Chemical Composition: The Mole, Molar Mass, and Empirical Formulas

Chemical Composition

The Mole and Counting Using the Mole

The mole is a fundamental unit in chemistry used to count particles such as atoms, molecules, or ions. It allows chemists to relate microscopic quantities to macroscopic measurements.

• Definition: One mole contains Avogadro's number of particles, which is $6.022 \times 10^{23}$.

• Application: The mole is used to convert between the number of particles and measurable quantities like mass.

• Example: One mole of carbon atoms contains $6.022 \times 10^{23}$ atoms and has a mass of 12.01 grams.

Molar Mass and Atomic Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Atomic mass is the mass of a single atom, usually given in atomic mass units (amu).

• Atomic Mass: The mass of an atom, found on the periodic table, measured in amu.

• Molar Mass: The mass of one mole of a substance, numerically equal to the atomic or molecular mass in grams.

• Formula: $\text{Molar mass} = \text{Sum of atomic masses of all atoms in the formula}$

• Example: The molar mass of water ($\mathrm{H_2O}$) is $2 \times 1.01 + 16.00 = 18.02$ g/mol.

Conversion from Mass to Moles to Molecules

Conversions between mass, moles, and molecules are essential for quantitative chemical calculations.

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

• Moles to Molecules: $\text{Number of molecules} = \text{Moles} \times 6.022 \times 10^{23}$

• Example: 36.04 g of water is $\frac{36.04}{18.02} = 2$ moles, which is $2 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$ molecules.

Conversion from Molecules to Moles to Mass

To convert from molecules to moles and then to mass, use Avogadro's number and molar mass.

• Molecules to Moles: $\text{Moles} = \frac{\text{Number of molecules}}{6.022 \times 10^{23}}$

• Moles to Mass: $\text{Mass (g)} = \text{Moles} \times \text{Molar mass (g/mol)}$

• Example: $3.011 \times 10^{23}$ molecules of $\mathrm{CO_2}$ is $\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} = 0.5$ moles, and $0.5 \times 44.01 = 22.01$ g.

Mass Percent

Mass percent expresses the proportion of each element in a compound as a percentage of the total mass.

• Formula: $\text{Mass percent} = \frac{\text{Mass of element in 1 mole}}{\text{Mass of 1 mole of compound}} \times 100\%$

• Example: In $\mathrm{H_2O}$, mass percent of hydrogen is $\frac{2.02}{18.02} \times 100\% = 11.2\%$.

Using Mass Percent as a Conversion Factor

Mass percent can be used to determine the mass of an element in a given mass of compound.

• Conversion: $\text{Mass of element} = \text{Mass percent} \times \text{Total mass} \div 100$

• Example: In 50 g of $\mathrm{H_2O}$, hydrogen mass is $11.2\% \times 50 \div 100 = 5.6$ g.

Calculating an Empirical Formula from Experimental Mass Data

The empirical formula represents the simplest whole-number ratio of elements in a compound. It can be determined from experimental mass data.

• Steps:

  1. Convert masses of each element to moles.

  2. Divide each by the smallest number of moles to get the ratio.

  3. Round or multiply to obtain whole numbers.

• Example: If a sample contains 2.00 g hydrogen and 16.00 g oxygen:

  • Hydrogen: $\frac{2.00}{1.01} = 1.98$ moles

  • Oxygen: $\frac{16.00}{16.00} = 1.00$ moles

  • Ratio: $1.98:1.00 \approx 2:1$

  • Empirical formula: $\mathrm{H_2O}$

Calculating Empirical Formula Using Mass Percent Composition

Mass percent composition can be used to determine the empirical formula by assuming a sample size (often 100 g).

• Steps:

  1. Assume 100 g sample; mass percent becomes mass in grams.

  2. Convert each element's mass to moles.

  3. Divide by the smallest number of moles to get the ratio.

  4. Adjust to whole numbers for the empirical formula.

• Example: A compound is 40% carbon, 6.7% hydrogen, 53.3% oxygen:

  • Carbon: $\frac{40}{12.01} = 3.33$ moles

  • Hydrogen: $\frac{6.7}{1.01} = 6.63$ moles

  • Oxygen: $\frac{53.3}{16.00} = 3.33$ moles

  • Ratio: $3.33:6.63:3.33 \approx 1:2:1$

  • Empirical formula: $\mathrm{CH_2O}$

Calculating Molecular Formulas Using Empirical Formulas and Molar Masses

The molecular formula shows the actual number of atoms in a molecule and is a multiple of the empirical formula. It is determined using the empirical formula and the compound's molar mass.

• Steps:

  1. Calculate the molar mass of the empirical formula.

  2. Divide the compound's molar mass by the empirical formula mass to find the multiplier.

  3. Multiply the subscripts in the empirical formula by the multiplier to get the molecular formula.

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

• Example: If empirical formula is $\mathrm{CH_2O}$ (molar mass 30.03 g/mol) and compound molar mass is 180.18 g/mol:

  • $n = \frac{180.18}{30.03} = 6$

  • Molecular formula: $\mathrm{C_6H_{12}O_6}$

Summary Table: Key Conversions in Chemical Composition

Conversion	Formula	Example
Mass to Moles	$\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}}$	18.02 g $\mathrm{H_2O}$ = 1 mole
Moles to Molecules	$\text{Molecules} = \text{Moles} \times 6.022 \times 10^{23}$	2 moles $\mathrm{H_2O}$ = $1.204 \times 10^{24}$ molecules
Molecules to Moles	$\text{Moles} = \frac{\text{Molecules}}{6.022 \times 10^{23}}$	$3.011 \times 10^{23}$ molecules $\mathrm{CO_2}$ = 0.5 moles
Mass Percent	$\frac{\text{Mass of element}}{\text{Mass of compound}} \times 100\%$	Hydrogen in $\mathrm{H_2O}$: 11.2%