<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 12.1: Mole Ratios

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Relate balanced chemical equations to everyday analogies, such as a recipe.
• Define stoichiometry.
• Use mole ratios to convert between amounts of substances in a chemical reaction.

## Lesson Vocabulary

• mole ratio
• stoichiometry

### Recalling Prior Knowledge

• Why is it necessary to balance chemical equations?
• What do the coefficients of a balanced chemical equation represent?

Chemical equations are balanced in order to satisfy the law of conservation of mass. The mass of each element in a chemical reaction must be conserved. A balanced equation is required for calculations that involve the quantities of reactants used and products generated in a chemical reaction.

## Everyday Stoichiometry

In the last chapter, Chemical Reactions, you learned about chemical equations and the techniques used to balance them. Chemists use balanced equations to obtain quantitative information about chemical reactions. Before we look at a chemical reaction, let’s reconsider the equation for the ideal ham sandwich.

As a reminder, our ham sandwich is composed of 2 slices of ham (H), a slice of cheese (C), a slice of tomato (T), 5 pickles (P), and 2 slices of bread (B). The equation for our sandwich is shown below:

2H + C + T + 5P + 2B → H2CTP5B2

Now let us suppose that you are having some friends over and need to make five ham sandwiches. How much of each sandwich ingredient do you need? You would take the number of each ingredient required for one sandwich (its coefficient in the above equation) and multiply by five. Using ham and cheese as examples and using a conversion factor, we can write:

5 H2CTP5B2×2 H1 H2CTP5B2=10 H\begin{align*}5 \ \text{H}_2\text{CTP}_5\text{B}_2 \times \frac {2 \ \text{H}}{1 \ \text{H}_2\text{CTP}_5\text{B}_2} = 10 \ \text{H}\end{align*}
5 H2CTP5B2×1 C1 H2CTP5B2=5 C\begin{align*}5 \ \text{H}_2\text{CTP}_5\text{B}_2 \times \frac {1 \ \text{C}}{1 \ \text{H}_2\text{CTP}_5\text{B}_2} = 5 \ \text{C}\end{align*}

The conversion factors contain the coefficient of each specific ingredient as the numerator and the formula of one sandwich as the denominator. The result is what you would expect. In order to make five ham sandwiches, you would need 10 slices of ham and 5 slices of cheese.

This type of calculation demonstrates the use of stoichiometry. Stoichiometry is the calculation of amounts of substances in a chemical reaction from the balanced equation. The sample problem below is another stoichiometry problem involving ingredients of the ideal ham sandwich.

Sample Problem 12.1: Ham Sandwich Stoichiometry

Kim looks in the refrigerator and finds that she has 8 slices of ham. In order to make as many sandwiches as possible, how many pickles does she need? Use the equation from the text.

Step 1: List the known quantities and plan the problem.

Known

• have 8 ham slices (H)
• 2 H = 5 P (conversion factor)

Unknown

• How many pickles (P) needed?

The coefficients for the two reactants (ingredients) are used to make a conversion factor between ham slices and pickles.

Step 2: Solve.

8 H×5 P2 H=20 P\begin{align*}8 \ \text{H} \times \frac{5 \ \text{P}}{2 \ \text{H}} = 20 \ \text{P}\end{align*}

Since 5 pickles combine with 2 ham slices in each sandwich, 20 pickles are needed to fully combine with 8 ham slices.

The 8 ham slices will make 4 ham sandwiches. Since there are 5 pickles per sandwich, 20 pickles would be used in these 4 sandwiches.

Practice Problems
1. How many sandwiches can be made from 30 pickles, assuming that you have enough of the rest of the ingredients?
2. You have 7 slices of cheese. How much of each of the other ingredients is required to make the maximum number of sandwiches possible? How many sandwiches could be made?

Watch this short video for an introduction to the concept of stoichiometry: http://www.youtube.com/watch?v=xGBtz-ihRbA.

