chapter 3 lec 1

Atomic Mass and Isotopes

Definition of Atomic Mass

Atomic mass (also called atomic weight) is the mass of an atom, measured in atomic mass units (amu). One amu is defined as exactly 1/12 the mass of a carbon-12 (12C) atom, which contains 6 protons, 6 neutrons, and 6 electrons. This standard allows for the comparison of atomic masses across all elements.

• 1 amu = 1/12 the mass of a 12C atom

• Example: 1 H atom = 1.0078 amu; 16O = 16.00 amu; Cl = 35.453 amu; Au = 197.0 amu

Isotopes and Average Atomic Mass

Most elements exist as a mixture of isotopes, which are atoms of the same element with different numbers of neutrons. The average atomic mass listed on the periodic table reflects the weighted average of all naturally occurring isotopes.

• Calculation: $\text{Average atomic mass} = \sum (\text{isotope mass} \times \text{fractional abundance})$

• Example (Carbon): $12.00 \times 0.9893 + 13.00 \times 0.0107 = 12.011$ amu

Example: Nitrogen Isotope Abundance

• Given: N-14 (14.0031 amu), N-15 (15.0001 amu), average = 14.0067 amu

• Set up: $(14.0031)X + (15.0001)(1-X) = 14.0067$

• Solve for X (fraction of N-14): $X = 0.99636$ (99.636% N-14), N-15 = 0.364%

Example: Copper Isotope Calculation

• Given: Cu-1 = 62.9 amu (69.1%), Cu-2 = X amu (30.9%), average = 63.5 amu

• Set up: $63.5 = (0.691)(62.9) + (0.309)(X)$

• Solve: $X = 64.8$ amu

Example: Average Atomic Weight from Multiple Isotopes

Mass Number	Mass (amu)	Abundance (%)
50	49.9461	4.35
52	51.9405	83.79
53	52.9407	9.50
54	53.9389	2.36

• Average atomic weight: $51.9959$ amu ≈ $52.0$ amu (Chromium, Cr)

The Mole Concept and Avogadro's Number

Definition of the Mole

The mole is a counting unit in chemistry, analogous to "dozen" or "gross." One mole contains Avogadro's number of entities (atoms, molecules, ions, etc.):

• Avogadro's Number: $N_A = 6.022 \times 10^{23}$

• 1 mole of O atoms = $6.022 \times 10^{23}$ atoms O

• 1 mole of Cl2 molecules = $6.022 \times 10^{23}$ molecules Cl2

• 1 mole of Cl2 contains $1.204 \times 10^{24}$ Cl atoms (since each molecule has 2 atoms)

Example: Atoms in a Given Mass

• Find atoms of Cl in 1.005 g Cl:

• $1.005 \text{ g} \times \left(\frac{1 \text{ mol}}{35.453 \text{ g}}\right) = 0.02835 \text{ mol}$

• $0.02835 \text{ mol} \times 6.022 \times 10^{23} = 1.707 \times 10^{22}$ atoms Cl

Molar Mass and Formula Mass

Definitions

• Molar mass: Mass of one mole of a substance (g/mol), numerically equal to its molecular or formula mass in amu.

• Formula mass: Used for ionic compounds, as they do not exist as discrete molecules.

Examples

• Fe: Atomic mass = 55.85 amu; Molar mass = 55.85 g/mol

• I: Atomic mass = 126.91 amu; Molar mass = 126.91 g/mol

• I2: Molar mass = $2 \times 126.91 = 253.82$ g/mol

Interconverting Mass, Moles, and Number of Particles

• Molar mass is used to convert between mass and moles.

• Avogadro's number is used to convert between moles and number of particles.

Example: Moles and Ions in NaCl

• Given: 233.772 g NaCl (MW = 58.443 g/mol)

• $233.772 \text{ g} \times \left(\frac{1 \text{ mol}}{58.443 \text{ g}}\right) = 4.000$ mol NaCl

• Each mole contains $6.022 \times 10^{23}$ Na+ and Cl- ions

• Total: $4.000 \times 6.022 \times 10^{23} = 2.409 \times 10^{24}$ ions of each

Example: Grams from Moles

• 1.000 mol I: $1.000 \times 126.91 = 126.91$ g

• 0.5 mol I: $0.5 \times 126.91 = 63.46$ g

• 0.0123 mol I: $0.0123 \times 126.91 = 1.56$ g

Example: Moles and Atoms in Silver

• 22.25 g Ag ($AW = 107.9$ g/mol): $22.25 \div 107.9 = 0.2062$ mol Ag

• Atoms: $0.2062 \times 6.022 \times 10^{23} = 1.242 \times 10^{23}$ atoms Ag

Example: Mass and Atoms in Compounds

• Given: 0.215 mol MgCl2 (FW = 95.216 g/mol)

• Mass of Mg: $0.215 \times 24.305 = 5.23$ g (using atomic mass of Mg)

• Mass of Cl: $0.215 \times 2 \times 35.453 = 15.25$ g

• Atoms: $0.215 \times 6.022 \times 10^{23} = 1.29 \times 10^{23}$ formula units; multiply by number of each atom per formula unit for total atoms

• Additional info: The above calculations are inferred based on standard atomic masses.

