Premed · Premed · General Chemistry 1
Lecture 3: Stoichiometry and Chemical Equations
General Chemistry I
Learning Objectives
By the end of this lecture, students will be able to:
- Define the mole and use Avogadro's number to convert between atoms/molecules and moles
- Calculate molar mass from a chemical formula
- Interconvert among mass, moles, and number of particles
- Balance chemical equations by inspection
- Use stoichiometric coefficients to relate amounts of reactants and products
- Identify the limiting reagent in a reaction and calculate theoretical yield
- Calculate percent yield of a reaction
Lecture Content
I. The Mole Concept
The mole (mol) is the SI unit for the amount of substance, and one mole contains exactly 6.022 x 10^23 representative particles, a quantity known as Avogadro's number (N_A). What constitutes a "representative particle" depends on the substance in question: atoms for elements, molecules for molecular compounds, and formula units for ionic compounds. The mole serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams that can be measured on a balance. Just as a "dozen" always means 12, a "mole" always means 6.022 x 10^23.
II. Molar Mass
The molar mass (M) is the mass of one mole of a substance, expressed in g/mol. For elements, the molar mass is numerically equal to the atomic mass from the periodic table -- carbon is 12.01 g/mol, oxygen is 16.00 g/mol, and iron is 55.85 g/mol. For compounds, the molar mass is found by summing the molar masses of all atoms in the formula. Water (H2O) has a molar mass of 2(1.008) + 16.00 = 18.02 g/mol, and calcium hydroxide (Ca(OH)2) has a molar mass of 40.08 + 2(16.00 + 1.008) = 74.10 g/mol. Molar mass serves as the essential conversion factor between grams and moles.
III. Interconverting Mass, Moles, and Particles
The relationships among mass, moles, and particle count are straightforward. To convert mass to moles, divide by the molar mass (n = mass / M). To convert moles to mass, multiply by the molar mass (mass = n x M). To convert moles to number of particles, multiply by Avogadro's number (N = n x N_A), and to go the other direction, divide by Avogadro's number. This creates a clear roadmap: grams connect to moles, which connect to number of particles. You can extend the conversion further to count atoms within a molecule -- for example, converting moles of H2O to molecules of H2O and then multiplying by 2 to find the number of hydrogen atoms.
<image>A conversion roadmap diagram showing the relationships between mass (grams), moles, and number of particles. Three boxes are connected by double-headed arrows. Box 1: "Mass (g)" connected to Box 2: "Moles (mol)" via arrows labeled "divide by molar mass (g/mol)" going right and "multiply by molar mass" going left. Box 2: "Moles (mol)" connected to Box 3: "Number of particles" via arrows labeled "multiply by 6.022 x 10^23" going right and "divide by 6.022 x 10^23" going left. Below, a worked example: 36.04 g H2O -> 2.000 mol H2O -> 1.204 x 10^24 molecules H2O, with each step's conversion factor shown.</image>
IV. Percent Composition
Percent composition by mass gives the mass percentage of each element in a compound. It is calculated as: % element = (number of atoms of that element x atomic mass of that element) / molar mass of compound x 100%. For water, the hydrogen content is 2(1.008) / 18.02 x 100% = 11.19%, and the oxygen content is 16.00 / 18.02 x 100% = 88.81%. Percent composition is a useful tool for verifying compound purity or determining the identity of an unknown substance.
V. Empirical and Molecular Formulas from Experimental Data
The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms and is always a whole-number multiple of the empirical formula. To determine an empirical formula from percent composition data, assume a 100 g sample so that percentages translate directly to grams. Convert grams of each element to moles by dividing by the respective atomic mass. Then divide all mole values by the smallest mole value to obtain the ratio. If the resulting ratios are not whole numbers, multiply through by an appropriate factor (for example, multiply by 2 if a ratio of 0.5 appears, or by 3 for 0.33). The molecular formula can then be found by comparing the molar mass of the compound to the empirical formula mass: molecular formula = n x (empirical formula), where n = molar mass / empirical formula mass.
