Chapter 3 Notes: Moles, Mass, Formulas, and Solutions

Avogadro's Number, the Mole, and the Concept of Amount of Substance

NA (Avogadro's number) = 6.022 \times 10^{23} particles per mole
A mole corresponds to NA particles (atoms, molecules, ions, etc.)
The mole provides a bridge between the microscopic world (atoms/molecules) and the macroscopic world (grams of material)
1 mole = NA particles
Molar mass is the same numeric value as atomic/molecular mass but with units of \text{g mol}^{-1}
The atomic weight (or average atomic weight) is the same numerical value as the molar mass but expressed in unified atomic mass units (u)
Examples:
- The molecular mass of water: M(\mathrm{H2O}) = 18.015\;\mathrm{u} and the molar mass: M(\mathrm{H2O}) = 18.015\;\mathrm{g\;mol^{-1}}
- The formula mass of sodium chloride: M(\mathrm{NaCl}) = 58.44\;\mathrm{u} and the molar mass: M(\mathrm{NaCl}) = 58.44\;\mathrm{g\;mol^{-1}}
The masses listed (molar mass, molecular/formula mass) all represent one mole of the substance
Avogadro’s number as a conversion factor: use NA to convert between number of particles and moles
One mole contains 6.02 \times 10^{23} formula units (atoms, ions, etc.)

Molecular Mass, Formula Mass, and Molar Mass

Molecular mass (molecular weight) = sum of the masses of all atoms in a molecule, in unified atomic mass units (u)
Formula mass = sum of masses of all atoms in a formula unit (applies to ionic compounds as well)
Molar mass (MM) = number of grams per mole corresponding to the molecular/formula mass, same numeric value as molecular/formula mass but with units \mathrm{g\;mol^{-1}}
Key relationships:
- M(M\mathrm{olecule}^*) = \text{molecular mass in u}
- MM = \text{Molecular/formula mass in u} = \text{molar mass in g/mol}
Examples:
- H₂O: molecular mass = 18.015\;\mathrm{u}; MM(H₂O) = 18.015\;\mathrm{g\;mol^{-1}}
- NaCl: formula mass = 58.44\;\mathrm{u}; MM(NaCl) = 58.44\;\mathrm{g\;mol^{-1}}
Note: MM and molecular/formula mass share the same numerical value; units differ

Conversions: Atoms, Molecules, Ions to Grams (Using NA and Molar Mass)

The mole relates microscopic entities to grams via conversion factors
General approach:
- Start with a number of particles or atoms; convert to moles using n = \dfrac{N}{NA} where NA = 6.022 \times 10^{23}
- Convert moles to grams using m = n \times M_\text{molar}
Example 1: Mass of 7.3 \times 10^{24} atoms Mg
- Molar mass of Mg = 24.305\;\mathrm{g\;mol^{-1}}
- Number of moles: n = \dfrac{7.3 \times 10^{24}}{6.022 \times 10^{23}} \approx 12.12\;\text{mol}
- Mass: m = 12.12 \times 24.305 \approx 294.7\;\mathrm{g}
- (The transcript lists ~290 g; depending on significant figures, around 290–295 g.)
Example 2: How many atoms are in 65.0 g Mg?
- Molar mass Mg = 24.305 g/mol
- Moles: n = \dfrac{65.0}{24.305} \approx 2.677\;\text{mol}
- Atoms: N = n \times N_A = 2.677 \times 6.022 \times 10^{23} \approx 1.61 \times 10^{24}

Empirical Formula vs Molecular Formula

Empirical formula = smallest whole-number ratio of atoms in a compound
Molecular formula = actual number of each type of atom in a molecule
Examples:
- Glucose has empirical formula \mathrm{CH_2O}
- Glucose actual molecular formula is \mathrm{C6H{12}O_6} (which is (CH₂O)×6)
Distinction:
- Empirical formula gives the simplest ratio
- Molecular formula gives the precise composition of the molecule

Percent Composition and Empirical Formula from Percent Composition

Percent composition (by mass) of an element in a compound:
- - \% \text{X} = \dfrac{nX \times AX}{M} \times 100\%
- where nX = number of atoms of element X in the empirical unit, AX = atomic mass of element X, and M = molar mass of the compound (or formula unit)
Example prompt (from transcript): "What is the percent composition of oxygen in aluminum dichromate?" → apply the method above to find the oxygen percentage and then determine empirical formula from the percent composition if needed
Transition to empirical formula: once you have percent composition, convert each percent to moles by dividing by the respective atomic masses, then reduce to the smallest whole-number ratio

