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}2\text{SO}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}6\text{H}{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}3)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}2 + \text{O}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}2 + \text{O}2 \rightarrow 2 \text{H}_2\text{O}
Example: \text{CH}4 + \text{O}2 \rightarrow \text{CO}2 + \text{H}2\text{O}
Unbalanced: \text{CH}4 + \text{O}2 \rightarrow \text{CO}2 + \text{H}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}3 \rightarrow \text{KCl} + \text{O}2
Example: \text{C}8 \text{H}{18} + \text{O}2 \rightarrow \text{CO}2 + \text{H}_2\text{O}
Chemical Symbols on Different Levels
Chemical symbols represent both microscopic and macroscopic levels.
Example: 2 \text{H}2(g) + \text{O}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}2 \rightleftharpoons 1 \text{ mol O}2 \rightleftharpoons 2 \text{ mol H}_2\text{O}
They are stoichiometrically equivalent.
Example: 2 \text{ H}2 + \text{ O}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}2\text{O} can be produced from 40.0 g \text{O}2?
Combustion of butane, \text{C}4 \text{H}{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}2\text{O}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}2\text{C}6\text{H}8\text{O}4, is used to produce nylon. The acid is made commercially by controlled reaction between cyclohexane (\text{C}6\text{H}{12}; mw = 84.14) and \text{O}_2:
2 \text{ C}6\text{H}{12} + 5 \text{ O}2 \rightarrow 2 \text{ H}2\text{C}6\text{H}8\text{O}4 + 2 \text{ H}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)