Formulas, Equations, and Moles

Chapter 3: Formulas, Equations, and Moles

Elements and the Periodic Table

• Periods: Horizontal rows in the periodic table.

• Groups: Vertical columns in the periodic table.

• Metals: Characterized by luster, malleability, ductility, and good conductivity of heat and electricity.

• Nonmetals: Non-lustrous, various colors, brittle (hard or soft), and poor conductors.

• Metalloids (Semimetals): Elements with properties intermediate between metals and nonmetals.

Ions

• Ions are formed when atoms gain or lose electrons, resulting in a charged particle.

• Cation: A positively charged ion formed by the loss of electrons.

• Anion: A negatively charged ion formed by the gain of electrons.

• The net charge is represented by a superscript (e.g., +, 2+, 3+ for loss of electrons; -, 2-, 3- for gain of electrons).

• General trend: Metal atoms tend to lose electrons, while nonmetal atoms tend to gain electrons.

  • Example: Sodium (Na) loses one electron to form Na+.

  • Example: Chlorine (Cl) gains one electron to form Cl–.

Predicting Ionic Charges

• Atoms gain or lose electrons to achieve the electron configuration of the nearest noble gas.

  • Group 1A atoms form 1+ ions (e.g., Li+).

  • Group 2A atoms form 2+ ions (e.g., Mg2+).

  • Group 7A atoms form 1– ions (e.g., F–).

  • Group 6A atoms form 2– ions (e.g., O2–).

Ionic Compounds

• Ionic compounds are composed of positively charged ions (cations) and negatively charged ions (anions).

• Typically, cations are metal ions, and anions are nonmetal ions.

• Ionic compounds are generally formed from combinations of metals and nonmetals.

• Molecular compounds are generally composed of nonmetals only.

Mass Relationships in Chemical Reactions

Atomic Mass and Molecular Weights

• Atomic Mass Scale:

  • $1 \text{ amu} = 1.66054 \times 10^{-24} \text{ g}$

  • $1 \text{ g} = 6.02214 \times 10^{23} \text{ amu}$

  • Example: $^{1}\text{H} = 1.6735 \times 10^{-24} \text{ g} = 1.0078 \text{ amu}$

The Mole and Molar Mass

• Mole Definition: The amount of substance that contains as many entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of carbon-12 ($^{12}\text{C}$).

• Avogadro's Number ($N_A$): $6.0221421 \times 10^{23}$, commonly used as $6.022 \times 10^{23}$.

• The mass of a single atom of an element in amu is numerically equal to the mass in grams of 1 mole of atoms of that element.

  • Example: One $^{12}\text{C}$ atom weighs 12 amu, so 1 mole of $^{12}\text{C}$ weighs 12 g.

  • Example: One $^{24}\text{Mg}$ atom weighs 24 amu, so 1 mole of $^{24}\text{Mg}$ weighs 24 g.

  • Example: One $^{197}\text{Au}$ atom weighs 197 amu, so 1 mole of $^{197}\text{Au}$ weighs 197 g.

• Molar Mass: The mass in grams of 1 mole of a substance.

Formula and Molecular Weights (Masses)

• The formula weight of a substance is the sum of the atomic weights of each atom in its chemical formula.

  • Example: Sulfuric acid ($\text{H}<em>2\text{SO}</em>4$) formula weight is calculated as:
    $\text{FW} = 2(\text{AW of H}) + (\text{AW of S}) + 4(\text{AW of O})$

• If the chemical formula is the molecular formula, then the formula weight is also called the molecular weight.

  • Example: Glucose ($\text{C}<em>6\text{H}</em>{12}\text{O}_6$) has a molecular weight of 180.16 amu.
    $\text{MW} = 6(12.01 \text{ amu}) + 12(1.008 \text{ amu}) + 6(16.00 \text{ amu}) = 180.16 \text{ amu}$

• For ionic substances like calcium nitrate ($\text{Ca(NO}<em>3)</em>2$), the term

Chemical Equations

• Reactants: The substances that react.

• Products: The substances that are produced.

• $+$: "reacts with"

• $\rightarrow$: "to produce"

• Coefficients: Indicate the amount of substance.

  • Example: $2 \text{H}<em>2 + \text{O}</em>2 \rightarrow 2 \text{H}_2\text{O}$

Balancing Equations

• Equations must be balanced to ensure equal amounts of each element on each side.

  • Example: $4 \text{H}_2\text{O}$ contains 4 H and 2 O.

• Important Considerations:

  • Subscripts should never be changed when balancing equations, as this changes the chemical identity.

  • Changing coefficients only changes the amount of a substance, not its identity.

• Example: $2 \text{H}<em>2 + \text{O}</em>2 \rightarrow 2 \text{H}_2\text{O}$

• Example: $\text{CH}<em>4 + \text{O}</em>2 \rightarrow \text{CO}<em>2 + \text{H}</em>2\text{O}$

  • Unbalanced: $\text{CH}<em>4 + \text{O}</em>2 \rightarrow \text{CO}<em>2 + \text{H}</em>2\text{O}$

• Guidelines for Balancing Equations:

  • Write the unbalanced equation.

