Ch. 12 Stoichiometry

AKA…… Chemistry Math….., Ooohhhh Scary!!

I. The Arithmetic of Equations A. Using Everyday Equations

• 1. Equations are like recipes that tell chemists the amount of reactants to mix and products to expect.

• 2. The calculation of quantities in chemical reactions is called stoichiometry. This is a form of book keeping for reactions.

I. The Arithmetic of Equations B. Interpreting Chemical Equations• 1.The information that can be

derived from an equationa. the types of particle (formulas)b. number of molesc. mass (law of conservation of

mass)d. volume ( only if at STP)

• 2. You can use this to solve several types of problems. Write the balanced equation. Then determine the mole to mole ratio.

What is Stoichiometry?• Stoichiometry is a powerful tool that

all chemist must utilize. It separates the scientists from everyone else. It allows a chemist to determine an amount of another compound that is needed or produced in a chemical reaction.

• Here is a four step process for solving problems.

How Do We Use It?• Question- How many grams of water can

be produced from 10.grams of H2 with excess O2?

• excess=you have more than enough O2 to complete this reaction, worry about H2

• 1. You need a balanced chemical reaction. – 2H2 + O2 ==> 2H2O– (remember that the coefficients are really

mole ratios)

How Do We Use It?• 2. FIND THE MOLES of your

compound. Sometimes it is already given, sometimes you have to calculate it.– Given==> 10. grams of H2

– Find the moles=> 10.g H2 (1mol / 2.0g

H2)= 5.0 mol H2

How Do We Use It?• 3. DO THE RATIO-Now you can find the moles of

any other compound in this reaction using the coefficients of the balanced reaction. Here are a few ways to do this.– Factor Label 5.0 mol H2

( 2 mol H2O /2 mol H2) =5.0 mol H2O

– or set up a proportion• (5.0 mol H2 / 2 H2 )= (x mol H2O / 2 H2O )

• x= 5.0 mol H2O

– or set up an equation proportion– (remember that the coefficients are really mole

ratios)• 5.0 mol H2 /2H2 + ? mol O2 / O2 ===> ? mol H2O/2H2O

• Now you solved for O2 also(2.5 mol) and H2O (5.0 mol)

How Do We Use It?• 4. ANSWER THE QUESTION-Now find the

mass from the number of moles– 5.0 mol H2O ( 18.0g/mol)= 90. g H2O

– If I wanted to know about the O2, I could just take the moles and find the mass.

– 2.5 mol O2 ( 32.0g/mol)= 80. g O2

– FYI the mass in equals the mass out.

– 10.g/ 2H2 + 80.g/ + O2 ===> 90.g /2H2O

• Conclusion- Find the moles, do the ratio, answer the question.

• 1. Write the balanced equation. • 2. Find the number of moles for the

given substance. • 3. Look at the balanced equation to

determine the ratio of moles of required substance to moles of given substance.

• 4. Convert the number of moles of required substance to grams.

Method 2 - Here is another method that puts it all together in one equation.

If Iron (III) Sulfide, Fe2S3 is removed from coal, it reacts with the oxygen in the air. How much Sulfur dioxide is formed from 100 g Fe2S3?1. Write a balance equations

2Fe2S3 + 9O2 ----> 2Fe2O3 +6SO2

2. Write down the mass given and unknown. 100 g ?

2Fe2S3 + 9O2 ----> 2Fe2O3 +6SO2

3.Calculate the number of moles and use the factor-label method to determine the mass or volume of your unknown.

(100gFe2S3)1 moleFe2S3 6moleSO2 64gSO2

160gFe2S3 2moleFe2S3

1moleSO2

=120g SO2

What is the mass of Oxygen produced by the decomposition of 30g of Potassium Chlorate?30g ?

2KClO3 ----> 2KCl +3O2

30 gKClO3 1moleKClO3 3moleO2 32g O2

122.5gKClO3 2mole KClO3

1moleO2=11.7 g O2

Example 2

II. Chemical CalculationB. Solving Other

Stoichiometric Problems• 1. Write the balanced equation. • 2. Find the number of moles for the

given substance. • 3. Compare the mole to mole ratio.

You can then solve for mass, volume, or number of particles.

III. Limiting Reagent and Percent YieldA. What is a Limiting Reagent?• 1. A limiting reagent determine

the amount of product that can be produced in a reaction.

• 2. The limiting reagent is the reactant that will be use up first in the reaction.

• 3. When solving a limiting reagent problem, the first step is to convert the reactants to moles. By comparing the number of moles, you sometimes can determine the limiting reagent.

III. Limiting Reagent and Percent Yield B. Percent Yield • 1. Percent yield is the comparison of

the actual amount of product by the amount of product calculated.

• 2. The formula for percent yield is the actual amount of product divided by the theoretical amount of product multiplied by 100.

• Percent Yield = actual amount of product x 100 theoretical amount of product

Percent Yield Calculations• Percent yield calculations differ from limiting

stoichiometry problems by only one extra step.• The question includes a mass recovered. Here

is the formula you will use at the end of the problem.

• When you read the question the mass of the product must not be used until the very end.

• So.....put it in your pocket for later.• You are allowed to use moles, grams, liters in

this equation, as long as actual and theoretical are both in these units.

