Chemical Quantities & Introduction to Equilibrium

Balancing Chemical Equations

A chemical equation must be balanced so that the same number of atoms of each element appears on both sides, directly reflecting the law of conservation of mass — atoms are rearranged in a reaction, never created or destroyed. Balancing is accomplished by adjusting coefficients (never subscripts, which would change the substances' identities) until each element's atom count matches on both sides. A systematic approach — balancing elements that appear in only one reactant and one product first, saving free elements and hydrogen/oxygen for last — makes balancing more efficient.

• •Only coefficients may be changed; subscripts define the compound's identity and are fixed.
• •Balance polyatomic ions as whole units when they appear unchanged on both sides.
• •Save elemental substances (like O₂) for last, since they can be fractionally adjusted or doubled at the end.
• •Always double-check every element's count after balancing to confirm conservation of mass.

The Mole and Avogadro's Number

The mole is the SI base unit for amount of substance and is defined as exactly 6.022 × 10²³ elementary entities (Avogadro's number), whether atoms, molecules, ions, or formula units. The mole allows chemists to bridge the invisible atomic scale to the measurable macroscopic scale, since counting individual atoms directly is impossible but weighing a mole-sized sample is routine. Moles, mass, and particle number are all interconvertible using molar mass and Avogadro's number as conversion factors.

• •Avogadro's number: 6.022 × 10²³ particles/mol.
• •Molar mass (g/mol) numerically equals the atomic/formula mass in amu, from the periodic table.
• •Conversion chain: mass (g) ⇌ moles (mol) ⇌ number of particles, using molar mass and Avogadro's number.

Mass Percent Composition and Empirical Formulas

Mass percent composition expresses the percentage of a compound's total mass contributed by each element, calculated by dividing each element's total mass in the formula by the compound's molar mass and multiplying by 100%. Working backward from experimentally measured percent composition data, chemists can determine a compound's empirical formula by converting each percentage to grams (assuming a 100 g sample), converting grams to moles, and dividing all mole values by the smallest one to find the simplest whole-number ratio.

• •Mass % element = (mass of element in formula / molar mass of compound) × 100%.
• •Assume a 100 g sample so percentages become grams directly.
• •Convert grams of each element to moles using its molar mass.
• •Divide all mole values by the smallest to get the simplest whole-number mole ratio (the empirical formula subscripts).
• •If ratios aren't whole numbers, multiply all subscripts by a small integer (2, 3, etc.) to reach whole numbers.

The Stoichiometry Road Map

Stoichiometry problems relate quantities of different substances in a balanced chemical equation, and nearly all such problems follow the same logical path: convert the given mass to moles, use the mole ratio from the balanced equation to convert to moles of the desired substance, then convert those moles to the desired final unit (usually mass). The balanced equation's coefficients provide the essential mole-to-mole conversion factor linking any two substances in the reaction.

• •Step 1: convert grams of substance A to moles of A using its molar mass.
• •Step 2: convert moles of A to moles of B using the coefficient ratio from the balanced equation.
• •Step 3: convert moles of B to grams of B using B's molar mass.
• •This 'grams → moles → moles → grams' path works for any two substances in a balanced equation.

Limiting Reactant and Excess Reactant

When reactants are not present in the exact stoichiometric ratio given by the balanced equation, one reactant — the limiting reactant — is completely consumed first and stops the reaction, while some of the other reactant, the excess reactant, remains unreacted. Two equivalent strategies identify the limiting reactant: calculating the moles of product each reactant could theoretically form (whichever produces less product is limiting), or comparing the actual mole ratio of reactants supplied to the required mole ratio from the balanced equation.

• •Method 1 (mol product method): calculate moles of product possible from each reactant separately; the smaller value identifies the limiting reactant and the actual yield achievable.
• •Method 2 (mole ratio method): compare the actual mol ratio of reactants to the coefficient ratio required by the equation.
• •Excess reactant remaining = initial moles of excess reactant − moles of excess reactant consumed by the limiting reactant.
• •Only the limiting reactant's initial amount should be used to calculate theoretical yield.