## Mole Ratios

Stoichiometry problems can be characterized by two things: (1) the information given in the problem, and (2) the information that is to be solved for, referred to as the unknown. The given and the unknown may both be reactants, both be products, or one may be a reactant while the other is a product. The amounts of the substances can be expressed in moles. However, in a laboratory situation, it is common to determine the amount of a substance by finding its mass in grams. The amount of a liquid or gaseous substance may also be expressed by its volume. In this lesson, we will focus on the type of problem where both the given and the unknown quantities are expressed in moles.

Given substance (mol) → Unknown substance (mol)

In the following lesson, “Stoichiometric Calculations,” we will expand our analysis of stoichiometry to include mass-based and volume-based problems.

Chemical equations express the relative amounts of reactants and products in a reaction. The coefficients of a balanced equation can represent either the number of molecules or the number of moles of each substance. The production of ammonia (NH3) from nitrogen and hydrogen gases is an important industrial reaction called the Haber process (Figure below), after German chemist Fritz Haber.

Nitrogen and hydrogen gases are combined under careful conditions to form ammonia in a reaction called the Haber process. Ammonia is a very important industrial chemical and is produced in very large quantities, mostly for fertilizers.

The balanced equation for the Haber process is:

N2(g) + 3H2(g) → 2NH3(g)

The reaction can be analyzed in several ways, as shown in Figure below.

This representation of the production of ammonia from nitrogen and hydrogen shows several ways to interpret the quantitative information given by a balanced chemical equation.

1 molecule of nitrogen reacts with 3 molecules of nitrogen to form 2 molecules of ammonia. These are the smallest possible relative amounts of the reactants and products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia.

As you have learned, the most useful quantity for counting very tiny particles is the mole. If each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation.

Notice that the number of moles of reactants is not conserved when they are converted to products. In the production of ammonia, 4 moles of reactant molecules are converted into 2 moles of product molecules. However, if each mole quantity is converted to grams by using the molar mass, we can see that the law of conservation of mass is followed. 1 mol of nitrogen has a mass of 28.02 g, 3 mol of hydrogen has a mass of 6.06 g, and 2 mol of ammonia has a mass of 34.08 g.

28.02 g N2 + 6.06 g H2 → 34.08 g NH3

Mass and the total number of each atom must be conserved in any chemical reaction. The number of molecules is not necessarily conserved.

A mole ratio is a conversion factor that relates the amounts in moles of any two substances in a chemical reaction. The numbers in a conversion factor come from the coefficients of the balanced chemical equation. The following six mole ratios can be written for the reaction above.

1 mol N23 mol H2\begin{align*}\dfrac{1 \ \text{mol N}_2}{3 \ \text{mol H}_2}\end{align*} or 3 mol H21 mol N2\begin{align*}\dfrac{3 \ \text{mol H}_2}{1 \ \text{mol N}_2}\end{align*}
1 mol N22 mol NH3\begin{align*}\dfrac{1 \ \text{mol N}_2}{2 \ \text{mol NH}_3}\end{align*} or 2 mol NH31 mol N2\begin{align*}\dfrac{2 \ \text{mol NH}_3}{1 \ \text{mol N}_2}\end{align*}
3 mol H22 mol NH3\begin{align*}\dfrac{3 \ \text{mol H}_2}{2 \ \text{mol NH}_3}\end{align*} or 2 mol NH33 mol H2\begin{align*}\dfrac{2 \ \text{mol NH}_3}{3 \ \text{mol H}_2}\end{align*}

In a mole ratio problem, the given amount, expressed in moles, is written first. The appropriate conversion factor is chosen in order to convert from moles of the given substance to moles of the unknown substance.

Sample Problem 12.2: Mole Ratio

How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?

Step 1: List the known quantities and plan the problem.

Known

• given: H2 = 4.20 mol

Unknown

• mol of NH3

The conversion is from mol H2 → mol NH3. The problem states that there is an excess of nitrogen, so we do not need to be concerned with any mole ratio involving N2. Choose the conversion factor that has moles of NH3 in the numerator and moles of H2 in the denominator so that we are left with the desired quantity after cancelling units.

