Chemical Composition & Reactions – Exam Review Notes

Chapter 6 – Chemical Composition and the Mole Concept

• Importance of Chemical Composition
– Determines nutritional guidelines (e.g., <2.4\,\text{g Na} per day) and mining yields (Fe in ore).
– Used in assessing environmental issues such as ozone depletion (Cl in CFCs).

6.1 How Much Sodium?

• FDA sodium limit <2.4\,\text{g Na} per day.
• Sodium consumed mainly as NaCl; need conversion NaCl → Na.
• Requires molar mass and mass-percent methods.

6.2 Counting Nails by the Pound: Why We Use Mass to Count Atoms

• Analogy: counting nails by weighing → counting atoms by mass.
• Example: 2.60\,\text{lb} nails, 0.150\,\text{lb/doz} → 208 nails.
• Two-step conversions: mass ↔ dozens ↔ number.

6.3 Counting Atoms by the Gram

• Mole (\text{mol}): 1\,\text{mol}=6.022\times10^{23} entities (Avogadro’s number).
• Defined via 12\,\text{g} of ^{12}\text{C}.
• Conversions
– Moles ↔ atoms: 1\,\text{mol}\,\text{He}=6.022\times10^{23}\,\text{He atoms}.
– Grams ↔ moles (molar mass =\text{atomic mass amu} expressed in g).
– Grams ↔ atoms (two-step).
• Example: 0.58\,\text{g C}\to4.8\times10^{-2}\,\text{mol C}\to2.9\times10^{22}\,\text{atoms}.

6.4 Counting Molecules by the Gram

• Molar mass of compounds equals formula mass in g mol⁻¹.
• Process: grams ⟶ moles (compound) ⟶ molecules (or f.u.).
• Example: 22.5\,\text{g CO}_2\to0.511\,\text{mol}\to3.08\times10^{23}\,\text{molecules}.

6.5 Chemical Formulas as Conversion Factors

• Subscripts give mole ratios (e.g., CO2 ⇒ 2\,\text{mol O} : 1\,\text{mol CO}2).
• Moles compound → moles element (use formula).
• Grams compound → grams element: 3-step (g ↔ mol compound ↔ mol element ↔ g element).
• Example: 15\,\text{g NaCl}\to5.9\,\text{g Na}.

6.6 Mass Percent Composition

• \%\,\text{X}=\dfrac{\text{mass X in sample}}{\text{mass sample}}\times100.
• Use as conversion factor (e.g., 39 % Na in NaCl).
• Example: 2.4\,\text{g Na}\to6.2\,\text{g NaCl} allowable.

6.7 Mass % from Formula

• X=\dfrac{n\,\text{(mol X)}\times M{X}}{M{\text{compound}}}\times100.
• Example: \%\,\text{Cl in CCl2F2}=58.6\%.

6.8 Empirical Formulas

Procedure: grams → moles → pseudo-formula → divide by smallest → multiply to whole numbers.
• Example water: 3.0 g H & 24 g O ⇒ \text{H}_2\text{O}.
• Example: Ti + O₂ to TiO₂.

6.9 Molecular Formulas

• n=\dfrac{M{\text{molar}}}{M{\text{empirical}}}; molecular = (empirical)×n.
• Example: fructose CH₂O, M=180.2 ⇒ n=6 ⇒ \text{C}6\text{H}{12}\text{O}_6.

Chapter 7 – Chemical Reactions

7.1 Everyday Reactions

• Volcano demo: \text{NaHCO}3+\text{HC}2\text{H}3\text{O}2\to\text{CO}2(g)+\text{H}2\text{O}+\text{NaC}2\text{H}3\text{O}2. • Combustion of octane \text{C}8\text{H}{18}+\text{O}2\to\text{CO}2+\text{H}2\text{O}.
• Hard-water curd: \text{Ca}^{2+}+\text{CO}3^{2-}\to\text{CaCO}3(s).

7.2 Evidence of Reaction

• Indicators: color change, precipitate, gas, light, heat (exo/endo).
• Caveat: physical changes (e.g., boiling) can mimic some signs.

