Watch this video to see what the stoichiometry road map is all about!


Introductory Steps to Use the Road Map

The Steps:

1. Identify the given
2. Identify the unknown
3. Locate on the road map the starting point (the given)
4. Locate on the road map the ending point (the unknown)

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Example Problem on Using the Road Map:

Given the following equation:

2 Na + 2 H₂O ==> 2 NaOH + H₂

How many grams of water does it take to react with 100.0 g of Na?

1. Identify the given:
·100.0 g of Na

2. Identify the unknown:
·g of water

3. Locate the starting point on the road map:
·The units of the given is "grams" (100.0 g of Na). Therefore, you would start at "Mass of A" on the road map.




4. Locate the ending point on the road map:
·The units of the unknown is "grams" (g of H₂O). In addition, water is a separate substance than Na. Therefore, you would end at "Mass of B" on the road map.



Now watch this video to see how to use the road map in an example.

Now try and do your own problem, using the reaction below. Make sure you are following the road map!

How many grams of lead (II) chloride (PbCl₂) would be produced from 100 g of sodium chloride?


Give your answer to the nearest 1g
The Road Map Gets Bigger....

So far, you have seen the road map useful for mass to mass calculations:



This road map works well for several problems, but it leaves a few concepts out. We need to expand the mass - mass road map to include the following:

• Particles of a substance
• Volume of a gas at STP (Standard Temperature-Pressure)

Therefore the expanded road looks like this:



Notes to keep in mind about the full road map:

1. It is just an expansion of the mass - mass road map. It CAN be used to perform mass - mass calculations!!!
   
   
2. You do not have to start and/or end at one of the end boxes!!! You CAN start ANYWHERE on the road map and end ANYWHERE on the road map!!!
   
   
3. Since the road map is more complex, it is important now (more than ever) to label your "given" and your "unknown" to determine your starting and stopping places on the full road map.

Here is a link to a presentation with some examples of how to follow the road map for different types of problems.

Where to start and stop on the road map...

Assume you are dealing with the following reaction:

S_(s) + 3 F_2(g) ==> SF_6(g)

(s) stands for solid
(g) stands for gas

You could be asked a number of questions to solve involving the reaction.

Here are some sample questions involving the above reaction. DO NOT SOLVE THE QUESTIONS!!! Instead, match the question to where you would start and where you would stop on the full road map.

MAKE SURE TO FIGURE OUT THE GIVEN & THE UNKNOWN!!!

1. How many mol of F₂ is needed to react with 10.0 mol of S?
   
2. How many g of S reacts to form 30.0 g of SF₆?
   
   
3. 30.0 g of S reacts to form how many molecules of SF₆?
   
4. At STP, how many L of F₂ is needed to form 10.0 x 10³⁰ molecules of SF₆?
   

Let's take some time to actually work through two of the examples from the previous page. One will be worked out for you, and you will do the other.



S_(s) + 3 F_2(g) ==> SF_6(g)

Example: At STP, how many L of F₂ is needed to form 1.00 x 10²⁴ molecules of SF₆?

To answer this question we need to start at Particles of A and go to Volume of B at STP. The process will be:

Now you try, using the same reaction above, figure out this problem.

30.0 g of S reacts to form how many molecules of SF₆?

Give your answer as _._ _ * 10²³ (you just fill in the blanks for example 1.23 if it is 1.23*10²³

Now that you are a master at calculating masses of products & reactants, it's time to apply them to a new concept. Look at the balloons in the picture. They were filled up with gas generated by a chemical reaction in between acid & metal in the flask. Each flask had the same amount of acid, but more metal was used in each subsequent reaction (going from left to right). Why are the 3rd and 4th balloons the same size if the 4th reaction used more metal reactant?

This topic may require some hard work...it is all about the concept of the limiting reagent. Start below by viewing a video that shows how the limiting reagent compares to cooking a recipe with a set amount of ingredients in your house. Then, read a brief introduction to the concept of the limiting reagent. It will then be time to start performing math.


The Limiting Reagent

Note: for the purposes of this unit the words "reactant" and "reagent" will be used interchangeably. They mean the EXACT SAME THING!!

Consider the following reaction:

N₂ + 3 H₂ ==> 2 NH₃

Recall:

• The ratio of N₂ to H₂ is 1:3
• This ratio means that for every 1 N₂ mole(cule) reacted, 3 H₂ mole(cules) react
• For every 1 N₂ mole(cule) that reacts, 2 NH₃ mole(cules) are produced
• For every 3 H₂ mole(cules) that react, 2 NH₃ mole(cules) are produced

What if I carried out this reaction by adding 5 N₂molecules and 15 H₂molecules to a reaction flask?

Answer: All of the N₂and H₂molecules would react producing 10 NH₃molecules (as shown in the picture below).




Notice that all of the N₂and all of the H₂is converted to NH₃in this example. Both reactants (reagents) being consumed entirely is extremely uncommon. Usually one runs out (which causes the reaction to stop) and the other reactant is left over.

• Limiting Reagent - The reagent (reactant) which is used up entirely in a reaction


• Excess Reagent - The reagent (reactant) which is left over after a reaction has stopped.

