Lesson 1: A Framework for Thinking Stoichiometrically

Part a: Recipes, Ratios, and Relationships

Bake a Cake

Let’s suppose you have plans to bake a cake for an upcoming event. How would you go about such a task? Would you simply toss several ingredients into a baking pan and put it in the oven? Or would you look for a recipe that called for specified amounts of certain ingredients to be measured and mixed before baking? Cooking, like Chemistry, occurs best when the ingredients are mixed together in specified amounts.
 
Here is a typical cake recipe, expressed as a cooking equation.

 
The recipe includes information about the ingredients that are needed and the relative amounts of each ingredient. Some recipes may even provide additional information about the state of the ingredients. For instance, if walnuts were included, it might specify whether the walnuts were “whole” or “halved” or “crushed”. If bananas were included as an ingredient, it might specify “ripe” or “mashed”. These extra specifications might be described as the state or status of the ingredient. The final bit of information has to do with the amount of cake produced. The recipe might state that the cake will be large enough to provide 12 servings.
 
Suppose the cook on a large ship wishes to use the above recipe to bake 20 cakes for the guests and crew. This is quite simple to do and doesn’t require a new recipe. It simply requires a scaling factor. To make 20 cakes, the cook would need to scale the recipe by a factor of 20. That is to say, the amount of each ingredient should be 20 times more than the amount listed in the recipe. That is what is meant by relative amounts. The cook would need 40 eggs, 20 cups of sugar, 10 cups of milk, etc. Recipes allow a cook to make how much? decisions. And those decisions are critical to the success of the recipe.
 
 
 

Chemical Equations are Like Recipes

A chemical equation is very much like a recipe for a chemical reaction. It lists the ingredients (formulae), the relative amounts (coefficients), and the state of the ingredients (state symbols). It also identifies the product(s) and product amounts. Chemical equations allow a Chemistry student to make how much? decisions.
 
Recipes indicate the amount of required ingredients using units such as cups, tablespoons, teaspoons, etc. Chemical equations list amounts using coefficients. Like recipes, the coefficients indicate the relative amounts of reactants and products in terms of particles. A particle could mean an atom (of an element), a molecule (of a molecular compound), or a formula unit (of an ionic compound).
 
Consider this reaction for the synthesis of ammonia (NH₃).

N₂(g)  +  3 H₂(g)   →  2 NH₃(g)

The reaction can be read as 1 molecule of nitrogen gas reacts with 3 molecules of hydrogen gas to produce 2 molecules of ammonia gas. The 1, 3, and 2 are the coefficients and indicate the ratios at which the reactants and products are involved in the reaction. The N₂-H₂ particle ratio is 1:3. The N₂-NH₃ particle ratio is 1:2. The H₂-NH₃ particle ratio is 3:2. These ratios help Chemistry students to make how much? decisions.
 
Like the cook on the ship, we could supersize this recipe by any whole number. If we wanted 20 molecules of ammonia, then we would simply use a scaling factor of 10:

10 N₂(g)  +  30 H₂(g)  →  20 NH₃(g)

Molecules are quite small and 20 molecules is hardly even measurable so perhaps we would prefer to scale by one million:

1 x10⁶ N₂(g)  +  3 x10⁶ H₂(g)  →  2 x10⁶ NH₃(g)

Wait! Molecules are really, really, really small. We need a scaling factor much bigger than one million. Can you think of a suitable scaling factor for particles that would scale the number of particles so that it becomes measurable? Read on.
 

The Ideal Superizing Factor

Since coefficients indicate relative amounts, they can be scaled by whatever factor is needed … as long as each coefficient is scaled by the same factor. The most convenient factor by which to scale the coefficients is Avogadro’s number – 6.022x10²³. It may seem quite strange to call this a convenient factor given both its enormity and the clumsiness of entering it on a calculator. But we are going to stick with our claim and here's our reasoning why:
 
Scaling the ammonia synthesis equation by Avogadro’s number results in

6.022 x 10²³ N₂(g)  +  18.066 x 10²³ H₂(g)  →  12.044 x 10²³ NH₃(g)

 
The unit associated with these numbers is molecules of N₂, H₂, and NH₃. As we learned in Chapter 7, Avogadro’s number tell us the number of things in 1 mole of things. The mole is the counting unit of Chemistry. So, the 6.022 x 10²³ molecules of N₂ is equivalent to 1 mole of N₂. The 18.066 x 10²³ molecules of H₂ is equivalent to 3 mole of H₂. And the 12.044 x 10²³ molecules of NH₃ is equivalent to 2 mole of NH₃. The above equation can be read to say

1 mole of N₂ gas reacts with 3 mole of H₂ gas to produce 2 mole of NH₃ gas.

