MCAT General Chemistry Review

Chapter 4: Compounds and Stoichiometry


We began our consideration of compounds with benzaldehyde. As a compound, it is made from constituent atoms of different elements in a set ratio defined by its empirical or molecular formula. Each molecule of a compound has a defined mass that is measured as its molecular weight. The mass of one mole of any compound is determined from its molar mass in the units of grams per mole. We reviewed the basic classifications of reactions commonly tested on the MCAT: combination, decomposition, combustion, single-displacement, double-displacement, and neutralization reactions. Furthermore, we are now confident in our understanding of the steps necessary to balance any chemical reaction; we are ready to tackle more stoichiometric problems in preparation for Test Day.

Before moving to the next chapters discussing chemical kinetics and thermodynamics, let us offer our congratulations to you. By completing these first four chapters, you have been introduced to the fundamental concepts of chemistry—everything from the structure of the atom and trends of the elements to bonding and the formation of compounds. The understanding you have gained so far will be the foundation for your comprehension of even the most difficult General Chemistry concepts tested on the MCAT. Keep moving forward with your review of General Chemistry; don’t get stuck in the details. Those details will be learned best through the application of the basic principles to MCAT practice passages and questions.

Concept Summary

Molecules and Moles

·        Compounds are substances composed of two or more elements in a fixed composition.

·        Molecular weight is the mass (in amu) of the constituent atoms in a compound as indicated by the molecular formula.

·        Molar mass is the mass of one mole (Avogadro’s number or 6.022 × 1023 particles) of a compound; usually measured in grams per mol.

·        Gram equivalent weight is a measure of the mass of a substance that can donate one equivalent of the species of interest.

·        Normality is the ratio of equivalents per liter; it is related to molarity by multiplying the molarity by the number of equivalents present per mole of compound.

·        Equivalents are moles of the species of interest; equivalents are most often seen in acid–base chemistry (hydrogen ions or hydroxide ions) and oxidation–reduction reactions (moles of electrons or other ions).

Representation of Compounds

·        The law of constant composition states that any pure sample of a compound will contain the same elements in the same mass ratio.

·        The empirical formula is the smallest whole-number ratio of the elements in a compound.

·        The molecular formula is either the same as or a multiple of the empirical formula; it gives the exact number of atoms of each element in a compound.

·        To calculate percent composition by mass, determine the mass of the individual element and divide by the molar mass of the compound.

Types of Chemical Reactions

·        Combination reactions occur when two or more reactants combine to form one product.

·        Decomposition reactions occur when one reactant is chemically broken down into two or more products.

·        Combustion reactions occur when a fuel and an oxidant (typically oxygen) react, forming the products water and carbon dioxide (if the fuel is a hydrocarbon).

·        Displacement reactions occur when one or more atoms or ions of one compound are replaced with one or more atoms or ions of another compound.

o   Single-displacement reactions occur when an ion of one compound is replaced with another element.

o   Double-displacement reactions occur when elements from two different compounds trade places with each other to form two new compounds.

·        Neutralization reactions are those in which an acid reacts with a base to form a salt (and, usually, water).

Balancing Chemical Equations

·        Chemical equations must be balanced to perform stoichiometric calculations.

·        Balanced equations are determined using the following steps in order:

o   Balancing the least common atoms.

o   Balancing the more common atoms (usually hydrogen and oxygen).

o   Balancing charge, if necessary.

Applications of Stoichiometry

·        Balanced equations can be used to determine the limiting reagent, which is the reactant that will be consumed first in a chemical reaction.

·        The other reactants present are termed excess reagents.

·        Theoretical yield is the amount of product generated if all of the limiting reactant is consumed with no side reactions.

·        Actual yield is typically lower than theoretical yield.

·        Percent yield is calculated by dividing actual yield by theoretical yield and converting to a percentage.