7.3 Chemical Equations

• Reactants → Products; include states (g, l, s, aq).
• Must be balanced— conserve atoms.

7.4 Balancing Procedure

1. Write skeletal formula.

2. Balance elements (metals → non-metals); free elements last.

3. Clear fractions.

4. Check atom counts.
   • Example combustion propane: \text{C}3\text{H}8+5\text{O}2\to3\text{CO}2+4\text{H}_2\text{O}.

7.5 Aqueous Solutions & Solubility Rules

• Dissociation of soluble ionic compounds → strong electrolytes.
• Key soluble ions: \text{Li}^+,\,\text{Na}^+,\,\text{K}^+,\,\text{NH}4^+,\,\text{NO}3^-,\,\text{C}2\text{H}3\text{O}2^- (always). • Halides soluble except with \text{Ag}^+,\text{Hg}2^{2+},\text{Pb}^{2+}.
• \text{SO}4^{2-} soluble except \text{Sr}^{2+},\text{Ba}^{2+},\text{Pb}^{2+},\text{Ca}^{2+}. • Insoluble: \text{CO}3^{2-},\text{PO}_4^{3-},\text{OH}^-,\text{S}^{2-} except with always-soluble cations.

7.6 Precipitation Reactions

• Mix two soluble salts → swap ions → check solubility.
• If insoluble product forms ⇒ precipitate.
• Example: \text{Pb}^{2+}+2\text{I}^-\to\text{PbI}_2(s) (yellow).

7.7 Molecular vs. Ionic Equations

• Molecular: full formulas.
• Complete Ionic: split all aq electrolytes into ions.
• Net Ionic: remove spectator ions.
• Example: \text{Ag}^+(aq)+\text{Cl}^-(aq)\to\text{AgCl}(s).

7.8 Acid–Base & Gas-Evolution

• Acid–Base (Neutralization): \text{H}^++\text{OH}^-\to\text{H}2\text{O} + salt. • Gas Evolution: certain anion–cation combos produce unstable intermediates (e.g., \text{H}2\text{CO}3\to\text{H}2\text{O}+\text{CO}2\uparrow). • Common gas formers: \text{H}2\text{S},\text{CO}2,\text{SO}2,\text{NH}_3.

7.9 Oxidation–Reduction (Redox)

• Oxidation: loss e⁻; Reduction: gain e⁻ (OIL RIG).
• Indicators: reaction with \text{O}2, metal + non-metal, electron transfer. • Combustion is a redox subtype: fuel + \text{O}2\to\text{CO}2+\text{H}2\text{O} + heat.

7.10 Classifying by Atom Patterns

• Synthesis: A+B\to AB
• Decomposition: AB\to A+B
• Single-Displacement: A+BC\to AC+B
• Double-Displacement: AB+CD\to AD+CB
• Precipitation, acid–base, and many gas-evolution reactions fit double-displacement pattern.

Key Equations & Numerical References

• Avogadro’s number: NA=6.022\times10^{23}\,\text{mol}^{-1}. • Molar mass ↔ mass ↔ moles conversions. • Mass-percent: \%\,\text{X}=\dfrac{mX}{m{\text{compound}}}\times100. • Empirical formula n: n=\dfrac{M{\text{molar}}}{M_{\text{empirical}}}.

Practical & Real-World Applications

• Health: Sodium restriction, fluoride in water (0.7 mg L⁻¹ using NaF mass %).
• Industry: Liming acid lakes (CaCO₃ neutralizes H₂SO₄/HNO₃), precipitation to remove toxic metals.
• Energy: Combustion of hydrocarbons powers vehicles; redox in batteries.
• Environmental: CFC chlorine quantification, smog (NO, NO₂, O₃) formation pathways.

Ethical & Safety Notes

• Excess sodium and fluoride have health risks (hypertension, fluorosis).
• Combustion contributes \text{CO}_2 to climate change.
• Acid rain results from SOₓ / NOₓ emissions; neutralization strategies are vital.

Study Tips

• Memorize solubility rules and strong acids/bases.
• Practice mole-mass-particle conversions systematically.
• Balance equations by atom type, leaving O & H last in combustion.
• For net ionic equations: split (aq) compounds, keep (s,l,g) intact.
• Identify reaction type quickly to anticipate products.