 
 
In equation form, this statement can be written as

N₂(g)  +  3 H₂(g)  →  2 NH₃(g)

where the unit for the coefficients 1, 3, and 2 is the mole. Now that’s convenient!

Using the supersizing factor of Avogadro’s number, the coefficients of any chemical equation indicate the relative number of moles of reactants and products involved in the reaction. The particle ratios are also mole ratios.
 
 
 

Mole Ratios and Mass Ratios

We learned in Chapter 7 that the atomic mass values of the periodic table indicate the mass of an element on a per mole basis. We refer to it as the molar mass. So, 1 mole of hydrogen atoms (H) has a mass of approximately 1.0 grams (more precisely, 1.0079 grams). And 1 mole of diatomic hydrogen (H₂) has a mass of approximately 2.0 grams. Using similar logic, we can assert that 1 mole of diatomic nitrogen (N₂) has a mass of approximately 28 grams. And 1 mole of ammonia (NH₃) has a mass of approximately 17 grams.

By supersizing the coefficients from molecules to moles, we are one short step away from relating the number of reactants and products in terms of grams. The molar masses of H₂, N₂, and NH₃ can be used to transition from moles to grams. The table below shows these relationships.

 
We are now equipped with three types sets of relationship for making make how much? decisions. The coefficients 1, 3, and 2 allow us to relate the relative number of particles involved in the reaction. They also allow us to relate the relative number of moles involved in the reaction. And finally, by combining the coefficients with molar mass information, we can relate the relative mass (in grams) that are involved in the reaction.
 

The Meaning of Stoichiometry

The title of this unit is Stoichiometry. It is a fancy word that is based on two Greek words joined together: Stoicheion + metron. The word stoicheion means element. And the word metron means to measure. Stoichiometry is concerned with measurements of elements involved in chemical reactions. Stoichiometry is the study of the quantitative relationship between the amounts of reactants and products involved in a chemical reaction. Stoichiometry focuses on the how much? questions of Chemistry. The Stoichiometry chapter will be one of the more heavily mathematical chapters in our Chemistry Tutorial.
 
<Pep Talk>
Chemistry students do best with this topic when they make sense of the logic behind the math and learn the skills for doing the math. In Lesson 1, we are making a strong effort to get you thinking about the relationships between quantities. We want you to understand coefficients, scaling factors, the concept of relative ratios of reactants and products, and mole-to-gram relationships. It’s indeed mathematical. But it is also sensible. Keep your sense-making ability active throughout the unit.
 
In Lesson 1c, we will review an important tool introduced in Chapter 1 of our Chemistry Tutorial – the factor label method. The factor label method (or as many students of Chemistry call it: conversion factors) is the skill that we will use for doing the math. It is a method that works for easy problems and (even more charmingly) for hard problems. It is a method that works in Chapter 1, in Chapter 7, in Chapter 9, and in every other chapter of our Tutorial. The sooner you become comfortable with it, the more you can take advantage of its usefulness. Do everything you can to learn this skill. We’ve made it to the end of this course enough times to confidently advise you that it is a skill that isn’t going away. The moral of the story is to invest in learning how to use conversion factors early and the return on your investment will be much appreciated.
</Pep Talk>
 
 
 

Example 1

We offer two multi-part examples in the form of table completion questions. Example 1 includes a table that pertains to moles. Use the coefficients of the equation to determine the missing cells of the table. Tap the View Answer button to view all missing cells and the logic by which their values were determined.

View Answer
Answers are in table. Detailed explanations are below the table.

 

Explanations:

The coefficients are 1, 3, and 2 respectively. The units associated with the coefficients are moles. For each row, compare the given amounts to the coefficients and determine the factor by which the given amounts have been scaled relative to 1, 3, and 2. Then scale the coefficient for the missing amount by the same scaling factor. The scaling factor is listed in blue in each row next to the given amount.
 
