Notes on Molecular Mass, Percent Composition, and Stoichiometry

Molecular Mass and Formula Mass

  • Key idea: Covalent compounds have a molecular mass (sometimes called molecular weight) calculated by summing the masses of all atoms in the molecule. Ionic compounds have formula units, but the same principle applies when you treat the formula as the count of atoms in the formula unit.
  • Definition and approach:
    • For a molecule, molecular mass (in atomic mass units, amu) is the sum of the atomic masses of its atoms:
      M{ ext{amu}} = \sumi ni \, mi,
      where $ni$ is the number of atoms of element $i$ in the molecule and $mi$ is the atomic mass of that element in amu.
  • Example: trichloromethane (CHCl$_3$) has molecular mass approximately 119.37\ ext{amu}.
  • Unit note: one atomic mass unit is an infinitesimally small mass; 1 amu ≈ 1.66054 \times 10^{-24} \text{ g}. (In practice we often use molar mass in g/mol, which numerically equals the same value as the molecular mass in amu for a given compound.)
  • From amu to grams per mole (molar mass): the same numeric value is used with units of g/mol. For example, the molar mass of NaCl is:
    M{ ext{NaCl}} = M{ ext{Na}} + M_{ ext{Cl}} \approx 22.99 + 35.45 = 58.44\ \text{g/mol}.
  • Hydrogen note: periodic table molar masses are typically given to four significant figures; hydrogen is often cited with fewer decimal places in some contexts. The transcript mentions: “molar masses are four sig figs except for hydrogen; there are two decimal places most of them, except hydrogen’s three.” When you combine masses, keep track of significant figures and round to the appropriate precision.
  • Ammonia as an example:
    • Formula: $\text{NH}_3$
    • $M{\text{NH}3} = MN + 3 MH = 14.01 + 3(1.008) \approx 17.034\ \text{amu}$ (approximate value used in the lecture).
    • The molecular mass of ammonia is about 17.03\ \text{g/mol} (rounded as appropriate for sig figs).
  • Polyatomic ions in formulas:
    • When a polyatomic ion appears in parentheses with a subscript outside, the subscript tells you how many of that entire ion you have (e.g., (SO$_4$)$^{2-}$ with a outside subscript multiplies the entire sulfate unit).
    • Do not break apart polyatomic ions when balancing equations; keep them together as units.
  • Percent composition (mass percent of each element in a compound):
    • Concept: percent composition is the mass fraction of an element in the compound, expressed as a percentage.
    • Formula (for element X in a compound):
      ext{wt%}X = \frac{nX \cdot MX}{M{ ext{compound}}} \times 100\%,
      where $nX$ is the number of atoms of element X in the formula, $MX$ is the atomic mass of X, and $M_{ ext{compound}}$ is the molar mass of the compound.
  • Ammonia percent composition example (NH$_3$):
    • $M{ ext{NH}3} \approx 17.034$ amu (or g/mol).
    • For nitrogen: $nN=1$, $MN=14.01$; for hydrogen: $nH=3$, $MH=1.008$.
    • $ ext{wt%}N = \frac{1 \times 14.01}{17.034} \times 100 \% \approx 82.3\%$; $ ext{wt%}H = \frac{3 \times 1.008}{17.034} \times 100 \% \approx 17.7\%$.
  • Practical note on rounding: differences in calculated percent compositions close to 0.01–0.1% can arise from rounding atomic masses; rounding to the least number of significant figures involved is standard practice.

