Lesson 2: Relating Stoichiometric Quantities

Part a: Stoichiometry Plus

Part a: Mole-to-Mole Relationships
Part b: Mole-to-Mass Relationships
Part c: Mass-to-Mass Relationships
Part d: Percent Yield
Part e: Stoichiometry Plus

 

What is Stoichiometry Plus?

Lesson 2 of our Stoichiometry Chapter has included a collection of problems which target student ability to convert between stoichiometric quantities. The typical problem provides either a mass or a number of moles of a reactant or product. Conversion factors are used to determine the mass or number of moles of another reactant or product.
 
Now we will introduce a slight twist in the routine. In addition to the above, the problems on this page will require an additional task. We call these Stoichiometry Plus problems because a student is combining stoichiometry with an additional skill. As examples, the additional skill could be …

• Writing and balancing a chemical equation,
• Relating the mass to the density and volume,
• Converting between metric and non-metric units,
• Including cost information in order to calculate a dollar figure.

 
We will provide an example and corresponding solution of each of the above skills. Of course, success requires that you have some understanding of that skill. If you find yourself a bit rusty, then we advise following the link to the page devoted to the particular skill.
 
 
 

Stoichiometry Plus Balancing Chemical Equations

We discussed writing balanced chemical equations in Chapter 8. We also discussed the predicting of products from a knowledge of the various types of reactions in Chapter 8. This skill of predicting products and writing balanced chemical equations is often embedded in a stoichiometry problem. A stoichiometry problem cannot be solved with a balanced chemical equation. Thus, it is either provided in the problem statement or assumed that the student possesses the skill to write and balance such an equation.
 
Here is an example of a problem in which a balanced equation must first be written from information given in the problem. Use the links below if you need to first review the skills of predicting products and writing balanced chemical equations.
   
 
 

Example 1:

 
This problem is based on a type of reaction known as a combustion reaction. A combustion reaction occurs when a substance reacts with oxygen gas (O₂). In this case, that substance is a hydrocarbon – C₃H₈. The products of hydrocarbon combustion are carbon dioxide (CO₂) and water vapor. The skeleton equation for the reaction is:
 
Skeleton Equation:  C₃H₈(g)   +   O₂(g)   →   CO₂(g)   +   H₂O(g)
 
By inserting coefficients in front of the formulae, the individual elements can be balanced. The resulting balanced chemical equation is …
 
Balanced Equation:  C₃H₈(g)   +   5 O₂(g)   →   3 CO₂(g)   +   4 H₂O(g)
 
Now that the balanced chemical equation is written, it is stoichiometry business as usual. The given quantity is the mass of reactant – C₃H₈. The desired quantity is the mass of CO₂. This a three-step gram-to-gram stoichiometry conversion. The molar mass of C₃H₈, the coefficients, and the molar mass of CO₂ are used in the conversion factors. Molar mass values are determined using a periodic table.
   
The three-step conversion is set up below with units and without numbers. Units are arranged to cancel grams C₃H₈ and in stepwise fashion to end with grams CO₂. The three pairs of colored diagonals show how the units are cancelling.
   
Numbers – molar masses and coefficients - can be substituted into the conversion factors. The first and the third factors include molar mass information. The “1” always goes with mole and the molar mass value always goes with the grams. The middle conversion factor uses coefficients from the balanced chemical equation. (Detailed information about gram-to-gram stoichiometry is included in Lesson 2c.)
   
 
 

Stoichiometry Plus Density Relationships

Density was the topic of Lesson 4 of Chapter 2 of our Chemistry Tutorial. Density refers to the mass per volume ratio of an object. The amount of a liquid and a gas is most easily measured by determining its volume. Density relates the volume to the mass of a sample. For these reasons, density is a common add-on to a stoichiometry problem. Example 2 illustrates the incorporation of the density concept into a stoichiometry problem.  
 

Example 2:

 
The given quantity is the volume of C₄H₁₀. The volume is related to the mass by the density. The density can be used to convert to grams C₄H₁₀. Once done, a three-step conversion can be performed to determine the desired quantity – the grams of CO₂. That’s a total of four conversion factors. We will string them together in consecutive fashion. The molar mass of C₄H₁₀ and CO₂ will be needed for two of the conversion factors. The CO₂ molar mass was calculated in Example 1.
   
The set-up of the four conversion factors is shown below. Observe the use of colored diagonals to cancel units. A unit in the denominator will cancel an identical unit in a numerator (or on the given quantity). They are arranged to cancel each other. The unit that does not get cancelled is the unit on the answer.
   
If you are having difficulty with conversion factors, we recommend our Mass-to-Mass Stoichiometry page or our Factor Label Method page:
   
 
The numbers for density, molar mass (twice), and coefficients can be inserted into the conversion factor and the mass of CO₂ can be calculated.
   