Row A: 2 mol N₂ is two times the coefficient of 1. Similarly, 6 mole of H₂ is two times the coefficieint of 3. And so the scaling factor is two. For the NH₃ amount, take the coefficient of 2 and multiply by 2. You get 4 mol NH₃.
 
Row B: 5 mol N₂ is five times the coefficient of 1. Similarly. 10 mole of NH₃ is five times the coefficieint of 2. And so the scaling factor is five. For the H₂ amount, take the coefficient of 3 and multiply by 5. You get 15 mol H₂.
 
Row C: 60 mol NH₃ is 30 times the coefficient of 2. And so the scaling factor is 30. For the N₂ amount, take the coefficient of 1 and multiply by 30. You get 30 mol N₂. For the H₂ amount, take the coefficient of 3 and multiply by 30. You get 90 mol H₂.
 
Row D: 9 mol N₂ is nine times the coefficient of 1. And so the scaling factor is nine. For the H₂ amount, take the coefficient of 3 and multiply by 9. You get 27 mol H₂. For the NH₃ amount, take the coefficient of 2 and multiply by 9. You get 18 mol NH₃.
 
Row E: 135 mol H₂ is 45 times the coefficient of 3. And so the scaling factor is 45. For the N₂ amount, take the coefficient of 1 and multiply by 45. You get 45 mol N₂. For the NH₃ amount, take the coefficient of 2 and multiply by 45. You get 90 mol NH₃.

 

Example 2

Example 2 includes a table that pertains to both moles and grams. Use the coefficients of the equation and molar mass values to determine the missing cells of the table. Tap the View Answer button to view all missing cells and the logic by which their values were determined.

View Answer
Answers are in table. Detailed explanations are below the table.

Explanations:

Example 2 reasoning is a bit more sophisticated than Example 1 reasoning. It also holds greater potential for error. The coefficients are 1, 3, and 2 respectively. The units associated with the coefficients are moles. So the mole amounts in each row can be related to each other by the coefficients.
 
For each row, compare the given or calculated amount of moles to the coefficients and determine the factor by which the mole amounts have been scaled relative to 1, 3, and 2. Then scale the coefficient for the missing amount of moles by the same scaling factor. The scaling factor is listed in blue in each row next to the given amount. The mass and the moles for any given substances is related to each other by the molar mass values. Those values are approximately 28 g/mol for N₂, 2 g/mol for H₂, and 17 g/mol for NH₃. DO NOT USE COEFFICIENTS TO RELATE THE MASS AMOUNTS.
 
Row A: 2 mol N₂ is two times the coefficient of 1. Similarly, 6 mole of H₂ is two times the coefficieint of 3. And so the scaling factor is two. For the NH₃ amount, take the coefficient of 2 and multiply by 2. You get 4 mol NH₃. To determine the grams of NH₃, multiply the 4 mol by the 17 g/mol to get 68 g of NH₃.
 
Row B: 12 mol H₂ is four times the coefficient of 3. And so the scaling factor is four. Use this factor to determine the missing mole amounts. For the N₂ mole amount, take the coefficient of 1 and multiply by 4. You get 4 mol N₂. For the NH₃ mole amount, take the coefficient of 2 and multiply by 4. You get 8 mol NH₃. To determine the grams of N₂, multiply the 4 mol by the 28 g/mol to get 112 g of N₂.To determine the grams of NH₃, multiply the 8 mol by the 17 g/mol to get 136 g of NH₃.
 
Row C: The only given is the 340 g of NH₃. You can divide by 17g/mol to determine the moles of NH₃. You get 20 moles of NH₃. You can use this to get the scaling factor for moles. The 20 moles of NH₃ is 10 times its coefficient of 2. So the scaling factor is 10.
 
For the N₂ mole amount, take the coefficient of 1 and multiply by 10. You get 10 mol N₂. For the H₂ mole amount, take the coefficient of 3 and multiply by 10. You get 30 mol H₂. You can determine the mass of N₂ by multiplying the 10 mol N₂ by 28 g/mol. You get 280 g of N₂. And you can determine the mass of H₂ by multiplying the 30 mole H₂ by 2 g/mol. You get 60 g of H₂.
 