Chapter 6 – Chemical Composition and the Mole Concept

• Importance of Chemical Composition

– Chemical composition is fundamental across various fields. For instance, in nutrition, it dictates daily limits for substances like sodium (e.g., less than 2.4\,\text{g Na} per day as recommended by the FDA) by informing us about the percentage of a specific element in a compound. In mining, understanding the composition of an ore (e.g., the percentage of Fe in iron ore) is crucial for calculating the yield and economic viability of extraction. In environmental science, it's used to quantify pollutants, such as the amount of chlorine present in chlorofluorocarbons (CFCs) that contribute to ozone depletion, enabling accurate assessment of their impact.

6.1 How Much Sodium?

• The FDA recommends a daily sodium limit of less than 2.4\,\text{g Na}.

• Since sodium is primarily consumed as sodium chloride (NaCl), it is essential to convert the mass of NaCl into the mass of Na. This conversion requires knowledge of the molar mass of both sodium and sodium chloride, along with the mass-percent composition method, which tells us the exact percentage of sodium within the NaCl compound.

6.2 Counting Nails by the Pound: Why We Use Mass to Count Atoms

• This section introduces an analogy to illustrate the concept of using mass to count microscopic particles like atoms, which are too small to count individually.

• Just as we can count a large number of nails by weighing a sample and knowing the average mass per nail (or per dozen nails), chemists count atoms by weighing a sample of an element or compound.

• Example: If you have 2.60\,\text{lb} of nails and you know that one dozen nails weighs 0.150\,\text{lb}, you can calculate the number of dozens (2.60\,\text{lb} \div 0.150\,\text{lb/doz} = 17.33\,\text{doz}) and then the total number of nails (17.33\,\text{doz} \times 12\,\text{nails/doz} = 208\,\text{nails}).

• This demonstrates a two-step conversion process: mass of bulk item ↔ dozens of items ↔ number of individual items. This is directly analogous to mass of substance ↔ moles of substance ↔ number of atoms/molecules.

6.3 Counting Atoms by the Gram

• Mole (\text{mol}): The mole is the SI unit for the amount of substance. It is defined as the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12\,\text{g} of pure carbon-12 (^{12}\text{C}). This number, known as Avogadro’s number (N_A), is 6.022\times10^{23} entities per mole (1\,\text{mol} = 6.022\times10^{23} entities).

• Conversions

– Moles ↔ atoms: Avogadro's number serves as the conversion factor. For example, 1\,\text{mol of He} = 6.022\times10^{23}\,\text{He atoms}. To convert from moles to atoms, multiply by NA. To convert atoms to moles, divide by NA.

– Grams ↔ moles: The molar mass (M) of an element is its atomic mass unit (\text{amu}), expressed in grams per mole (\text{g/mol}). For example, the atomic mass of Carbon-12 is 12.01\,\text{amu}, so its molar mass is 12.01\,\text{g/mol}. To convert grams to moles, divide by the molar mass. To convert moles to grams, multiply by the molar mass.

– Grams ↔ atoms: This is a two-step conversion process, combining the previous two. First, convert grams to moles using molar mass, then convert moles to atoms using Avogadro's number.

• Example conversion: To find the number of atoms in 0.58\,\text{g C}:
1. Convert grams to moles: \frac{0.58\,\text{g C}}{12.01\,\text{g C/mol C}} \approx 4.8\times10^{-2}\,\text{mol C}
2. Convert moles to atoms: (4.8\times10^{-2}\,\text{mol C}) \times (6.022\times10^{23}\,\text{atoms/mol}) \approx 2.9\times10^{22}\,\text{C atoms}

6.4 Counting Molecules by the Gram

• Just as elements have molar masses, compounds also have molar masses, which are equal to their formula mass (sum of atomic masses of all atoms in the chemical formula) expressed in grams per mole (\text{g/mol}). For molecules, this is often called molecular mass.

