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

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 (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

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 = 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 (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

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

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 = $\mathrm{C3H4O3} \times 2 = \mathrm{C6H8O6}$

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.

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

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: $Mi Vi = Mf Vf$