Row D: The only given is the 42 g of N₂. You can divide by 28 g/mol to determine the moles of N₂. You get 1.5 moles of N₂. You can use this to get the scaling factor for moles. The 1.5 moles of N₂ is 1.5 times its coefficient of 2. So the scaling factor is 1.5.
 
For the H₂ mole amount, take the coefficient of 3 and multiply by 1.5. You get 4.5 mol H₂. For the NH₃ mole amount, take the coefficient of 2 and multiply by 1.5. You get 3 mol NH₃. You can determine the mass of H₂ by multiplying the 4.5 mol H₂ by 2 g/mol. You get 9 g of H₂. And you can determine the mass of NH₃ by multiplying the 3 mole NH₃ by 17 g/mol. You get 51 g of NH₃.
 
Row E: 18 mol H₂ is six times the coefficient of 3. And so the scaling factor is six. Use this factor to determine the missing mole amounts. For the N₂ mole amount, take the coefficient of 1 and multiply by 6. You get 6 mol N₂. For the NH₃ mole amount, take the coefficient of 2 and multiply by 6. You get 12 mol NH₃. To determine the grams of N₂, multiply the 6 mol by the 28 g/mol to get 168 g of N₂.To determine the grams of NH₃, multiply the 12 mol by the 17 g/mol to get 204 g of NH₃.

 

Before You Leave

• Download our Study Card on Stoichiometry Ratios. Save it to a safe location and use it as a review tool. (Coming Soon.)
• Our Calculator Pad tool is a great place to practice any skill involving calculations. Try our Problem Set ST1, Introduction to Stoichiometry.
• Try our Stoichiometry: Relationships Concept Builder. Demonstrate your understanding of the relationships between moles and mass by completing a table similar to the above.
• The Check Your Understanding section below include questions with answers and explanations. It provides a great chance to self-assess your understanding.

Check Your Understanding

Use the following questions to assess your understanding. Tap the Check Answer buttons when ready.
 
1. For the following chemical equations, provide a verbal description in terms of particles (atoms, molecules, and formula units).
Example:  N₂(g)  +  3 H₂(g)  →  2 NH₃(g)
is described by the statement: 1 molecule of nitrogen gas reacts with 3 molecules of hydrogen gas to produce 2 molecules of ammonia gas.

1. 2 Al(s)   +   3 Br₂(l)   →   2 AlBr₃(s) 
   
   Check Answer
   Answer: 
   2 atoms of solid aluminum react with 3 molecules of liquid bromine to produce 2 formula units of aluminum bromide.
   (essential words in red)
   
    
   
    
2. 4 Na(s)   +   O₂(g)   →   2 Na₂O(s) 
   
   Check Answer
   Answer: 
   4 atoms of solid sodium react with 1 molecule of oxygen gas to produce 2 formula units of sodium oxide.
   (essential words in red)
   
    
   
    
3. 2 NaCl(aq) +   F₂(g)   →   Cl₂(g)   +   2 NaF(aq) 
   
   Check Answer
   Answer: 
   2 formula units of aqueous sodium chloride react with 1 molecule of fluorine gas to produce 1 molecule of chlorine gas and 2 formula units of aqueous sodium fluoride.
   (essential words in red)
   
    
   
    

1. For every 4 mol of Na that react, _____ mol of Na₂O will be produced. 
   
   Check Answer
   Answer: 2
   
    
   
    
2. For every 2 mol of Na that react, _____ mol of Na₂O will be produced. 
   
   Check Answer
   Answer: 1
   
    
   
    
3. For every 8 mol of Na that react, _____ mol of Na₂O will be produced. 
   
   Check Answer
   Answer: 4
   
    
   
    
4. For every 20 mol of Na that react, _____ mol of Na₂O will be produced. 
   
   Check Answer
   Answer: 10
   
    
   
    
5. For every 10 mol of Na that react, _____ mol of Na₂O will be produced. 
   
   Check Answer
   Answer: 5
   
    
   
    
6. For every 3 mol of Na that react, _____ mol of Na₂O will be produced. 
   
   Check Answer
   Answer: 1.5
   
    
   
    

1. Determine the molar mass of CH₄. 
   
   Check Answer
   Answer: ~16 g/mol 
   
   12 g/mol + 4•(1 g/mol)
   