Reaction Equations and Stoichiometry Basics

  • What happens in a chemical reaction: atoms are neither created nor destroyed; the count of each element is conserved from reactants to products.
  • Reactants vs. products:
    • Left side: reactants (ingredients in a recipe).
    • Right side: products (what you get).
  • Physical states:
    • s = solid, l = liquid, g = gas, aq = aqueous (in water).
    • Example: NaCl(s) vs NaCl(aq) (dissolved in water).
  • Balance by inspection or by an inventory method:
    • Inventory method: tally the number of each element on both sides and balance element by element.
    • Inspection: balance “easy” elements first (often metals/cations) and move to troublesome ones.
  • Role of coefficients:
    • The coefficients in front of species multiply the number of atoms in that species.
    • Subscripts in formulas are fixed and should not be altered when balancing; subscripts come from the chemical formulas themselves.
    • Balance to achieve the lowest whole-number ratio of coefficients.
  • Tricks and guidelines:
    • Gases (especially O$2$, H$2$, N$_2$) are often saved for last when balancing.
    • If you have an ionic compound, you can break it into its constituent ions and balance via net ionic perspective, but keep polyatomic ions intact when possible (they behave as units in solution).
    • Do not balance by changing subscripts; use coefficients instead.
    • If a fractional coefficient is needed, temporarily use fractions to balance, then multiply the entire equation by a common factor to clear fractions (you must convert to whole numbers).
  • Example 1: direct synthesis of NaOH from Na and water (simplified ionic/potential redox view):
    • Reactants: solid Na and H$_2$O(l).
    • Products: NaOH(aq) and H$_2$(g).
    • Balanced form (typical simplified version):
      2\ Na + 2\ H2O \rightarrow 2\ NaOH + H2.
    • Note: when balancing, focus on the sodium cation as a starting point and then check hydrogen and oxygen, adjusting with coefficients as needed; polyatomic ions (like sulfate) stay together if they appear as a unit.
  • Example 2: acid–base neutralization with sulfuric acid and sodium hydroxide:
    • Reactants: $\text{NaOH}$ (aqueous) and $\text{H}2\text{SO}4$ (aqueous).
    • Products: $\text{Na}2\text{SO}4$ (aqueous) and $\text{H}_2\text{O}$ (liquid).
    • Balanced equation:
      2\ NaOH(aq) + H2SO4(aq) \rightarrow Na2SO4(aq) + 2\ H_2O(l).
  • Example 3: decomposition of sodium bicarbonate upon heating (thermal decomposition):
    • Reactant: $\text{NaHCO}_3$.
    • Products: $\text{Na}2\text{CO}3$, CO$2$, and H$2$O.
    • Balanced equation:
      2\ NaHCO3(s) \rightarrow Na2CO3(s) + CO2(g) + H_2O(g).
  • Example 4: hydrocarbon combustion (ethane or hexane as examples):
    • General rule for hydrocarbons (C and H only) combusting in O$_2$:
    • Products are always CO$2$ and H$2$O.
    • Ethane example (C$2$H$6$): balanced form:
      2\ C2H6(g) + 7\ O2(g) \rightarrow 4\ CO2(g) + 6\ H_2O(g).
    • Hexane example (C$6$H${14}$): a standard balanced form is:
      2\ C6H{14}(g) + 19\ O2(g) \rightarrow 12\ CO2(g) + 14\ H_2O(g).
    • Note on fractional coefficients: sometimes a fractional coefficient like $\frac{19}{2}$ O$_2$ appears mid-balancing, but you should multiply all coefficients by 2 to clear fractions and keep whole numbers.
  • Example 5: balancing a combustion with a word problem context (to illustrate practice):
    • If a hydrocarbon (e.g., C$x$H$y$) burns in O$2$, the products are CO$2$ and H$_2$O.
    • Stepwise approach: balance carbon first (CO$2$), then hydrogen (H$2$O), then balance the remaining oxygen with O$_2$; if needed, use the fractional-coefficient trick and clear fractions by multiplying all coefficients.
  • Important conceptual point: balanced equations ensure conservation of atoms; the coefficients give the smallest whole-number ratios of species involved.

The Mole, Avogadro’s Number, and Molar Mass

  • The mole as a counting unit:
    • A mole represents the amount of substance containing exactly N_A = 6.022\times 10^{23} particles (atoms, molecules, formula units, etc.).
    • One mole of any substance contains the same number of particles: $N_A$.
  • Relationship between atomic/molar masses:
    • The average atomic weight on the periodic table (in amu) equals the molar mass in g/mol for that element or compound. Example: one mole of carbon weighs about 12.01 g.
    • 1 amu ≈ 1.66054 \times 10^{-24} \text{ g}. Therefore, the mass of one mole of carbon is about 12.01 g.
  • Avogadro’s concept in practice:
    • If you know the number of particles, you can find moles by dividing by $NA$: n = \frac{N}{NA}.
    • If you know the mass, you can find moles by dividing by molar mass:
      n = \frac{m}{M},
      where $M$ is the molar mass in g/mol.
  • Example: density-free discussion used to illustrate the mole concept; a drop of water may contain about $8.00\times 10^{20}$ water molecules, corresponding to roughly n \approx \frac{8.00\times 10^{20}}{6.022\times 10^{23}} \approx 1.33\times 10^{-3}\,\text{mol} of water, i.e., about a millimole. The mass of that amount of water is
    m = nM = (1.33\times 10^{-3}\text{ mol})(18.016\text{ g/mol}) \approx 0.024\text{ g}.
  • Molar mass and calculations: the lecture emphasizes that many problems boil down to interconversions among grams, moles, and numbers of particles, using:
    • m \to n via n = \dfrac{m}{M}
    • n \to m via m = nM
    • n \to N via N = nN_A
    • N \to n via n = \dfrac{N}{N_A}
  • Example conversions mentioned:
    • 1 L of air contains about 9.2\times 10^{-4}\text{ mol of Ar (Argon)}. The Ar molar mass is M_{Ar} = 39.95\ \text{g/mol}. Mass of Ar in that liter: m = nM = (9.2\times 10^{-4}\text{ mol})(39.95\ \text{g/mol}) \approx 0.0367\ \text{g}.
  • Gram-to-mole problems (and their inverses) are the bread-and-butter of these exercises:
    • Grams to moles: divide by molar mass.
    • Moles to grams: multiply by molar mass.
    • Moles to molecules: multiply by $N_A$.
    • Molecules to moles: divide by $N_A$.
  • Example: moles to molecules for saccharin (C$7$H$5$NO$_3$S).
    • Step 1: compute molar mass of saccharin: $M{ ext{C}7 ext{H}5 ext{NO}3 ext{S}} = 7(12.01) + 5(1.008) + 14.01 + 3(16.00) + 32.06 \approx 183.18\ \text{g/mol}$.
    • Step 2: convert mass to moles: given mass $m = 40\ \text{mg} = 0.040\ \text{g}$, so n = \dfrac{0.040}{183.18} \approx 2.18\times 10^{-4}\ \text{mol}.
    • Step 3: convert moles to molecules: N = nN_A = (2.18\times 10^{-4}\ \text{mol})(6.022\times 10^{23}) \approx 1.31\times 10^{20}\ \text{molecules}.
    • Step 4: carbon atoms per saccharin molecule: 7 C per molecule, so moles of carbon atoms: n{C} = 7n = 7 \times 2.18\times 10^{-4} \approx 1.526\times 10^{-3}\ \text{mol}, and corresponding number of carbon atoms: NC = nC NA \approx 9.19\times 10^{20}.
  • The speaker notes about practicality:
    • Once you set up the molar-mass-based conversions, most problems are interconversions among grams, moles, and numbers of particles; subsequent steps in many chapters build on mastering these interconversions.

