CH101 CH102 Course Notes (Pages 1–7)

Appendix A: Stoichiometry (Overview)

• A.1
  – A.2: Ratios and unit training using simple ratio examples (e.g., money units as analogies).

• A.5
  – Avogadro’s Number and The Mole:

  • $1~\text{amu} = 1.66 \times 10^{-24}~\text{g}$

  • Avogadro’s number $N_A = 6.022 \times 10^{23}$

  • Average atomic mass concept: $(1.66 \times 10^{-24}~\text{g}) \times (6.022 \times 10^{23}) \approx 1.000~\text{g per mole}$

  • The masses on the periodic table can serve as masses per mole for atoms (e.g., C: $12.01~\text{amu per atom}$; $12.01~\text{g per mole of C}$

  • Example calculations:

  • How many Fe atoms in $65.0~\text{g}$ of Fe?

    $65.0~\text{g Fe} \times \frac{6.022 \times 10^{23}~\text{atoms}}{55.85~\text{g Fe}} \,=\ 7.01 \times 10^{23}~\text{atoms Fe}$

  • How many moles of Fe in $65.0~\text{g Fe}$?

    $65.0~\text{g Fe} \times \frac{1~\text{mol}}{55.85~\text{g Fe}} \,=\ 1.16~\text{mol Fe}$

  • Atoms in a Molecule: formula and subscripts determine the ratio of atoms (e.g., HNO$<em>2$ has 1 H, 1 N, 2 O per molecule; in one mole of HNO$</em>2$, you have 1 mole H, 1 mole N, 2 moles O).

  • The concept extends to more complex molecules like Ca(NO$<em>3$)$</em>2$ (1 Ca, 2 N, 6 O per formula unit).

  • Q&A style questions (Q1-1 to Q1-4) involve converting grams to moles and counting atoms in molecules.

• Mass and Molecules: Mass of a Molecule (A.7)

  • Example calculation: 2 N atoms $\times$ 14.007 amu = 28.014 amu; 1 Ca atom $\times$ 40.078 amu = 40.078 amu; 6 O atoms $\times$ 15.999 amu = 95.994 amu; Total = 164.086 amu for Ca(NO$<em>3$)$</em>2$.
  • Masses per mole alignment: 2 N $\times$ 14.007 g = 28.014 g; 1 Ca $\times$ 40.078 g = 40.078 g; 6 O $\times$ 15.999 g = 95.994 g; total mass per mole = 164.086 g.
  • Q1-5 (Practice): $0.784~\text{moles}$ of Mg$<em>3$(PO$</em>4$)$_2$ weighs how much? Options: A) 206 g, B) 263 g, C) 263 amu, D) 206 amu

• Balancing Equations (Section 1.4)

  • Steps to balance a chemical equation:
    1) Add coefficients to the most complex species first.
    2) Balance atoms on the other side using coefficients.
    3) Iterate until all atoms balance and charges balance (if needed).
    4) Check that atoms and charges balance.
  • Example: Balance C$<em>2$H$</em>6$ + O$<em>2$ $\\rightarrow$ CO$</em>2$ + H$_2$O (use whole-number coefficients):
  • Balanced form: $2~C<em>2H</em>6 + 7~O<em>2 \rightarrow 4~CO</em>2 + 6~H_2O$
  • WebAssign requires whole-number coefficients.
  • Example: Balance S$<em>8$ + O$</em>2$ $\\rightarrow$ SO$_3$:
  • Balanced coefficients: $1~S<em>8 + 12~O</em>2 \rightarrow 8~SO_3$
  • (Answer: D)
  • Balance H$<em>2$SO$</em>4$ + NaOH $\\rightarrow$ Na$<em>2$SO$</em>4$ + H$_2$O:
  • Balanced form: $H<em>2SO</em>4 + 2~NaOH \rightarrow Na<em>2SO</em>4 + 2~H_2O$

• Mass to Moles to Mass (Section A.8)

  • Concept: Convert mass of one substance to another via stoichiometric ratios.

  • Breakfast example (eggs and bacon): two eggs per three slices of bacon.

  • Given: 1043.68 g of eggs; 12 eggs weigh 782.76 g; 1 dozen slices of bacon weigh 453.58 g.

  • Calculation approach (clean version):

  • Convert grams of eggs to number of eggs:

    $1043.68~\text{g eggs} \times \frac{12~\text{eggs}}{782.76~\text{g}} = 16.0~\text{eggs}$

  • Use ratio 2 eggs : 3 bacon to find number of bacon slices:

    $16~\text{eggs} \times \frac{3~\text{bacon}}{2~\text{eggs}} = 24~\text{slices bacon}$

  • Convert slices to grams using weight per dozen slices (12 slices per dozen):

    $24~\text{slices} \times \frac{453.58~\text{g}}{12~\text{slices}} = 907.16~\text{g bacon}$

  • Final result: approximately 907.16~\text{g bacon}} for the given eggs-to-bacon ratio.

• Page 7: Additional walk-through of the Mass to Moles to Mass example (reiterating the same breakfast-platter calculation with the given data). Final result remains approximately $907.16~\text{g bacon}$.

• Quick references and formulas

  • Atomic mass unit: $1~\text{amu} = 1.66 \times 10^{-24}~\text{g}$
  • Avogadro’s number: $N_A = 6.022 \times 10^{23}$
  • Relationship: average atomic mass in amu $\approx$ grams per mole for a given element (e.g., C has 12.01 amu per atom, so 1 mole of C weighs 12.01 g).
  • Mass to moles to mass uses stoichiometric coefficients to relate amounts of reactants and products via a balanced equation.

• Important takeaways

  • The course emphasizes attendance tracking, problem sessions, and a structured assessment plan (final, four semester tests, homework, participation).
  • Stoichiometry basics are foundational: atomic masses, Avogadro’s number, mole concept, balancing equations, and mass-to-mass conversions.
  • Real-world practice problems (e.g., kitchen/baking analogies) help illustrate proportion and ratio concepts in chemistry.