MOLE-TO-MOLE CALCULATIONS

10.1. MOLE-TO-MOLE CALCULATIONS

In chemical work, it is important to be able to calculate how much raw material is needed to prepare a certain quantity of products. It is also useful to know if a certain reaction method can prepare more product from a given quantity of material than another reaction method. Analyzing material means finding out how much of each element is present. To do the measurements, often we convert parts of the material to compounds that are easy to separate, and then we measure those compounds. All these measurements involve chemical stoichiometry, the science of measuring how much of one thing can be produced from certain amounts of others.

From the practical viewpoint of a student, this chapter is extremely important. The calculations introduced here are also used in the chapters on gas laws, solution chemistry, equilibrium, and other topics. In Chap. 8, the balanced chemical equation was introduced. The equation expresses the ratios of numbers of formula units of each chemical involved in the reaction. Thus, for the reaction of aluminum with oxygen to produce aluminum oxide

4 Al + 3 O 2 −→ 2 Al 2 O 3

the equation states that the chemicals react in the ratio of four atoms of aluminum with three molecules of oxygen ( O 2 ) to produce two formula units of aluminum oxide. Thus, if eight atoms of aluminum react, they will react with six molecules of oxygen, and four formula units of aluminum will be produced. The balanced chemical equation may also be used to express the ratios of moles of reactants and products involved. Thus, for the reaction whose equation is given above, 4 mol of Al reacts with 3 mol of O 2 to produce

2 mol of Al 2 O 3 . It is also true that 8 mol of aluminum can react with 6 mol of oxygen to produce 4 mol of Al 2 O 3 , and so on.

EXAMPLE 10.1. How many moles of Al 2 O 3 can be prepared from the reaction of 0.450 mol of O 2 plus sufficient Al? Ans.

The first step in any stoichiometry problem is to write the balanced chemical equation:

4 Al + 3 O 2 −→

2 Al 2 O 3

Then the coefficients in the balanced chemical equation can be used as factors in the factor-label method to convert from moles of one chemical to moles of any other in the equation:

0.450 mol O

2 = 3 mol O 0.300 mol Al 2 O 3

In the problem given before Example 10.1 with 4 mol of aluminum, it was not necessary to use the factor-label method; the numbers were easy enough to work with. However, when the numbers get even slightly complicated, it is useful to use the factor-label method. Note that any of the following factors could be used for this equation, In the problem given before Example 10.1 with 4 mol of aluminum, it was not necessary to use the factor-label method; the numbers were easy enough to work with. However, when the numbers get even slightly complicated, it is useful to use the factor-label method. Note that any of the following factors could be used for this equation,

EXAMPLE 10.2. How many moles of oxygen does it take to react completely with 1.48 mol of aluminum? Ans.

According to the equation in Example 10.1, it takes

1.48 mol Al

1.11 mol O

4 mol Al

EXAMPLE 10.3. How many moles of NO 2 are produced by the reaction at high temperature of 1.50 mol of O 2 with sufficient N 2 ?

Ans. The balanced equation is

N 2 + 2O 2 −→

2 NO 2

1.50 mol O

1.50 mol NO

2 mol O

A simple figure linking the quantities, with the factor label as a bridge, is shown in Fig. 10-1.

Balanced chemical

Moles A

Moles B

equation

Fig. 10-1. The conversion of moles of one reagent to moles of another, using a ratio of the coefficients of the balanced chemical equation as a factor label

In all the problems given above, a sufficient or excess quantity of a second reactant was stated in the problem. If nothing is stated about the quantity of a second (or third, etc.) reactant, it must be assumed to be present in sufficient quantity to allow the reaction to take place. Otherwise, no calculation can be done.

EXAMPLE 10.4. How many moles of NaCl are produced by the reaction of 0.750 mol Cl 2 (with Na)? Ans.

We must assume that there is enough sodium present. As long as we have enough sodium, we can base the calculation on the quantity of chlorine stated.

2 Na + Cl 2 −→

2 NaCl

0.750 mol Cl 2 =

1.50 mol NaCl

1 mol Cl 2