Theoretical, Actual, and Percent Yield

Theoretical yield is the maximum amount of product predicted by stoichiometric calculation from the limiting reactant, assuming the reaction goes to completion with perfect efficiency. Actual yield is the amount of product actually collected in the laboratory, which is almost always less than theoretical due to real-world imperfections. Percent yield expresses actual yield as a percentage of theoretical yield and is a key measure of a reaction's practical efficiency.

• •Percent yield = (actual yield / theoretical yield) × 100%.
• •Common reasons actual yield is less than theoretical: side reactions, incomplete reactions/equilibrium limits, loss during transfer/filtration/purification, impure reactants.
• •Percent yield cannot exceed 100% under normal circumstances; values above 100% usually indicate impure product or measurement error.

Particulate (Box) Diagrams in Stoichiometry

Particulate diagrams represent reactant and product molecules as simple shapes or symbols inside a box before and after a reaction, allowing students to visually apply stoichiometric reasoning without doing formal calculations. By counting the number of each type of molecule before and after reaction, one can directly identify the limiting reactant (the species that runs out completely) and confirm the correct product ratio predicted by the balanced equation.

• •Count each reactant type before reaction and match to the balanced equation's coefficient ratio.
• •The reactant present in a proportionally smaller amount relative to its coefficient is limiting.
• •Leftover (excess) reactant particles remain unchanged in the 'after' box.
• •Product particle count should match the mole ratio implied by the balanced equation, scaled to how much limiting reactant reacted.

Reversible Reactions and Dynamic Equilibrium

Many chemical reactions are reversible, meaning products can react with each other to reform the original reactants; this is denoted with a double arrow (⇌) instead of a single arrow. When a reversible reaction is allowed to proceed in a closed system, the forward reaction rate (reactants → products) gradually decreases while the reverse reaction rate (products → reactants) increases, until the two rates become equal — this state is dynamic equilibrium. At equilibrium, concentrations of reactants and products remain constant over time, but the forward and reverse reactions continue to occur at equal, non-zero rates.

• •Double arrow (⇌) indicates a reversible reaction capable of reaching equilibrium.
• •Dynamic equilibrium: forward rate = reverse rate; concentrations remain constant but are not necessarily equal.
• •Equilibrium is a dynamic, not static, condition — reactions have not stopped, they simply have equal opposing rates.

The Equilibrium Constant, K

For a general reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant expression K equals the concentrations of products raised to their coefficients divided by the concentrations of reactants raised to their coefficients, all measured at equilibrium. Pure solids and pure liquids are omitted from the expression because their 'concentration' (density-based activity) does not change during the reaction; only gases and aqueous species appear. The magnitude of K at a given temperature reveals the extent to which a reaction proceeds toward products before reaching equilibrium.

• •K = [C]^c[D]^d / [A]^a[B]^b, using equilibrium concentrations (or partial pressures) only.
• •Pure solids and pure liquids are excluded from the K expression entirely.
• •K >> 1: equilibrium lies far toward products (product-favored reaction).
• •K << 1: equilibrium lies far toward reactants (reactant-favored reaction).
• •K ≈ 1: significant amounts of both reactants and products are present at equilibrium.

Le Chatelier's Principle

Le Chatelier's principle states that if a system at equilibrium is disturbed by a change in concentration, pressure/volume, or temperature, the system shifts in the direction that partially counteracts the disturbance and re-establishes equilibrium. This is a conceptual, predictive tool rather than a calculation: it tells us the direction of the shift, not new numeric concentrations. Temperature changes are unique among these disturbances because they actually change the value of K itself, while concentration and pressure changes shift the position of equilibrium without changing K.

• •Adding reactant: shifts equilibrium toward products (forward direction) to consume the extra reactant.
• •Removing product: shifts equilibrium toward products to replace what was removed.
• •Adding product: shifts equilibrium toward reactants (reverse direction).
• •Removing reactant: shifts equilibrium toward reactants.
• •Adding heat to an endothermic reaction shifts equilibrium toward products (treat heat like a reactant); for an exothermic reaction, adding heat shifts equilibrium toward reactants (treat heat like a product).

Balance the equation: C₃H₈ + O₂ → CO₂ + H₂O.

