Untitled - Kelly Walsh High School

32 CHEMISTRY FOR THE UTTERLY CONFUSED
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This is a critical chapter in your study of chemistry. Our goal is to help you master
the mole concept. You will learn about balancing equations and the
mole/mass relationships (stoichiometry) inherent in these balanced equations.
You will learn, given amounts of reactants, how to determine which one limits
the amount of product formed. You will also learn how to determine the empirical
and molecular formulas of compounds. All of these will depend on the mole
concept. Make sure that you can use your calculator correctly. If you are unsure
about setting up problems, refer back to Chapter 1 of this book and go through
Section 1-4, on using the Unit Conversion Method. Review how to find atomic
masses on the periodic table. Practice, Practice, Practice.
3-1 Balancing Chemical Equations
A chemical equation represents a chemical change that is taking place. On the
left side of the reaction arrow are the reactants, the chemical substances that
are changed. On the right of the reaction arrow are the reaction products, the
new substances formed. Sometimes additional information appears above or
below the reaction arrow.
The Law of Conservation of Mass states that the total mass remains unchanged.
This means that the total mass of the atoms of each element represented in the
reactants must appear as products. In order to indicate this, we must balance
the reaction. When balancing chemical equations, it is important to realize that
you cannot change the formulas of the reactants and products; the only things
you may change are the coefficients in front of the reactants and products. The
coefficients indicate how many of each chemical species react or form. A balanced
equation has the same number of each type of atom present on both sides
of the equation and the coefficients are present in the lowest whole number
ratio. For example, iron metal reacts with oxygen gas to form rust, iron(III)
oxide. We may represent this reaction by the following balanced equation:
4 Fe(s) 3 O 2(g) l 2 Fe 2O 3(s)
This equation tells us that 4 iron atoms react with 3 oxygen molecules to form
2 rust compounds. Note that the number of iron atoms and oxygen atoms are
the same on both sides of the equation (2 Fe 2O 3 4 Fe and 6 O). Let’s see
how we went about arriving at the balanced equation for the rusting of iron.