Step 2: Solve

4.20 mol H2×2 mol NH33 mol H2=2.80 mol NH3\begin{align*}4.20 \ \text{mol H}_2 \times \frac{2 \ \text{mol NH}_3}{3 \ \text{mol H}_2} = 2.80 \ \text{mol NH}_3\end{align*}

The reaction of 4.20 mol of hydrogen with excess nitrogen produces 2.80 mol of ammonia.

The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation. Note that a mole ratio represents exact quantities, since the coefficients in a balanced equation are considered to have an infinite number of significant figures. The number of significant figures in the result is determined directly from the number of significant figures in the given quantity. In this case, both have three significant figures.

Practice Problems
1. How many moles of hydrogen are required to fully react with 0.26 moles of nitrogen?
2. A reaction produces 3.12 moles of ammonia. How many moles of nitrogen and hydrogen were reacted?

Watch an animation showing the mole ratio of hydrogen to oxygen when they react to form water at http://www.dlt.ncssm.edu/core/Chapter6-Stoichiometry/Chapter6-Animations/OneLiterH2O.html.

Practice mole ratios using both sandwiches and real chemical reactions using the simulation at http://phet.colorado.edu/en/simulation/reactants-products-and-leftovers. Several activities associated with this lesson are available by scrolling down.

## Lesson Summary

• A balanced chemical equation provides the same information as a recipe. Recipes can be manipulated to account for different amounts of various ingredients.
• Stoichiometry is the branch of chemistry that involves relationships between the amounts of substances in a balanced equation. The coefficients represent the moles of each reactant and product in the reaction.
• Mole ratios can be constructed between any two of the substances in a balanced equation. Moles of a given substance can be converted to moles of an unknown substance with the appropriate mole ratio.

## Lesson Review Questions

### Reviewing Concepts

1. The balanced equation for the combustion of ethene (C2H6) is given below:
1. Interpret the equation in terms of the number of molecules of each substance that are involved in the reaction.
2. Describe the reaction again in terms of moles of each substance.
3. Determine the mass of each reactant and product in the equation and demonstrate that the law of conservation of mass is obeyed.
2. For the reaction given in question 1, write the following mole ratios:
1. ethene to oxygen
2. carbon dioxide to water
3. ethene to carbon dioxide
3. Write all of the possible mole ratios for the equation below, in which calcium reacts with nitrogen to form calcium nitride:

### Problems

1. A certain automobile contains 4 tires, 2 headlights, 1 steering wheel, and 6 spark plugs. How many of each part will be required to build 17 of these cars?
2. The reaction of sodium with water produces sodium hydroxide and hydrogen gas:
1. How many moles of water are required to react with 0.89 moles of sodium?
2. The complete reaction of 1.92 moles of sodium will produce how many moles of hydrogen?
3. If 0.00482 moles of H2 were produced by this reaction, how many moles of NaOH were also produced?
3. Ammonia will react with nitrogen monoxide to produce nitrogen gas and water vapor.
1. How many moles of N2 will be produced by the complete reaction of 3.20 moles of NH3?
2. In order to completely react with 0.810 moles of NO, how many moles of NH3 are needed?
3. How many moles of NH3 were consumed in a reaction that produced 9.04 moles of H2O?
4. You have 2.28 moles of NH3. How many moles of NO are needed to react with this amount of ammonia? How many moles of each of the products will be formed?
4. The reaction of aluminum metal with an aqueous solution of silver nitrate produces aqueous aluminum nitrate and solid silver metal.
1. Write the balanced chemical equation for this reaction.
2. If 0.612 moles of aluminum are reacted, how many moles of silver will be produced?

## Points to Consider

In the laboratory, there is no way for a chemist to directly measure out moles of a substance. He or she instead must determine amounts either by mass or by volume. It is thus important to be able to perform stoichiometric calculations with mass and volume.

• How is mass converted to moles?
• How are moles converted to mass?
• How many conversion factors are required to convert from the mass of a given substance to the mass of an unknown substance?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: