Unit 1: Back to the Basics

Section 6: Determining What Is Reacting

Introduction

Finding the Formula of a Compound

Chemical Equations

Balancing Equations

Determining How Much Is Produced

Limiting Reagents

Introduction

In Chemistry, the need to know what is reacting and how much is reacting is very important. This section deals with how to derive the formulas of the compounds in the reaction and how to balance chemical reactions. The next section deals with how much of each substance reacts in the reaction.

Finding the Molecular Formula of a Compound

Many times, the amount of a substance (whether it be in grams or moles) is given along with its name, but what happens if the name of the substance isn't given? To answer this, we must review (Section 5), because it contains many of the basic principles that will be applied in this section. Below is an example which will explain how the mole, molecular weight, and percent composition are intertwined. However, some vocabulary should be defined to you before reading the example. The first word is empirical formula, and it is defined as the simplest whole number ratio of the various types of atoms in a compound. The second word is molecular formula, and it is simply the formula of a molecule. Here is an example which might explain things better. Examples of empirical formulas are C0₂, H₂O, and CH₄. Examples of molecular formulas are C₂O₄, H₄O₂, and C₂H₈.

In an unknown molecule, there is 4.15 g carbon and 1.38 g hydrogen. Determine the empirical formula for the substance.

1. Step 1 - Divide mass of each substance by atomic mass to determine number of moles of each substance

4.15 g C/12.0 g C = .346 mol C

1.38 g H/1 g H = 1.38 mol H

2. Step 2 - Divide number of moles of each substance by smallest number of moles to determine what the ratio of carbon is to hydrogen

.346 mol C/.346 mol C = 1

1.38 mol H/.346 mol C = 4

The empirical formula would be CH₄.

The molecular weight of the above unknown substance is 32 g.

1. Step 1 - Find the molecular weight of CH₄.

1 mol C (12 g C) = 12 g C

4 mol H (1 g H) = 4 g H

Total mass of CH₄ = 12 g C + 4 g H = 16 g CH₄

32 g (unknown substance molecular weight) / 16 g (empirical formula) = 2

2 (CH₄) = C₂H₈

Chemical Equations

A chemical reaction is an equation that shows what happens in a reaction. All chemical reactions are composed of reactants (chemicals that are present before the reaction) and products (chemicals that are left after the reaction takes place.

Na⁺ + Cl^- ==> NaCl

In the chemical equation above, the Na⁺ and the Cl^- are the reactants. The NaCl is the product. The symbol between the Cl^- and the NaCl, (==>), is the yields sign. The arrow points towards the products and is used to show how the reaction takes place.

Balancing Equations

In a chemical reaction, matter cannot be created nor destroyed. For example in the reaction...

H₂ + Cl₂ ==> HCl

there are 2 atoms of hydrogen and 2 atoms of chlorine on the reactants side, but only one atom of hydrogen and chlorine on the products side. To fix this problem and to follow the rule that matter cannot be created nor destroyed, we must balance the equation.

By balancing the equation, you must have equal numbers of each of the different atoms on both the reactants and the products sides. To balance equations, there are a few rules that must be followed. First, locate the most complex compound and start balancing each of the different atoms (saving oxygen and hydrogen for last). For example,

NaCl + H₂ ==> HCl + Na

1. Start with NaCl. (This is the most complex compound because HCl has hydrogen in it and we save that for last.)

2. There is one atom of sodium on each side so move on to the chlorine.

3. There is one atom of chlorine on each side, so move on to the HCl.

4. There is one atom of hydrogen in HCl and two atoms of hydrogen on the reactants side, so put a 1/2 coefficient in front of H₂, so there is only one atom of hydrogen on each side.

NaCl + 1/2 H₂ ==> HCl + Na

5. There can't be fractions in the final answer so multiply all the coefficients by 2 as to eliminate the 1/2 coefficient.

2 NaCl + H₂ ==> 2 HCl + 2 Na

Determining How Much is Produced

Now that you know how to balance equations, you can predict how many moles, grams, etc will be produced in a reaction. The below example will show you how do accomplish this.

Example:

It is given that 92 grams of Na are produced by the reaction of NaCl + H₂ ==> HCl + Na; and they want to know how many grams of HCl are produced.

1. 1st: Balance the equations

2 NaCl + H₂ ==> 2 HCl + 2 Na

2. 2nd: Divide grams by the molar mass of the element to determine the number of moles produced

92 grams / 23 grams Na = 4 mol Na

3. 3rd: Using the balanced equation, determine the number of moles of product that you are looking for.

4 mol Na produced / 2 moles Na in equation = x moles HCl produced / 2 moles HCl in equation

4. 4th: Multiply the number of moles of product that you are looking for times the molar mass of the substance

4 moles HCl * 36.5 grams = 146.0 grams HCl produced

Limiting Reagents

Limiting reagents are the substance that limits the full reaction of all the substances. For example if some one handed you 12 pen caps and 8 pens and asked you to make as many complete pen and cap combinations, how many could be made? Obviously, you can only make eight complete pens and there would be four caps left over so the eight main pen pieces were the limiting reagents in this reaction because there weren't enough pens to allow all the caps to be used up.

For example, if the reaction:

Ba⁺² + F^- ==> BaF₂

1. 1st: Balance the reaction.

Ba⁺² + 2F^- ==> BaF₂

2. 2nd: Divide the number of moles of substance given by the number of moles needed in the reaction.

4 mol Barium given / 1 mol Barium in equation = 4

6 mol Fluorine given / 2 mol Fluorine in equation = 3

3. 3rd: Choose the smallest number and this is the limiting reagent.

4. 4th: Multiply the smallest number by the coefficient of the substance that you are looking for:

3 * 1 (coefficient of BaF2) = 3 mol BaF₂ produced