Quantitative Composition of Compounds: The Mole, Molar Mass, Percent Composition, Empirical and Molecular Formulas

Quantitative Composition of Compounds

The Mole

The concept of the mole is fundamental in chemistry for quantifying atoms, molecules, or ions. Because individual atoms are extremely small and have minuscule masses, chemists use the mole as a convenient counting unit, similar to how items like fruit are counted by weighing.

• Definition: 1 mole = $6.022 \times 10^{23}$ objects (Avogadro's number).

• Applications: Moles can describe elements, molecules, or electrons.

• Conversion: Avogadro’s number is used as a conversion factor between the number of particles and moles.

• Example: 1 mol of atoms = $6.022 \times 10^{23}$ atoms; 1 mol of molecules = $6.022 \times 10^{23}$ molecules.

Analogy: Counting candies by weighing is similar to counting atoms by mass.

Moles and Mass

One mole of any element contains the same number of atoms, but the mass varies depending on the element. The mass of 1 mole of an element is defined as the amount containing the same number of particles as exactly 12 g of carbon-12 ($^{12}C$).

• Atomic Mass: Different elements have different atomic masses, so 1 mole of each element has a unique mass.

• Example: 12 quarters do not weigh the same as 12 dimes, just as 1 mole of lead does not weigh the same as 1 mole of hydrogen.

Molar Mass of Compounds

Molar Mass

Molar mass is the mass (in grams) of one mole of a substance, whether an element or a compound. It is calculated by summing the atomic masses of all atoms in the formula unit.

• Formula: Molar mass (MM) = sum of atomic masses of all atoms in the compound.

• Example: For water ($H_2O$): $\text{Molar Mass} = 1 \times 16.00\,\text{g} + 2 \times 1.008\,\text{g} = 18.02\,\text{g}$

• Significant Figures: Molar masses are typically expressed to four significant figures.

Molar Mass in the Laboratory

Different elements have different molar masses, which correspond to the mass of $6.022 \times 10^{23}$ atoms of each element.

Element	Molar Mass (g)	Number of Atoms
Sulfur	32.07	$6.022 \times 10^{23}$
Iodine	126.9	$6.022 \times 10^{23}$
Iron	55.85	$6.022 \times 10^{23}$
Mercury	200.6	$6.022 \times 10^{23}$

Conversions Using Molar Mass

Molar mass is used as a conversion factor between grams and moles. The process involves:

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

• Converting moles to grams: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

• Converting moles to number of particles: $\text{number of particles} = \text{moles} \times 6.022 \times 10^{23}$

Example: To find the number of atoms in a given mass, use two conversion factors: grams to moles, then moles to atoms.

Percent Composition of Compounds

Percent Composition

Percent composition is the mass percent of each element in a compound. It is independent of sample size and can be determined from the compound’s formula or experimental data.

• Formula: $\text{Percent composition of element} = \frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100$

• Example: For $H_2O_2$, calculate the percent composition of oxygen:

• Step 1: Calculate molar mass of $H_2O_2$

• Step 2: $\text{Percent O} = \frac{2 \times 16.00}{34.02} \times 100 = 94.06\%$

Percent Composition from Experimental Data

Percent composition can also be determined experimentally by measuring the masses of elements that react to form a compound.

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

• Example: If 1.63 g Zn reacts with 0.40 g O to form ZnO, total mass = 2.03 g. Percent Zn = $\frac{1.63}{2.03} \times 100$; Percent O = $\frac{0.40}{2.03} \times 100$.

Empirical and Molecular Formulas

Empirical Formula

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms in one molecule or formula unit and is a whole-number multiple of the empirical formula.

• Example: Acetylene ($C_2H_2$) and Benzene ($C_6H_6$) both have the empirical formula CH.

• Compounds with the same empirical formula have the same percent composition.

Calculating Empirical Formulas

To determine the empirical formula:

1. Assume a starting mass (usually 100.0 g) and express the mass of each element.

2. Convert grams to moles using molar mass.

3. Divide each mole amount by the smallest value to get subscripts.

4. If fractions result, multiply by a common factor to obtain whole numbers.

Example: A compound with 11.19% H and 88.79% O yields the empirical formula $H_2O$.

Calculating Molecular Formulas

If the molar mass is known, the molecular formula can be calculated from the empirical formula.

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

• Find $n$ by dividing the molar mass of the compound by the molar mass of the empirical formula: $n = \frac{\text{molar mass of compound}}{\text{molar mass of empirical formula}}$

• Example: If empirical formula is $NO_2$ and molar mass is 92.00 g, then $n = \frac{92.00}{46.01} = 2$, so molecular formula is $N_2O_4$.

Summary Table: Empirical and Molecular Formulas

Compound	Empirical Formula	Molecular Formula
Acetylene	CH	C2H2
Benzene	CH	C6H6
Propylene	CH2	C3H6
Nitrogen Compound	NH2	N2H4
Carbon Compound	CH3	C2H6

Learning Objectives

• Apply the concept of the mole, molar mass, and Avogadro’s number to solve chemistry problems.

• Calculate the molar mass of a compound.

• Calculate the percent composition of a compound from its chemical composition and from experimental data.

• Determine the empirical formula for a compound from its percent composition.

• Compare an empirical formula to a molecular formula and calculate a molecular formula from an empirical formula, using the molar mass.