·        Like organic chemistry, ions in general chemistry have a system of nomenclature:

o   Roman numerals are used for nonrepresentative elements to denote ionic charge.

o   –ous endings can also be used to indicate lesser charge, while –ic endings indicate greater charge.

o   All monatomic anions end in –ide.

o   Oxyanions are given a suffix indicating how oxidized the central atom is. Those that contain a lesser amount of oxygen are given the suffix –ite, and those with a greater amount are given the suffix –ate.

o   Oxyanion series with more than two members are given an additional level of nomenclature. The species with the fewest oxygens is given the prefix hypo–, and the species with the most oxygens is given the prefix per–.

o   Polyatomic ions containing hydrogen denote the number of hydrogens using hydrogen or bi– to denote one, or dihydrogen to denote two.

·        Ionic charges are predictable by group number and type of element (metal or nonmetal) for representative elements, but are generally unpredictable for nonrepresentative elements.

o   Metals form positively charged cations based on group number.

o   Nonmetals form negatively charged anions based on the number of electrons needed to achieve an octet.

·        Electrolytes contain equivalents of ions from molecules that dissociate in solution. The strength of an electrolyte depends on its degree of dissociation or solvation.

Answers to Concept Checks

·        4.1

1.    NaBr:  CuCl2 C6H12O6

2.    NaBr: 0.97 mol; CuCl2: 0.74 mol; C6H12O6: 0.56 mol

3.    Both values equal one mole of the given substance. The number of entities in a mole is always the same (Avogadro’s number, 6.022 × 1023 mol–1).


95 g PO43−  in 100 mL solution = 950 g PO43− per liter solution =

·        4.2

1.    Both molecular and empirical formulas contain the same elements in the same ratios. They differ in that molecular formulas give the actual number of atoms of each element in the compound; empirical formulas give only the ratio and therefore may or may not give the actual number of atoms.

2.    The molar mass of sodium carbonate is 106  The percent compositions are:

3.    Assuming a 100 g compound: 28.5 g Fe, 24.0 g S, and 49.7 g O. Using the atomic weight, there are about 0.5 mol Fe, 0.75 mol S, and 3 mol O. Divide by lowest number of moles to find the Fe:S:O ratio of 1:1.5:6. Multiply by 2 to find the smallest whole number ratio of Fe2S3O12.

·        4.3

1.    Ammonium cations swap places with (or displace) zinc cations yielding ammonium nitrate and zinc(II) sulfide. Zinc(II) sulfide then precipitates out of solution as a solid salt.





Reaction Type

2 H2 + O2

2 H2O


Al(OH)3 + H3PO4

3 H2O + AlPO4

Neutralization (a type of double-displacement)

2 H2O

2 H2 + O2


NaNO3 + CuOH

NaOH + CuNO3

Double-displacement (metathesis)

Zn + AgCl

ZnCl + Ag


·        4.4

·        2 Fe + 3 Cl2 → 2 FeCl3

·        Zn + 2 HCl → ZnCl2 + H2

·        C5H12 + 8 O2 → 5 CO2 + 6 H2O

·        3 Pb(NO3)2 + 2 AlCl3 → 3 PbCl2 + 2 Al(NO3)3

·        4.5

·        4 Na (s) + O2 (g) → 2 Na2O (s)


Because 4 sodium atoms are needed for every oxygen molecule, sodium will run out first. To determine the amount of Na2O formed:

·        The limiting reagent is Na because 4 sodium atoms are needed for every oxygen molecule.

will be used, so 0.94 – 0.54 mol O2 = 0.40 mol O2 will remain, representing 0.40 mol  excess O2

·        Reaction: BeO + H2O → Be(OH)2

Theoretical yield:

·        4.6

1.    Electrolytes: HCl, MgBr2; Nonelectrolytes: sucrose, CH4



Cation or Anion























Equations to Remember

(4.1) Moles from mass:

(4.2) Gram equivalent weight:

(4.3) Equivalents from mass:

(4.4) Molarity from normality:

(4.5) Percent composition:

(4.6) Percent yield:

Shared Concepts

·        General Chemistry Chapter 2

o   The Periodic Table

·        General Chemistry Chapter 3

o   Bonding and Chemical Interactions

·        General Chemistry Chapter 9

o   Solutions

·        General Chemistry Chapter 10

o   Acids and Bases

·        General Chemistry Chapter 11

o   Oxidation–Reduction Reactions

·        Physics and Math Chapter 5

o   Electrostatics and Magnetism