    
   
    
2. Determine the molar mass of O₂. 
   
   Check Answer
   Answer: ~32 g/mol
   
   2•(16 g/mol)
   
    
   
    
3. Determine the molar mass of CO₂. 
   
   Check Answer
   Answer: ~44 g/mol
   
   12 g/mol + 2•(16 g/mol)
   
    
   
    
4. Determine the molar mass of H₂O. 
   
   Check Answer
   Answer: ~18 g/mol
   
   2•(1 g/mol) + 16 g/mol
   
    
   
    
5. Determine the mole ratio of CH₄:O₂. 
   
   Check Answer
   Answer: 1:2
   
   The CH₄:O₂ mole ratio is simply the ratio of coefficients in the balanced chemical equation. It indicates that for every 1 mole of CH₄ that react, 2 moles of O₂ will react.
   
    
   
    
6. Determine the mass ratio of CH₄:O₂. 
   
   Check Answer
   Answer: 16 g:64 g
   
   The CH₄:O₂ mole ratio is 1:2. It indicates that for every 1 mole of CH₄ that react, 2 moles of O₂ will react. The mass ratio accounts for the fact that 1 mole of CH₄ has a mass of 16 g and 2 moles of O₂ has a mass of 64 g. These numbers are determined by multiplying the moles by the molar mass.
   
    
   
    
7. Determine the mole ratio of CH₄:CO₂. 
   
   Check Answer
   Answer: 1:1
   
   The CH₄:CO₂ mole ratio is simply the ratio of coefficients in the balanced chemical equation. It indicates that for every 1 mole of CH₄ that react, 1 mole of CO₂ will be produced.
   
    
   
    
8. Determine the mass ratio of CH₄:CO₂. 
   
   Check Answer
   Answer: 16 g:44 g
   
   The CH₄:CO₂ mole ratio is 1:1. It indicates that for every 1 mole of CH₄ that react, 1 moles of CO₂ will be produced. The mass ratio accounts for the fact that 1 mole of CH₄ has a mass of 16 g and 1 mole of CO₂ has a mass of 44 g. These numbers are determined by multiplying the moles by the molar mass.
   
    
   
    
9. Determine the mole ratio of CH₄:H₂O. 
   
   Check Answer
   Answer: 1:2
   
   The CH₄:H₂O mole ratio is simply the ratio of coefficients in the balanced chemical equation. It indicates that for every 1 mole of CH₄ that react, 2 moles of H₂O will be produced.
   
    
   
    
10. Determine the mass ratio of CH₄:H₂O. 
    
    Check Answer
    Answer: 16 g:36 g
    
    The CH₄:H₂O mole ratio is 1:2. It indicates that for every 1 mole of CH₄ that react, 2 moles of H₂O will be produced. The mass ratio accounts for the fact that 1 mole of CH₄ has a mass of 16 g and 2 moles of H₂O has a mass of 36 g. These numbers are determined by multiplying the moles by the molar mass.
    
     
    
     

 
 
4. Use the mole ratios to complete the blanks in the following table.

Check Answer
Answer: Explanations are below table.

Explanations:
The coefficients are 1, 2, 1, and 2 respectively. The units associated with the coefficients are moles. For each row, compare the given amounts to the coefficients and determine the scale factor by which the given amounts have been scaled relative to 1, 2, 1, and 2. This scaling factor is displayed in blue. Then scale the coefficient for the missing amount by the same scaling factor. 

 

Check Answer
Answer: Explanations are below the table.

 

Explanations:

Question 5 reasoning is a bit more sophisticated than Question 4 reasoning. It also holds greater potential for error. The coefficients are 1, 2, 1, and 2 respectively. The units associated with the coefficients are moles. So the mole amounts in each row can be related to each other by the coefficients. Determine the scaling factor first (in blue on table). Then for any missing number of moles, multiply the coefficient by the scaling factor for that row.
 
The mass and the moles for any given substances is related to each other by the molar mass values. Those values were calculated in Question #3 and are approximately 16 g/mol, 32 g/mol, 44 g/mol, and 18 g/mol. They can be used to convert from the number of moles to the mass in grams or vice versa. But DO NOT USE COEFFICIENTS TO RELATE THE MASS AMOUNTS.