Which one is the limiting reactant and why? Before we move on to look at a slightly more difficulty limiting reactant problem, here are some tips to recognize when you need to be doing a limited reactant problem:


Tips for Identifying Limiting Reactant Problems

One of the many difficulties that arises with limiting reagent problems is that students often have trouble distinguishing them from typical mol-mol or mass-mass problems. Here are some tips:


Clues that it IS a limiting reagent problem:

1. The question asks you to find the limiting reagent (reactant).
   
2. TWO OR MORE REACTANT amounts are given.

Clues that it IS NOT a limiting reagent problem:

1. The question only gives you the amount of one reactant or it does not give you the amount of any of the reactants.
   
2. The question mentions that all but one of the reactants are present in excess.

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Examples:

C + O₂ ==> CO₂

A) 20.0 mol of C is mixed with 10.0 mol of O₂. What is the limiting reagent?
This is a limiting reagent problem, because 1) it asks for the limiting reagent, and 2) the amounts of two reactants (20.0 mol C & 10.0 mol O₂) are given.


B) 20.0 g of C is mixed with 100.0 g of O₂. How many grams of CO₂ will form?
This is also a limiting reagent problem, because the amounts of two reactants (20.0 g of C & 100.0 g of O₂) are given.


C) If 90.0 mol of O₂ is reacted, how many mol of C are needed?
This is NOT a limiting reagent problem. Only the amount of one of the reactants (90.0 mol of O₂) was given.


D) If 200.0 g of O₂ is reacted with excess C, how many g of CO₂ will be produced?
This is NOT a limiting reagent, because one of the two reactants (carbon) is present in excess. This leaves only one reactant (O₂) that has a finite amount.

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Now, let's move on to working with limiting reactant problems that require you to use more than moles. One example will be worked out for you on this video, one will be worked out for you here, and the other you will do.

For the following balanced chemical reaction, if 38.35grams of H₂Oare reacted with 107.63grams of CO₂, how many grams of C₁₂H₂₂O₁₁will be formed?

11H₂O + 12CO₂ => C₁₂H₂₂O₁₁ + 12O₂
The steps for doing this problems are as follows:

1. Figure out how many moles of each of the reactants you have.
2. Figure out how many moles of the product each can form.
3. Determine what amount is formed by picking the smallest value.
   
4. Use what you picked in step 3 to determine the mass of the product needed.

Mathematically the process looks like this:

Give your answer to the nearest ones place. Enter 10 if you are ready to give up and move on - will not get credit.
The next topic is bound to be easier than the last...

But before we move on, here are a few videos with examples worked out showing how to find the remaining amount of the excess reactant.

Video one: mol-mol problem

Video two: mass-mas problem


Topic 7 - % Yield

The test tubes on the right appear to contain more thanone substance. Each tube is likely a mixture of products & reactants. It is uncommon for a reaction to use up all of even one reactant. Chemists like to compare how far the reaction "proceeds" to how far "it could have proceeded." That is called the percent yield....and it is your last topic in stoichiometry!!


Percent Yield Introduction

Ted Williams (a baseball player for the Boston Red Sox) had a career batting average of 0.344. Your friend's GPA is 3.50. It is rumored that you only use around 10% of your brain power.

What do these things have to do with percent yield??!??

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Batting Average:

Ted Williams's batting average of 0.344 meant that Ted Williams got a hit 344 times out of every 1000 at bats. It also means that Ted Williams got a hit 34.4 times out of every 100 at bats, or 34.4% of the time.

In theory, Ted Williams could have had a batting average of 1.000. That would have meant that he was a god! No person who batted enough times would every come close to having a career batting average of 1.000 (meaning that he got a hit on EVERY AT BAT!!!) BUT IN THEORY, IT IS POSSIBLE FOR SOMEONE TO GET AN AT BAT EVERY TIME! IN OTHER WORDS, "100% of the time."

A batting average of 1.000 would be 100% yield!

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Brain Power:

It is often rumored that you only use about 10% of your brain power. This would correspond to 10% yield. Even though you only use 10% of your brain power, IN THEORY, YOU COULD USE 100% OF YOUR BRAIN POWER.It doesn't matter that no person ever uses 100% of their brain power, it is still possible in theory.

Percent Yield and Chemistry

Percent Yield relates how much is actually accomplished to how much could have been accomplished in theory. For example, in 1000 at bats:

Ted Williams actually gets a hit only 344 times.

In theory, Ted Williams could have gotten a hit 1000 times.

His percent yield is (344/1000) = 0.344

0.344 is the Fraction Form of percent yield. Multiply by 100 to get it into a percent form:

(344/1000 * 100) = 34.4 %

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Percent Yield is given by the following formula:



All of the mol - mol and mass - mass problems that we have performed so far ARE THEORETICAL PROBLEMS!!!The tell you the MAXIMUM POSSIBLE YIELD if you were to carry out the reaction.

Rarely does some one get 100% yield out of a reaction. % YIELD IS ALMOST ALWAYS LESS THAN 100%!!!

The actual yieldis what result someone gets when they perform the reaction in the laboratory. It is the amount that they measure. ACTUAL YIELD SHOULD NEVER BE GREATER THAN THE THEORETICAL YIELD!!!

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Summary:



Actual Yield-The amount of product that actually forms when someone measures it in the laboratory.

Theoretical Yield-The maximum amount of product that can be performed as calculated by stoichiometry.

Here is a videoto show you an example of a full percent yield calculation.

Last question!

So, if you have calculated that you should be able to make 10g of product, but the reaction only produced 8g, what is your percent yield?