  • Use coefficients to indicate how many formula units are required to balance the equation.

  • Balance species that occur in the fewest formulas on each side.

  • Reduce coefficients to the smallest whole number values.

  • When balancing reactions involving organic compounds, balance in the order: C, H, O.

  • Example: $\text{KClO}<em>3 \rightarrow \text{KCl} + \text{O}</em>2$

  • Example: $\text{C}<em>8 \text{H}</em>{18} + \text{O}<em>2 \rightarrow \text{CO}</em>2 + \text{H}_2\text{O}$

Chemical Symbols on Different Levels

• Chemical symbols represent both microscopic and macroscopic levels.

  • Example: $2 \text{H}<em>2(g) + \text{O}</em>2(g) \rightarrow 2 \text{H}_2\text{O}(g)$

  • 2 molecules + 1 molecule → 2 molecules

  • $2(6 \times 10^{23} \text{ molecules}) + 6 \times 10^{23} \text{ molecules} \rightarrow 2(6 \times 10^{23} \text{ molecules})$

  • 2 moles + 1 mole → 2 moles

• Coefficients in a balanced chemical equation can be interpreted as the relative numbers of molecules (or formula units) and as the relative numbers of moles involved in the reaction.

Stoichiometry: Chemical Arithmetic

• Example: $2 \text{ mol H}<em>2 \rightleftharpoons 1 \text{ mol O}</em>2 \rightleftharpoons 2 \text{ mol H}_2\text{O}$

• They are stoichiometrically equivalent.

• Example: $2 \text{ H}<em>2 + \text{ O}</em>2 \rightarrow 2 \text{ H}_2\text{O}$

• Steps for converting grams of A to grams of B:

  • Grams of A → Moles of A → Moles of B → Grams of B

  • Use molar mass of A as a conversion factor.

  • Use coefficients in the balanced equation to find mole ratios.

  • Use molar mass of B as a conversion factor.

  • For the balanced equation: $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$

• Example: How many grams of $\text{H}<em>2\text{O}$ can be produced from 40.0 g $\text{O}</em>2$?

• Combustion of butane, $\text{C}<em>4 \text{H}</em>{10}$. If we burn 1.00 g of butane, what mass of $\text{CO}_2$ is produced?

Reactions with Limiting Amounts of Reactants

• The limiting reactant (or limiting reagent) is the reagent that is completely consumed in a reaction and determines the amount of product formed.

• The reagent that gives the lesser amount of product is the limiting reagent.

• The amount the limiting reagent produces is the maximum amount of product.

• The other reactants are sometimes called excess reactants or excess reagents.

• Analogy: Making cheese sandwiches.

  • Rule: 1 slice of cheese + 2 slices of bread → 1 sandwich

  • If we have 6 slices of cheese and 20 slices of bread, how many sandwiches can we make?

Example

• Assume that 5.52 g of sodium reacts with 5.10 g of aluminum oxide.

  • a) Which reactant is limiting, and which reactant is in excess?

  • b) How many grams of aluminum are produced?

  • c) What mass of excess reactant remains at the end of the reaction?

• Important: Always make sure the chemical equation is balanced!

  • $\text{Na}(l) + \text{Al}<em>2\text{O}</em>3(s) \rightarrow \text{Al}(l) + \text{Na}_2\text{O}(s)$

Yields of Chemical Reactions

• Theoretical Yield: The quantity of product calculated to form when all of the limiting reactant reacts. It is the maximum quantity expected based on the stoichiometry of the chemical equation.

• Actual Yield: The amount actually obtained in a reaction.

• Percent Yield: Relates the actual yield to the theoretical yield:

  • $\text{Percent Yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

• The percent yield is the percentage of the theoretical yield that is actually achieved.

Example

• Adipic acid, $\text{H}<em>2\text{C}</em>6\text{H}<em>8\text{O}</em>4$, is used to produce nylon. The acid is made commercially by controlled reaction between cyclohexane ($\text{C}<em>6\text{H}</em>{12}$; mw = 84.14) and $\text{O}_2$:

  • $2 \text{ C}<em>6\text{H}</em>{12} + 5 \text{ O}<em>2 \rightarrow 2 \text{ H}</em>2\text{C}<em>6\text{H}</em>8\text{O}<em>4 + 2 \text{ H}</em>2\text{O}$

• Assume you carry out this reaction starting with 25.0 g of cyclohexane and that cyclohexane is the limiting reagent.

  • a) What is the theoretical yield of adipic acid (mw = 146.14)?

  • b) If you obtain 37.6 g of adipic acid from your reaction, what is the percent yield of adipic acid?

Additional practice problem

How many moles of O₂ would be required to generate 13.0 mol of NO₂ in the reaction below assuming the reaction has only 83.1% yield?
$2 NO (g) + O₂ (g) → 2 NO₂ (g)$