EXAMPLE QUESTION #1 (not limiting)• 2CO(g) + O2(g) --> 2CO2(g)

• Calculate the % yield if 69.1g of CO combines with excess O2 to form an experimental yield of 48.3L of CO2 @STP.– See how it says yields 48.3L of CO2 save

that for later (in your pocket).– Now determine the volume of of CO2 from

69.1g of CO using stoichiometry

Step 2• Find the moles, the mol ratio, then find

the volume of CO2

– 69.1g CO(1 mol CO/28.0g)= 2.57mol CO– 2.57mol CO(2mol CO2/2mol CO)= 2.57mol

CO2

– 2.57mol CO2(22.4L/1 mol)= 55.3L of CO2

– (69.1 g CO)(1 mol CO/28.0g)(2mol CO2/2mol CO)(22.4L/1 mol) =

– = 55.3 L CO2

• Now take out that volume recovered and plug into the % yield equation.– (48.3L/55.3L) x 100%= 77.3% yield

Initial-Change-End Box Method• Here are the Rules from the other Page

– STEP1= SET UP the ICE Box– STEP 2- Find the moles, This is where you have

to problem solve.– STEP 3- Find X, find the moles of everything– STEP 4- Answer the questions, convert moles to

mass• The question-

– You have 20.0 g of elemental sulfur, S, and 160.0 grams of O2. What mass of SO2 can be formed? How much reactant is left over?

– S(s) + O2(g) ==> SO2(g)

STEP1= SET UP the ICE Box below the balanced reaction. (I-Initial C-Change E-End).

S(s) + O2(g) ==> SO2(g)

I-Initial

C-Change -X -X +X

E-End

Using the coefficients, (the #'s in front of each compound) to determine the relative change. If there is a 2 in front, use 2X, if there is a 3 use 3X. Also reactants decrease in their amounts (-X), products will increase (+X).

Step 2- Moles are the only thing allowed in the ice box. Find the moles of each compound and insert that value in its “I” box.

• Moles S– 20.0g S X 1 mole/ 32.1g– =0.623 moles S

• Moles O2

– 160.0g O2 X 1 mole/32.0g

– =5.000 moles O2

S(s) + O2(g) ==> SO2(g)

I-Initial 0.623 5.00

C-Change -X -X +X

E-End

Step 3- Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction.

– if S runs out ==> 0.623 mol -X =O ; X is therefore 0.623 mol

– if O2 runs out ==> 5.000 mol -X=O ; X is therefore 5.000 mol

• Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX.

Solve the Problem• (If you applied the incorrect, larger

X value, will get a negative amount at the end, so go back and change it.)Now solve for everything (add or subtract down each column).S(s) + O2(g) ==> SO2(g)

I-Initial 0.623 5.000

C-Change -0.623 -0.623 +0.623

E-End 0 4.377 0.623

Step 4- Answer the questions, you have them moles of everything. Convert to grams.

• What mass of SO2 can be formed? – 0.623mol SO2X 64.1g/1 mol

– =39.9g SO2

• How much reactant is left over?– 4.377mol O2X 32.0g/ 1 mol

– =140.1g SO2

Level 2- The equation now has coefficients.

• If you are provided 200.g of sodium and 250. grams of iron (III) oxide, which substance is the limiting reactant? How many grams of Iron are produced? How much reactant is in excess?–6Na + Fe2O3 --> 3Na2O + 2Fe

STEP1= SET UP the ICE Box below the balanced reaction. (I-Initial C-Change E-End). • Using the coefficients, (the #'s in

front of each compound) to determine the relative change. If there is a 2 in front, use 2X, if there is a 3 use 3X. Also reactants decrease in their amounts (-X), products will increase (+X).

6Na + Fe2O3 ==> 3Na2O + 2Fe

I-Initial

C-Change -6X -X +3X + 2X

End

Step 2- Moles are the only thing allowed in the ice box. Find the moles of each compound and insert that value in its I box.

• Moles Na– 200.g Na X 1 mole/ 23.0g– =8.70 moles Na

• Moles Fe2O3

– 250.g Fe2O3 X 1 mole/ 159.7g

– =1.57 moles Fe2O36Na + Fe2O3 ==> 3Na2O + 2Fe

I-Initial 8.70 1.57

C-Change -6X -X +3X +2X

End

Step 3- Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction.

• if Na runs out ==> 8.70 mol -6X =O ; X is therefore 1.45 mol

• if Fe2O3 runs out ==> 1.57mol -X=O ; X is therefore 1.57 mol

6Na + Fe2O3 ==> 3Na2O + 2Fe

I-Initial 8.70 1.57

C-Change -6X -X +3X +2X

End

Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX.

• Na is therefore limiting.– (If you applied the incorrect, larger X value, will get a

negative amount at the end, so go back and change it.)

– Now solve for everything (add or subtract down each column).

6Na + Fe2O3 ==> 3Na2O + 2Fe

I-Initial 8.70 1.57

C-Change -6(1.45) -1.45 +3(1.45) +2(1.45)

End 0 .12 4.35 2.90

Step 4- Answer the questions, you have them moles of everything. Convert to grams.

• How many grams of Iron are produced? – 2.90 mol Fe X 55.85g/1 mol– =162g Fe

• How much reactant is in excess?– 0.12 mol Fe2O3 X 159.7g/1 mol

– =19.2g Fe2O3

Work Cited

• http://kentchemistry.com/• Textbook: Chemistry ,

Wilbraham, Staley, Matta, Waterman; Addison-Wesley