AP CHEMISTRY
Chapter 3 – Stoichiometry: Quantitative Chemical Relationships
Objectives:
Know and define Avogadro’s number
Be able to convert from grams of a substance to moles of that substance and vice versa.
Be able to determine percent of any element within a compound.
Be able to calculate an empirical formula for any compound, given mass of one element and
mass of the compound.
Balance chemical equations given the reactants and products.
Use chemical equations to convert from moles of one species in a reaction to any other species.
Understand the concept of a limiting reagent.
Know what a theoretical yield is.
Determine the percent yield of a reaction given an actual yield of one product.
3.1– The Mole Concept
A. Introduction
1. stoichiometry – branch of chemistry concerned with the ratios by atoms of the
elements in compounds and with the ratios by atoms, molecules, or formula units
of the substances in chemical reactions
Greek stoicheion – element and metron – measure
a. ratios of atoms in chemical formulas are expressed in whole numbers; cannot
have a fraction of an atom
b. formula for water, H2O, shows combining ratio of 2 atoms H : 1 atom O; to
make one molecule of water; same ratio applies for any quantity
2. stoichiometric law – every pure substance has a ratio of elements given by its
formula
B. Avogadro’s Number and the Mole
1. mole (mol) – number of formula units as there are atoms in exactly 12 g of
carbon-12; SI unit for measuring the amount of substance
2. Avogadro’s Number – number of particles or formula units in one mole of a
substance: 6.0221367  1023 particles; atoms, ions, molecules, formula units
a. usually rounded to three or four significant digits: 6.02 or 6.022
b. equal number of moles contain equal number of fundamental particles;
atoms, ions, molecules, or formula units
c. moles of atoms always combine in the same ratio as the individual atoms
C. Stoichiometric Equivalences
1. stoichiometric equivalence – ratio by moles between two elements in a formula or
two substances in a chemical reaction; stoichiometrically equivalent to 
a. compound: P4O10
(1) 1 mol P4O10  4 mol P  10 mol O
(2) make conversion factors with equivalences
b. reaction: 2 H2 + O2  2 H2O: 2 mol H2  1 mol O2  2 mol H2O
3.2 – Measuring Moles of Elements and Compounds
A. Carbon – 12
1. 1 mol of Carbon-12 has a mass of exactly 12 g
a. average atomic mass of carbon is 12.011 u; mixture of isotopes
AP Chemistry 3.2,3,4
b. mass of 1 mol of average atoms is 12.011 g
2. One mole of any element has a mass in grams that is numerically equal to the atomic
mass of that element in u
B. Rounding Atomic Masses
1. periodic table has masses to several places beyond the decimal point
2. round atomic mass to have enough significant digits to accommodate the number of
significant digits of the other data in the problem
3. Using Avogadro’s number in calculations
a. rarely used in actual laboratory calculations
b. relate a macroscopic lab-size property like mass to number of particles
C. Formula Masses and Molecular Masses
1. formula mass – sum of the atomic masses (in u) of all of the atoms represented in
the chemical formula of an ionic compound
2. molecular mass – sum of the atomic masses of all of the atoms present in one
molecule of a molecular compound
D. Applications of Stoichiometry
1. relate mass of one substance to the mass of another in a compound
2. convert each mass to moles and use mole equivalent to compare moles
3.3– Percentage Composition
A. Discovery or Synthesis of New Substance
1. qualitative analysis – use of experimental procedures to determine what elements
are present in a substance
2. quantitative analysis – use of experimental procedures to determine the percentage
composition of a compound or the percentage of a component of a mixture
3. percentage composition – list of the percentages by mass of the elements in a
compound
4. percentage by mass – number of grams of an element combined in 100 g of a
compound
B. Theoretical Percentage Composition
1. calculated from the formula of the compound
2. calculate formula mass and compare the contribution by each element
3.4 – Empirical and Molecular Formulas
A. Introduction
1. molecular formula – chemical formula that gives the actual composition of one
molecule; P4O 10
2. empirical formula – chemical formula that uses the smallest whole-number
subscripts to give the proportions by atoms of the different elements; P2O5
B. Empirical Formulas from Percentage Compositions
1. difficult to use the same sample of a compound to obtain masses of all the elements
2. results are then converted to percentage compositions
C. Indirect Analysis
1. compound is seldom broken down completely to its elements in a quantitative
analysis; instead it is changed into other compounds
2. reactions separate the elements by capturing each one entirely in a separate
compound in which the formula is known
AP Chemistry 3.4,5,6,8
3. complete combustion of methyl alcohol in oxygen according to following reaction
2CH3OH + O2  2CO2 + 4H2O
a. carbon dioxide and water can be separated and masses determined
b. all carbon atoms in original compound result as CO2 and all hydrogen as H2O
(1) from masses of CO2 and H2O and mass of original sample,
percentages of C and H in original compound can be calculated
(2) percentage of O is simply the difference between 100 % and sum of
C and H
D. Molecular Formulas from Empirical Formulas and Molecular Masses
1. if experimental molecular mass equals the calculated empirical formula mass, then
empirical formula is the molecular formula
2. if not, the experimental molecular mass will be some whole number multiple of the
value calculated from the empirical formula; whole number is common
multiplier for subscripts of the empirical formula
3.5 – Writing and Balancing Chemical Equations
A. Chemical Equations
1. chemical equation – a before-and-after description that uses formulas and
coefficients to represent a chemical reaction
2. balanced equation – chemical equation that has on opposite sides of the arrow the
same number of each atom
3. coefficient – number in front of formulas in chemical formulas
B. Balancing Equations; two-step process
1. Step 1: Write the unbalanced equation; correct formulas are essential
2. Step 2: Adjust the coefficients to get equal numbers of each kind of atom on each
sides of the arrow; make no changes to the formulas; symbols or subscripts
3.6 – Using Chemical Equations in Calculations
A. Using Mole Ratios
1. coefficients in a balanced equation provide the ratios by moles of the reactants and
products;
3NaOH + H3PO4  Na3PO4 + 3H2O
2. stoichiometric equivalencies result from the coefficients; make conversion factors
B. Stoichiometric Mass Calculations
1. calculate mass in grams of reactants and products
2. Fig. 3.3; p. 109, mass-mass flow chart
C. Limiting Reactant Calculations
1. limiting reactant – reactant that is completely consumed in a chemical reaction
when nonstoichiometric amounts of reactants are used
2. must always base the calculation of the maximum yield of product on the
stoichiometric equivalency between it and the limiting reactant
3.8 – Theoretical Yield and Percentage Yield
A. Chemical Experiments Designed for Chemical Synthesis
1. amount of product actually isolated falls short of the calculated maximum amount
2. losses of product occur for several reasons
(1) mechanical; materials sticking to the sides of glassware
AP Chemistry 3.8
(2) volatile – description of a liquid that has a low boiling point, a high vapor
pressure at room temperature, and therefore evaporates easily
(3) precipitate that forms that is not totally insoluble
B. Competing Reaction
1. competing reaction – reaction that reduces the yield of the main product by forming
by-products
2. by-product – substances formed by side reactions
3. main reaction – desired reaction between the reactants as opposed to competing
reactions that give by-products
Main reaction: 2P + 3Cl2  2PCl3
Competing reaction: 2PCl3 + 3Cl2  PCl5
4. actual yield – amount of desired product isolated from a reaction
5. theoretical yield – maximum amount of desired product that could be isolated from a
reaction if no losses occurred
6. percentage yield – ratio of the mass of product obtained to the mass calculated from
the reaction’s stoichiometry
Percentage yield =
actual yield  100 %
theoretical yield