Relationships:

• Particles (atoms, molecules, ions) $\leftrightarrow$ Moles: $\text{Number of particles} = \text{moles} \times 6.022 \times 10^{23}$

• Mass $\leftrightarrow$ Moles: $\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

Example: How many atoms are in 2.5 mol of copper?

• $2.5 \text{ mol Cu} \times \frac{6.022 \times 10^{23} \text{ atoms}}{1 \text{ mol Cu}} = 1.5 \times 10^{24}$ atoms

Percent Composition

Definition and Calculation

• Percent Composition: The percentage by mass of each element in a compound.

• Formula: $\%\text{ composition} = \frac{\text{mass of element}}{\text{total mass of compound}} \times 100$

Example: Find the percent composition of copper in Cu2S.

• Molar mass of Cu2S: $2 \times 63.55 + 32.06 = 159.16$ g/mol

• Mass of Cu: $2 \times 63.55 = 127.10$ g

• Percent Cu: $\frac{127.10}{159.16} \times 100 = 79.86\%$

Empirical and Molecular Formulas

Empirical Formula

• The simplest whole-number ratio of atoms in a compound.

• Steps to determine:

  1. Find the mass (or percent) of each element.

  2. Convert mass to moles for each element.

  3. Divide each by the smallest number of moles to get subscripts.

  4. If necessary, multiply subscripts to get whole numbers.

Example: A compound is 25.9% N and 74.1% O by mass.

• Moles N: $\frac{25.9}{14.01} = 1.85$ mol

• Moles O: $\frac{74.1}{16.00} = 4.63$ mol

• Divide by smallest: N: $1.85/1.85 = 1$, O: $4.63/1.85 = 2.5$

• Multiply by 2: N2O5

Molecular Formula

• The actual number of atoms of each element in a molecule.

• Steps to determine:

  1. Find the empirical formula.

  2. Calculate the empirical formula mass.

  3. Divide the molecular mass by the empirical formula mass.

  4. Multiply each subscript in the empirical formula by this ratio.

• Formula: $\text{Molecular Formula Mass} = n \times \text{Empirical Formula Mass}$

Example: Empirical formula for propylene is CH2, molecular mass is 42.1 g/mol.

• Empirical mass: $12.01 + 2 \times 1.008 = 14.03$ g/mol

• Ratio: $42.1/14.03 = 3$

• Molecular formula: (CH2)3 = C3H6

Summary Table: Key Terms and Formulas

Term	Definition	Formula/Example
Mole (mol)	SI unit for amount of substance	1 mol = $6.022 \times 10^{23}$ particles
Molar Mass	Mass of 1 mole of substance	H2O: 18.02 g/mol
Percent Composition	Percent by mass of each element	$\% = \frac{\text{mass of element}}{\text{total mass}} \times 100$
Empirical Formula	Simplest ratio of atoms	CH2O
Molecular Formula	Actual number of atoms	C6H12O6

Additional info: These notes are based on lecture slides and introductory materials for a General Chemistry course, focusing on quantitative analysis of compounds.