Determining the Empirical Formula from Percent Composition (Ibuprofen Example)

Percent composition given: 75.69% C, 8.80% H, 15.51% O
Steps:
- Convert each percent to moles:
- n_C = \dfrac{75.69}{12.01}
- n_H = \dfrac{8.80}{1.008}
- n_O = \dfrac{15.51}{16.00}
- Determine the smallest ratio by dividing all by the smallest value among nC, nH, n_O to obtain the empirical formula
Note: The transcript presents the prompt as a question to perform this calculation; the final empirical formula is derived from these steps

Molecular Formula from Empirical Formula and Molar Mass (Vitamin C Example)

Given: empirical formula for Vitamin C is \mathrm{C3H4O_3} with empirical molar mass
- M_{emp} = 3(12.01) + 4(1.008) + 3(16.00) = 36.03 + 4.032 + 48 = 88.062\;\mathrm{g\;mol^{-1}}
Provided molar mass of Vitamin C = 176\;\mathrm{g\;mol^{-1}}
Ratio: \dfrac{M}{M_{emp}} = \dfrac{176}{88.062} \approx 2.0
Molecular formula = empirical formula × ratio = \mathrm{C3H4O3} \times 2 = \mathrm{C6H8O6}

Solutions and Concentration: Molarity (M)

Solution components:
- Solute = substance dissolved
- Solvent = dissolving medium (e.g., water)
- Example: NaCl(aq) → NaCl is the solute, water is the solvent
Molarity (M) definition:
- M = \dfrac{n{\text{solute}}}{V{\text{solution}}} where n{\text{solute}} is moles of solute and V{\text{solution}} is volume of solution in liters
- Unit: \mathrm{mol\;L^{-1}}
Common shorthand: 2.0 M NaCl means 2.0 moles of NaCl per liter of solution
Interconversions with M:
- Molarity can be used as a conversion factor: to go from moles to liters and from liters to moles
- General relationships:
- n = M \times V
- V = \dfrac{n}{M}
- Also, to relate mass of solute: m = n \times M_{\text{molar}}
Example: Interpret interconversion: 2.0 M NaCl with a given volume is used to compute moles; given moles, compute volume; given mass, compute moles, etc.

Dilution: Preparing Solutions of Known Molarity

Concept: Concentrated solution + solvent → dilute solution
Key idea: Dilution changes concentration but not the number of moles of solute
Dilution (equation): Mi Vi = Mf Vf
- Where Mi, Vi are the initial molarity and volume, and Mf, Vf are the final molarity and volume
Example 1: Diluting a glucose solution
- Given: initial concentration Mi = 3.50\;\mathrm{M}, initial volume Vi = 75.0\;\mathrm{mL}
- Final volume V_f = 400.0\;\mathrm{mL}
- Find final concentration: Mf = \dfrac{Mi Vi}{Vf} = \dfrac{3.50 \times 75.0}{400.0} = 0.65625\;\mathrm{M}
Example 2: Preparing 250.0 mL of 0.500 M H2SO4 from a concentrated stock of 18.0 M
- Use dilution formula: Vi = \dfrac{Mf Vf}{Mi}
- Compute: V_i = \dfrac{0.500 \times 250.0}{18.0} \approx 6.94\;\mathrm{mL}
Practical notes:
- When diluting, the amount of solute remains the same; only the volume changes, so concentration changes inversely with volume
- The process is used to prepare solutions of a desired molarity from concentrated stocks or by dissolving a solid directly into solvent

Quick Reference: Common Formulas (Summary)

Avogadro's number: N_A = 6.022 \times 10^{23}
Molarity: M = \dfrac{n}{V} with n in moles and V in liters
Moles from molarity: n = M \times V
Volume from molarity: V = \dfrac{n}{M}
Mass from moles: m = n \times M_{\text{molar}}
Mass of a number of particles: m = \dfrac{N}{NA} \times M{\text{molar}}
Empirical formula from percent composition: convert each percent to moles, find the smallest whole-number ratio
Molecular formula from empirical formula: multiply empirical formula by the factor \dfrac{M{\text{actual}}}{M{\text{empirical}}}
Dilution: Mi Vi = Mf Vf