 
 

Stoichiometry Plus Non-Metric Conversions

Managing units of measurement is a topic of constant attention in a Chemistry course. Part of unit management involves converting from one unit to another unit. While most all Chemistry courses use the metric system of measurements, not all countries do (or at least, two or three don’t). And so, there is a frequent need to convert between metric and non-metric units. Such conversions are often added to a stoichiometry problem. This means additional conversion factors are needed beyond the three that allow one to do a gram-to-gram conversion. Example 3 is a problem that fits into this category.
 
 
 

Example 3:

 
Example 3 has a lot of details to give attention to. These details include scientific notation, percent yield, a gram-to-gram conversion, a gram-to-kilogram conversion, and a kilogram to pound conversion. The solution strategy involves using the given quantity – grams of N₂ – to determine the theoretical amount of NH₃ produced in grams. This will be done in the manner demonstrated in Lesson 2c (and as done in Examples 1 and 2) by means of a three-step conversion. The percent yield value is used to calculate the actual yield of NH₃ in grams. Then a couple of conversions are required to convert grams to kilograms and then to pounds.
 
Molar masses of N₂ and NH₃ are needed for the three-step stoichiometry conversion.
   
The gram-to-gram conversion set up (units only, no numbers) is shown below. All units cancel except for the grams NH₃.
   
Numbers for the molar mass and coefficients are inserted into the conversion factors and the mass of NH₃ produced is calculated.
   
The above number is the theoretical yield. It is based on a stoichiometry calculation using the mass of reactant. The actual yield is 42.5% of this value. That is what percent yield means. The unrounded number is used to calculate the actual yield. Rounding is done once the last calculation is completed.
   
Now the grams are converted to kilograms and then to pounds using two conversion factors. The final answer is rounded to three significant digits.
   
 
 

Stoichiometry Plus Cost Calculations

Our last example (for now) of a stoichiometry plus problem involves a cost analysis. Upon calculating an amount of product, you might be required to determine some dollar figures associated with the product. A good example of this is provided below.
 
 
 

Example 4:

 
Like Example 3, this problem has a lot of details to give attention to. These details include gram-kilogram relationships, percent yield, a gram-to-gram conversion, and a mass-to-dollar conversion. The solution strategy involves using the given quantity – kg of Cl₂ – to determine the theoretical amount of NaCl produced in grams. This is done in the manner demonstrated in Lesson 2c (and as done in Examples 1 and 2) by means of a three-step conversion (plus an additional kg-to-g conversion). The percent yield value is used to calculate the actual yield of NaCl in grams. Then a couple of conversions are required to convert grams to kilograms and then to dollars.
 
Molar masses of Cl₂ and NaCl are needed for the gram-to-gram stoichiometry conversion.
   
The kilogram-to-gram-to-gram conversion set up (units only, no numbers) is shown below. All units cancel except for the grams NaCl.
   
Numbers for the molar mass and coefficients are inserted into the conversion factors and the mass of NaCl produced is calculated.
   
The above number is the theoretical yield. It is based on a stoichiometry calculation using the mass of reactant. The actual yield is 87.5% of this value. That is what percent yield means. The unrounded number is used to calculate the actual yield. Rounding is done once the last calculation is completed.
   
Now the grams are converted to kilograms and then to dollars using two conversion factors. The final answer is rounded to three significant digits.
   
 
 

Summary

Stoichiometry Plus problems are more common in honors and advanced Chemistry courses. The sophistication of the problems increases with the level of course expectations. In general, stoichiometry is such a fundamental skill in Chemistry that it emerges in nearly every unit of study. As such, students at any course level are likely to experience some stoichiometry mixed with other skills. Expect to see more stoichiometry plus problems in the coming chapters of this Chemistry Tutorial.
 
 
 

Before You Leave

• CalcPad - Stoichiometry Problem Set ST10: This set of 5 problems provides feedback and chances to correct errors. It is great practice for stoichiometry plus problems.
• CalcPad – Stoichiometry Problem Set ST16: This Balance and Solve problem set includes six stoichiometry problems for which a balanced chemical equation must first be written.
• The Check Your Understanding section below includes 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. Solve the following two step problem.

1. Write the balanced chemical equation for the combustion of ethanol (C₂H₆O).
2. Use conversion factors to determine how many liters of ethanol must be burned to produce 1.00 L of water. (D_water = 1.00 g/mL; D_ethanol = 0.789 g/mL)

 

2. An electricity generating power plant burns crude oil that contains 1.08% sulfur by mass. The sulfur reacts with oxygen to produce SO₂. The density of SO₂ is 2.60 g/L.
a. Write the balanced chemical equation for the combustion of sulfur.
b. Use conversion factors to determine the volume (in L) of SO₂ released by the combustion of 1000.0-kg of crude oil (and the sulfur it contains).