VI. Writing and Balancing Chemical Equations
A chemical equation represents a chemical reaction using chemical formulas and symbols. Reactants appear on the left side of the arrow and products on the right. Coefficients placed before the formulas indicate the mole ratios in which substances react and form, and state symbols denote the phase of each species: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solution.
The Law of Conservation of Mass requires that matter cannot be created or destroyed, so every balanced equation must have the same number of each type of atom on both sides. To balance an equation by inspection, begin by writing the correct formulas, then balance elements that appear in only one reactant and one product first. If polyatomic ions appear intact on both sides, balance them as a unit. Balance hydrogen and oxygen last, as they often appear in multiple compounds. Always use the smallest set of whole-number coefficients, and verify by counting atoms on each side. It is critical never to change subscripts to balance an equation -- only coefficients may be adjusted.
VII. Stoichiometric Calculations
The coefficients in a balanced equation provide mole ratios between all reactants and products, forming the basis of stoichiometric calculations. The general strategy is: first, write and balance the equation; second, convert the given quantity to moles (using molar mass if given in grams); third, use the appropriate mole ratio to convert to moles of the desired substance; and fourth, convert those moles to the requested units. For the reaction 2 H2 + O2 -> 2 H2O, the coefficients tell us that 2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O. If given 4.0 g of H2, dividing by the molar mass (2.016 g/mol) yields 1.98 mol H2, which produces 1.98 mol H2O, corresponding to 35.7 g of water.
<image>A step-by-step stoichiometry problem-solving flowchart. Step 1 (top): "Given quantity" in a box (e.g., grams of A). Arrow labeled "divide by molar mass of A" leads to Step 2: "Moles of A." Arrow labeled "multiply by mole ratio (coefficients from balanced equation: mol B / mol A)" leads to Step 3: "Moles of B." Arrow labeled "multiply by molar mass of B" leads to Step 4: "Desired quantity" (e.g., grams of B). A worked example runs parallel: 10.0 g CH4 -> 0.624 mol CH4 -> (x2 ratio) 1.248 mol H2O -> 22.5 g H2O, using the reaction CH4 + 2 O2 -> CO2 + 2 H2O.</image>
VIII. Limiting Reagent (Limiting Reactant)
In most real reactions, reactants are not present in exact stoichiometric amounts. The limiting reagent is the reactant that is completely consumed first, and it determines the maximum amount of product that can form. The excess reagent is whatever remains after the limiting reagent is used up. There are two common methods for identifying the limiting reagent. In the first, convert each reactant's amount to moles of a chosen product; the reactant that produces the least product is limiting. In the second, divide the moles of each reactant by its stoichiometric coefficient; the smallest quotient identifies the limiting reagent. Once the limiting reagent is identified, its moles are used to calculate the theoretical yield.
IX. Theoretical Yield, Actual Yield, and Percent Yield
The theoretical yield is the maximum amount of product that stoichiometry predicts based on the limiting reagent. The actual yield is the amount of product actually obtained from an experiment, and it is always less than or equal to the theoretical yield. Percent yield expresses the efficiency of the reaction: percent yield = (actual yield / theoretical yield) x 100%. Several factors explain why actual yields fall short of theoretical predictions, including incomplete reactions (those that reach equilibrium before going to completion), side reactions that produce unwanted byproducts, losses during transfer, purification, or isolation, and the use of impure reagents. A percent yield exceeding 100% is not physically possible for a properly conducted experiment and suggests measurement error or product contamination.
X. Combustion Analysis
Combustion analysis is a technique for determining the empirical formulas of organic compounds containing carbon, hydrogen, and sometimes oxygen or nitrogen. The sample is burned in excess O2, and the combustion products are captured: CO2 is absorbed by a NaOH trap (providing the mass of carbon), while H2O is absorbed by a Mg(ClO4)2 trap (providing the mass of hydrogen). If the compound contains oxygen, its mass is determined by difference: mass of O = mass of sample - mass of C - mass of H. The masses are then converted to moles, and the mole ratios yield the empirical formula.