Example: Atoms in a Penny

• 3.10 g Cu ($AW = 63.55$ g/mol): $3.10 \div 63.55 = 0.0488$ mol Cu

• Atoms: $0.0488 \times 6.022 \times 10^{23} = 2.94 \times 10^{22}$ atoms Cu

Example: Mass from Number of Atoms

• 2.25 x 1022 W atoms ($AW = 183.9$ g/mol):

• Moles: $2.25 \times 10^{22} \div 6.022 \times 10^{23} = 0.0374$ mol

• Mass: $0.0374 \times 183.9 = 6.88$ g

Example: Volume and Radius of an Al Sphere

• Given: $8.55 \times 10^{22}$ Al atoms, density = 2.70 g/cm3

• Moles: $8.55 \times 10^{22} \div 6.022 \times 10^{23} = 0.142$ mol

• Mass: $0.142 \times 26.98 = 3.83$ g

• Volume: $3.83 \div 2.70 = 1.42$ cm3

• Volume of sphere: $V = \frac{4}{3}\pi r^3$

• Solve for r: $r = \left(\frac{3V}{4\pi}\right)^{1/3}$

Molecular Mass and Formula Mass

Calculating Molecular Mass

The molecular mass is the sum of the atomic masses of all atoms in a molecule. For ionic compounds, the term formula mass is used.

• Formula: $\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass})$

Example: C9H16O3

• 9 C: $9 \times 12.011 = 108.099$ amu

• 16 H: $16 \times 1.008 = 16.128$ amu

• 3 O: $3 \times 16.00 = 48.00$ amu

• Total: $108.099 + 16.128 + 48.00 = 172.227$ amu (rounded to 173.12 amu in example)

Example: Large Molecule (C56F64FeClN6O)

• Sum atomic masses: $672.616 + 1216.00 + 55.85 + 35.453 + 84.042 + 16.00 = 2079.961$ amu

• 1 mole = 2079.961 g; contains Avogadro's number of molecules

• Number of each atom: multiply Avogadro's number by the subscript for each element

Example: Hydrated Compound (CuSO4·5H2O)

• Molar mass: $63.54 + 32.06 + (4 \times 16.00) + (10 \times 1.008) = 249.68$ amu (Note: original text used 185.66 amu, which omits some O atoms; correct calculation includes all atoms)

• To find mass of H in 13.58 g: $(13.58 \text{ g}) \times \left(\frac{10.08 \text{ g H}}{249.68 \text{ g compound}}\right) = 0.547$ g H

• Number of H atoms: calculate moles of compound, multiply by 10 (H atoms per molecule), then by Avogadro's number

• Additional info: The calculation above corrects the molar mass to include all atoms in the hydrate.

The Mass Spectrometer

Principle and Use

The mass spectrometer is an instrument used to measure atomic and molecular masses. It works by ionizing a sample, accelerating the ions, and deflecting them in a magnetic field. The degree of deflection depends on the mass-to-charge ratio, allowing separation and detection of isotopes.

• Heavier ions are deflected less; lighter ions are deflected more.

• Detector counts ions, providing data on relative abundance of isotopes.

• First developed by F. W. Aston in the 1920s.

Empirical and Molecular Formulas

Percent Composition

The percent composition of a compound is the percentage by mass of each element in the compound.

• Formula: $\% \text{ element} = \frac{\text{(number of moles of element)} \times \text{atomic mass}}{\text{molar mass of compound}} \times 100$

Example: C7H12O2

• Molar mass = 128.173 g/mol

• %C = $\frac{7 \times 12.011}{128.173} \times 100 = 65.60\%$

• %H = $\frac{12 \times 1.008}{128.173} \times 100 = 9.44\%$

• %O = $\frac{2 \times 16.00}{128.173} \times 100 = 24.97\%$

Determining Empirical Formulas from Percent Composition

The empirical formula is the simplest whole-number ratio of atoms in a compound. It can be determined from percent composition data.

1. Assume 100 g of compound (so % = grams).

2. Convert grams to moles for each element.

3. Divide all mole values by the smallest number of moles to get ratios.

4. Round to nearest whole number (multiply all by a common factor if necessary).

Example: Empirical Formula Calculation

Element	Mass (g)	Moles	Ratio
C	35.51	2.956	4.993
H	4.77	4.732	7.993
O	37.85	2.366	3.997
N	8.29	0.592	1.000
Na	13.60	0.592	1.000

• Empirical formula: C5H8NNaO4

Summary Table: Key Concepts and Formulas

Concept	Definition/Formula	Example
Atomic Mass Unit (amu)	1 amu = 1/12 mass of 12C atom	1 H = 1.0078 amu
Average Atomic Mass	$\sum (\text{isotope mass} \times \text{fractional abundance})$	Carbon: 12.011 amu
Mole	6.022 x 1023 entities	1 mol O = 6.022 x 1023 atoms
Molar Mass	g/mol; numerically equal to molecular mass in amu	NaCl: 58.443 g/mol
Percent Composition	$\% = \frac{\text{mass of element}}{\text{molar mass}} \times 100$	C in C7H12O2: 65.60%
Empirical Formula	Smallest whole-number ratio of atoms	C5H8NNaO4