• The process for converting between grams, moles, and molecules (or formula units for ionic compounds) of compounds is similar to that for elements:

*   **Grams compound ⟶ moles compound:** Divide by the molar mass of the compound.
*   **Moles compound ⟶ molecules (or formula units):** Multiply by Avogadro's number.

• Example: To find the number of molecules in 22.5\,\text{g CO}2: 1. Calculate molar mass of ext{CO}2: (1 \times 12.01) + (2 \times 16.00) = 44.01\,\text{g/mol}
2. Convert grams to moles: \frac{22.5\,\text{g CO}2}{44.01\,\text{g CO}2\text{/mol CO}2} \approx 0.511\,\text{mol CO}2
3. Convert moles to molecules: (0.511\,\text{mol CO}2) \times (6.022\times10^{23}\,\text{molecules/mol}) \approx 3.08\times10^{23}\,\text{CO}2\,\text{molecules}

6.5 Chemical Formulas as Conversion Factors

• The subscripts in a chemical formula provide the mole ratios of elements within a compound. For example, in the formula ext{CO}2, the subscript '2' for oxygen and the implicit '1' for carbon indicate that there are 2\,\text{mol of O atoms} for every 1\,\text{mol of C atoms} or 1\,\text{mol of CO}2\,\text{molecules}. These ratios act as conversion factors in stoichiometric calculations.

• Moles compound → moles element: Use the mole ratio derived from the chemical formula. For instance, to find moles of O in ext{CO}2, you would multiply moles of ext{CO}2 by \frac{2\,\text{mol O}}{1\,\text{mol CO}_2}.

• Grams compound → grams element: This is a three-step conversion that combines molar mass conversions with the mole ratio:
1. Convert grams of compound to moles of compound (using molar mass of compound).
2. Convert moles of compound to moles of element (using the mole ratio from the formula).
3. Convert moles of element to grams of element (using molar mass of element).

• Example: To find the mass of Na in 15\,\text{g NaCl}. Molar mass of NaCl is 58.44\,\text{g/mol}; molar mass of Na is 22.99\,\text{g/mol}.
1. Grams NaCl to moles NaCl: \frac{15\,\text{g NaCl}}{58.44\,\text{g/mol NaCl}} \approx 0.2567\,\text{mol NaCl}
2. Moles NaCl to moles Na: Since there is 1 mol Na per 1 mol NaCl, this is 0.2567\,\text{mol Na}.
3. Moles Na to grams Na: (0.2567\,\text{mol Na}) \times (22.99\,\text{g/mol Na}) \approx 5.9\,\text{g Na}

6.6 Mass Percent Composition

• Mass percent (\,\%\,\text{X}) of an element X in a sample or compound is calculated as the mass of element X divided by the total mass of the sample, multiplied by 100:
\%\,\text{X}=\dfrac{\text{mass X in sample}}{\text{mass sample}}\times100

• This mass percent can be used as a conversion factor to calculate the mass of an element in a given mass of a compound, or vice versa. For example, if NaCl is 39% Na by mass, you can write this as \frac{39\,\text{g Na}}{100\,\text{g NaCl}}.

• Example: If the daily allowable sodium is 2.4\,\text{g Na}, and NaCl is 39% Na by mass, how much NaCl can be consumed?
(2.4\,\text{g Na}) \times \left(\frac{100\,\text{g NaCl}}{39\,\text{g Na}}\right) \approx 6.2\,\text{g NaCl}

6.7 Mass % from Formula

• The mass percent of an element X in a compound can also be calculated directly from its chemical formula and molar masses. The formula is:
\%\,\text{X}=\dfrac{n\,\text{(mol X)}\times M{\text{X}}}{M{\text{compound}}}\times100
Where:
* n is the number of moles of element X in one mole of the compound (from the subscript).
* M{\text{X}} is the molar mass of element X. * M{\text{compound}} is the molar mass of the entire compound.