Quick Practice Problems (from the transcript context)

  • Problem idea: How many oxygen atoms are in one mole of CO$_2$?
    • Each CO$2$ contains 2 O atoms; thus, 1 mole of CO$2$ contains 2 \times NA = 2NA oxygen atoms.
  • Problem idea: If you have 1 mole of NO$_2$, how many oxygen atoms are in that amount?
    • Each NO$2$ has 2 O; thus, 1 mole NO$2$ contains 2N_A oxygen atoms as well.

Summary of Key Concepts and Takeaways

  • Always balance chemical equations to conserve atoms; use coefficients, not subscripts, to balance.
  • Distinguish between molecular mass (amu) and molar mass (g/mol). Numerically the same, but units differ.
  • Sig figs matter in chemistry calculations; propagate the least precise measurement through your calculation.
  • Percent composition expresses how much of each element is present by mass in a compound.
  • The mole is a counting unit tied to Avogadro’s number; conversions between mass, moles, and number of particles are foundational to quantitative chemistry.
  • For hydrocarbon combustion, expect CO$2$ and H$2$O as products; adjust coefficients to balance the number of C and H first, then finish with O$_2$.
  • Polytomic ions (e.g., SO$_4^{2-}$) should be treated as single units when balancing; ionic compounds can be represented as ions in aqueous solution, but avoid breaking polyatomic ions apart when they remain intact.
  • Fractional coefficients can appear during balancing; multiply the entire equation by an integer to clear fractions and obtain the lowest whole-number ratio.

Notation and Formulas (LaTeX recap)

  • Molecular/Formula Mass:
    M{ ext{amu}} = \sumi ni \, mi
  • Molar Mass:
    M = M_{ ext{amu}} \,( ext{g/mol})
  • Percent Composition:
    ext{wt%}X = \frac{nX \cdot MX}{M{ ext{compound}}} \times 100\%
  • Mole concept and conversions:
    • n = \frac{m}{M}
    • m = nM
    • N = nNA\quad\text{with}\quad NA = 6.022\times 10^{23}
  • Example molar masses (for reference):
    • NaCl: M_{ ext{NaCl}} = 22.99 + 35.45 \approx 58.44\ \text{g/mol}
    • NH$3$: M{ ext{NH}_3} = 14.01 + 3(1.008) \approx 17.034\ \text{amu}\approx 17.034\ \text{g/mol}
  • Combustion balance exemplars:
    • H$2$ + O$2$ → H$2$O → balanced: 2\ H2 + O2 \rightarrow 2\ H2O
    • Na + H$2$O → NaOH + H$2$ → balanced: 2\ Na + 2\ H2O \rightarrow 2\ NaOH + H2
    • NaOH + H$2$SO$4$ → Na$2$SO$4$ + H$2$O → balanced: 2\ NaOH(aq) + H2SO4(aq) \rightarrow Na2SO4(aq) + 2\ H2O(l)
  • Final note: Practice with these techniques to streamline problem-solving for later chapters that build on stoichiometry and reaction calculations.