1. 1Balance carbon first: 3 carbons in C₃H₈ require 3 CO₂.
2. 2Balance hydrogen next: 8 hydrogens in C₃H₈ require 4 H₂O (4 × 2 = 8).
3. 3Now count oxygens needed on the product side: 3 CO₂ (6 O) + 4 H₂O (4 O) = 10 O atoms total.
4. 4Since O₂ supplies oxygen in pairs, use 5 O₂ to supply the needed 10 oxygen atoms.
5. 5Write the final balanced equation and verify all atoms balance: 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

How many moles are in 45.0 g of CO₂ (molar mass 44.01 g/mol), and how many CO₂ molecules is that?

1. 1Convert grams to moles: mol = 45.0 g / 44.01 g/mol = 1.0225 mol.
2. 2Round to 3 sig figs: 1.02 mol CO₂.
3. 3Convert moles to molecules using Avogadro's number: 1.0225 mol × 6.022 × 10²³ molecules/mol.
4. 4Multiply: 1.0225 × 6.022 × 10²³ = 6.16 × 10²³ molecules.

1.02 mol CO₂, or 6.16 × 10²³ molecules

A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

1. 1Assume a 100 g sample, so the percentages become grams: 40.0 g C, 6.7 g H, 53.3 g O.
2. 2Convert each to moles: C: 40.0/12.01 = 3.33 mol; H: 6.7/1.008 = 6.65 mol; O: 53.3/16.00 = 3.33 mol.
3. 3Divide all values by the smallest (3.33): C = 1.00, H = 2.00, O = 1.00.
4. 4These are already whole numbers, giving the ratio C₁H₂O₁.

CH₂O (empirical formula)

For N₂ + 3H₂ → 2NH₃, if 4.00 mol N₂ reacts with 9.00 mol H₂, identify the limiting reactant and calculate mol NH₃ produced.

1. 1Determine mol NH₃ possible from N₂: 4.00 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 8.00 mol NH₃.
2. 2Determine mol NH₃ possible from H₂: 9.00 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 6.00 mol NH₃.
3. 3Compare the two results: H₂ produces less product (6.00 mol vs. 8.00 mol), so H₂ is the limiting reactant.
4. 4The actual amount of NH₃ produced is governed by the limiting reactant: 6.00 mol NH₃.
5. 5Confirm N₂ is in excess: only 3.00 mol N₂ (from 9.00 mol H₂ × 1/3) is actually consumed, leaving 1.00 mol N₂ unreacted.

H₂ is limiting; 6.00 mol NH₃ produced (1.00 mol N₂ left over)

A reaction has a theoretical yield of 25.8 g of product. The chemist actually collects 21.4 g. Calculate the percent yield.

1. 1Write the percent yield formula: % yield = (actual / theoretical) × 100%.
2. 2Substitute values: % yield = (21.4 g / 25.8 g) × 100%.
3. 3Divide: 21.4 ÷ 25.8 = 0.8295.
4. 4Multiply by 100%: 82.95%.
5. 5Round to 3 sig figs: 82.9%.

82.9% yield

Q1. Why must chemical equations be balanced?

Q2. How many atoms are present in 2.5 moles of iron?

Q3. What is the molar mass of H₂O (H = 1.008, O = 16.00 g/mol)?

Q4. A compound has an empirical formula of CH₂ and a molar mass of 42.08 g/mol. What is its molecular formula?

Q5. In the stoichiometry road map, what is the correct order of steps to convert grams of A to grams of B?

Q6. Which method can be used to identify the limiting reactant?

Q7. If a reaction's theoretical yield is 50.0 g and the percent yield is 88.0%, what is the actual yield?

Q8. Which of the following is NOT a common reason actual yield is less than theoretical yield?

Q9. A reversible reaction is best represented by which symbol?

Q10. At dynamic equilibrium, which statement is true?

Q11. In the equilibrium expression for aA + bB ⇌ cC + dD, which substances are excluded from K?

Q12. If K for a reaction is 4.5 × 10⁻⁸, the reaction is best described as:

Q13. According to Le Chatelier's principle, adding more reactant to a system at equilibrium will:

Q14. For an exothermic reaction at equilibrium, adding heat will shift the equilibrium:

Q15. For the equation N₂ + 3H₂ ⇌ 2NH₃, which is the correct equilibrium constant expression?