• Example: Calculate \,\%\,\text{Cl} in dichlorodifluoromethane (\text{CCl}2\text{F}2).
* Molar mass of C: 12.01\,\text{g/mol}
* Molar mass of Cl: 35.45\,\text{g/mol}
* Molar mass of F: 19.00\,\text{g/mol}
* Molar mass of ext{CCl}2\text{F}2: 12.01 + (2 \times 35.45) + (2 \times 19.00) = 120.91\,\text{g/mol}
* Mass of Cl in one mole of ext{CCl}2\text{F}2: 2 \times 35.45 = 70.90\,\text{g}
* \%\,\text{Cl}=\dfrac{70.90\,\text{g Cl}}{120.91\,\text{g CCl}2\text{F}2}\times100\% \approx 58.6\%

6.8 Empirical Formulas

• An empirical formula represents the simplest whole-number ratio of atoms in a compound. It is determined from experimental data, typically the masses of elements obtained from decomposition or combustion analysis.

• Procedure to determine empirical formula from mass data:
1. Convert the mass (grams) of each element to moles of each element (using their respective molar masses).
2. Form a pseudo-formula using these mole values as subscripts.
3. Divide all mole values by the smallest mole value obtained. This step attempts to convert the mole ratios into integers.
4. If the resulting ratios are not whole numbers, multiply all numbers by the smallest integer that will convert them into whole numbers (e.g., if you have 1.5, multiply by 2; if you have 0.33, multiply by 3).

• Example (water): A sample contains 3.0\,\text{g H} and 24\,\text{g O}.
1. Moles H: \frac{3.0\,\text{g H}}{1.008\,\text{g/mol H}} \approx 2.976\,\text{mol H}
2. Moles O: \frac{24\,\text{g O}}{16.00\,\text{g/mol O}} = 1.500\,\text{mol O}
3. Pseudo-formula: ext{H}{2.976}\text{O}{1.500}
4. Divide by smallest (1.500): H: 2.976/1.500 \approx 1.98 (round to 2); O: 1.500/1.500 = 1
5. Empirical Formula: ext{H}_2\text{O}

• Example (Titanium Oxide): If titanium reacts with oxygen to form an oxide, and you determine the mass of Ti and O in the product, you can find its empirical formula. For example, if 1.0\,\text{mol Ti} combines with 2.0\,\text{mol O} to form ext{TiO}_2.

6.9 Molecular Formulas

• A molecular formula provides the actual number of atoms of each element in a molecule. It is always a whole-number multiple of the empirical formula. To determine the molecular formula, you need both the empirical formula and the molar mass of the compound (which is typically determined experimentally, e.g., via mass spectrometry).

• The relationship is given by:
n=\dfrac{M{\text{molar}}}{M{\text{empirical}}}
Where:
* n is the whole-number multiplier.
* M{\text{molar}} is the experimentally determined molar mass of the compound. * M{\text{empirical}} is the empirical formula molar mass (calculated from the empirical formula).

• Once n is found, the molecular formula is derived by multiplying all subscripts in the empirical formula by n: molecular formula = (empirical formula) \times n.

• Example: Fructose has an empirical formula of ext{CH}2\text{O} and an experimentally determined molar mass (M{\text{molar}}) of 180.2\,\text{g/mol}.
1. Calculate empirical formula molar mass (M{\text{empirical}}) for ext{CH}2\text{O}: 12.01 + (2 \times 1.008) + 16.00 = 30.03\,\text{g/mol}.
2. Calculate n: n=\dfrac{180.2\,\text{g/mol}}{30.03\,\text{g/mol}} \approx 6
3. Multiply empirical formula by n: ( ext{CH}2\text{O}) \times 6 = \text{C}6\text{H}{12}\text{O}6
4. Molecular Formula: ext{C}6\text{H}{12}\text{O}_6

Chapter 7 – Chemical Reactions

7.1 Everyday Reactions

• Chemical reactions are ubiquitous in our daily lives, transforming substances into new ones with different properties.

• Volcano demo: A classic example is the reaction between baking soda (sodium bicarbonate, NaHCO3) and vinegar (acetic acid, ext{HC}2\text{H}3\text{O}2) to simulate a volcano eruption:
NaHCO3(s) + \text{HC}2\text{H}3\text{O}2(aq) \to \text{CO}2(g) + \text{H}2\text{O}(l) + \text{NaC}2\text{H}3\text{O}_2(aq)
This reaction produces carbon dioxide gas, which causes the bubbling and