The Mole Concept and Percent Composition in General Chemistry

The Mole Concept

Definition and Importance

The mole (mol) 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. This allows chemists to count particles by weighing them.

• Avogadro's Number: $6.022 \times 10^{23}$ particles/mol

• Molar Mass: The mass (in grams) of one mole of a substance, numerically equal to the substance's average atomic or molecular mass in amu.

Counting Atoms and Molecules

To relate the number of particles to mass, use the following relationships:

• Number of particles in 1 mole: $6.022 \times 10^{23}$

• Mass of 1 mole: Equal to the molar mass in grams

Example Table: Atoms and Mass in CO2

	# C atoms	# O atoms	Mass (g)
1 molecule CO2	1	2	$7.31 \times 10^{-23}$
1 mol CO2	$6.022 \times 10^{23}$	$2 \times (6.022 \times 10^{23})$	44.01

Atoms in a Molecule and a Mole

• Example: In one molecule of (NH4)2HPO4 (ammonium hydrogen phosphate), there are 2 N, 9 H, 1 P, and 4 O atoms.

• In one mole: Multiply the number of each atom by Avogadro's number to get the total atoms per mole.

Calculating Moles from Mass

To find the number of moles from a given mass:

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

• Example: $6.46\ \text{g He} \times \frac{1\ \text{mol He}}{4.003\ \text{g He}} = 1.61\ \text{mol He}$

Calculating Mass from Moles

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

• Example: $12.4\ \text{mol Pb} \times \frac{207.2\ \text{g Pb}}{1\ \text{mol Pb}} = 2.57 \times 10^3\ \text{g Pb}$

Atoms in a Given Number of Molecules or Moles

• To find atoms in molecules: Multiply the number of molecules by the number of atoms per molecule.

• To find atoms in moles: Multiply moles by Avogadro's number and the number of atoms per molecule.

• Example: Number of H atoms in 10.0 mol urea ($\text{(NH}_2)_2\text{CO}$): $10.0\ \text{mol urea} \times \frac{6.022 \times 10^{23}\ \text{molecules}}{1\ \text{mol}} \times \frac{4\ \text{H atoms}}{1\ \text{molecule}} = 2.41 \times 10^{25}\ \text{H atoms}$

Converting Between Atoms, Moles, and Mass

• Atoms to grams: Atoms $\rightarrow$ moles $\rightarrow$ grams

• Grams to atoms: Grams $\rightarrow$ moles $\rightarrow$ atoms

• Example: How many grams of K in $5.51$ atoms? $5.51\ \text{atoms} \times \frac{1\ \text{mol}}{6.022 \times 10^{23}\ \text{atoms}} \times 39.10\ \text{g/mol} = 3.58 \times 10^{-21}\ \text{g K}$

Percent Composition

Definition and Formula

Percent composition expresses the mass percentage of each element in a compound.

• Formula: $\frac{n \times \text{mass of element}}{\text{mass of compound}} \times 100$

• Where n is the number of atoms of the element in the formula unit.

Example Calculation

Given a 12.04 g sample of a compound with 7.34 g C, 1.85 g H, and 2.85 g N:

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

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

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

Percent Composition from Molecular Formula

• Calculate the total molar mass of the compound.

• For each element, multiply the number of atoms by its atomic mass, divide by the total molar mass, and multiply by 100.

• Example: For aspirin ($\text{C}_9\text{H}_8\text{O}_4$): - Molar mass = $9(12.011) + 8(1.00794) + 4(15.9994) = 180.159$ g/mol - %C: $\frac{9 \times 12.011}{180.159} \times 100 = 60.000\%$ - %H: $\frac{8 \times 1.00794}{180.159} \times 100 = 4.4760\%$ - %O: $\frac{4 \times 15.9994}{180.159} \times 100 = 35.520\%$

Additional info: These calculations are foundational for chemical analysis, stoichiometry, and understanding the composition of substances in chemistry.