stoichiometry moles, masses, reactions, oh my!. stoikheion + metria from the greek for “element”...

StoichiometryStoichiometryMoles, masses, reactions, oh my!

StoikheionStoikheion + + metriametriaFrom the Greek for “element”

and “the process of measuring”Quantitative analysis of reactants

and productsMolesMolar massesBalanced equations

Grinch SandwichesGrinch SandwichesIngredients per sandwich

◦Three pieces of bread◦Sauerkraut (10 g)◦Toadstool (15 g)◦Arsenic sauce (5 g)

Balanced EquationsBalanced Equations2H2 + O2 2H20

◦Coefficients = moles◦2 moles of hydrogen react with 1 mole

of oxygen to produce 2 moles of water◦What if you only had 1 mole of

hydrogen?◦What if you had 10 moles of both

hydrogen and oxygen?◦What if you needed to make 5 moles

of water?

Law of Conservation of Law of Conservation of MassMassConvert each of the moles in the

previous equation to gramsAdd the reactants’ masses

together.Compare to the product’s mass.Be in awe!

Mole RatioMole Ratio ssEquation must be balancedRelate reactants to each other,

products to each other, or reactants to products

Molar MassesMolar MassesIf mass is mentioned, find that

substance’s molar mass

Making Stoichiometry Making Stoichiometry WorkWorkDimensional Analysis

◦Number, unit, substance…always!

Problem #1Problem #1When 2.50 g of hydrogen gas are

mixed with an excess of oxygen gas, how many grams of water should be produced?

Problem #1—step 1Problem #1—step 1Write a balanced equation.

◦2H2 + O2 2H20

Problem #1—step 2Problem #1—step 2Begin with your given information

and make a plan of conversions.◦Grams of hydrogen moles of

hydrogen moles of water grams of water

Write it as dimensional analysis.

g H2 x mol H2 x mol H2O x g H2O =

g H2 mol H2 mol H20

Problem #1—step 3Problem #1—step 3Fill in your numbers.

2.5 g H2 x 1 mol H2 x 2 mol H2O x 2.02 g H2 2 mol H2

18.02 g H2O =1 mol H20 • Cross out what divides out.• Calculate your answer.

Problem #1—step 4Problem #1—step 4

Problem #2Problem #2The thermite reaction: aluminum

foil reacts with ferric oxide

Problem #2Problem #2Wow! Now let’s assume that

0.010 g of aluminum reacted with the excess ferric oxide on the ball bearing. What mass of aluminum oxide should be produced?

Problem #2—Step 1Problem #2—Step 1Write a balanced equation.

2Al + Fe2O3 Al2O3 + 2Fe

Problem #2—Step 2Problem #2—Step 2Write your plan.

g Al x mol Al x mol Al2O3 x g Al2O3 =

g Al mol Al mol Al2O3

Problem #2—Step 3Problem #2—Step 3Place the proper numbers and

calculate.

0.010 g Al x 1 mol Al x 1 mol Al2O3 x

26.98 g Al 2 mol Al101.96 g Al2O3 =

1 mol Al2O3

Problem #2—Answer Problem #2—Answer 0.0189 g of Al2O3 should be

formed

Problem #3Problem #3What if only 0.0150 g of Al2O3

were actually produced in the previous problem? What is the percent yield?

So, you can calculate the % yield◦% yield = actual amt x 100

theoretical amt

Problem #3Problem #3Thus, the % yield calculation

0.0150 g x 100 = 79.4 % yield0.0189 g

Limiting ReagentLimiting ReagentIdeally, the perfect amount of

each reactantOften one reactant limits how

much product can be formed—limiting reagent (LR)

Often the other reactant(s) is(are) in excess (ER)

You will be given at least 2 quantities of reactants

Grinch SandwichesGrinch SandwichesIngredients per sandwich

◦Three pieces of bread◦Sauerkraut (10 g)◦Toadstool (15 g)◦Arsenic sauce (5 g)

Problem #4Problem #40.500 g of silver nitrate in

solution reacts with 0.750 g of tin (IV) chloride in solution. If 0.400 g of solid silver chloride are retrieved, what is the percent yield?

Problem #4—Step 1Problem #4—Step 1Write a balanced equation.

4AgNO3 + SnCl4 4AgCl + Sn(NO3)4

Write shorthand, to save time. You’ll write the real stuff in the answer.

A B C D

Problem #4—Step 2Problem #4—Step 2Ask yourself, when I have 0.500 g

of AgNO3, how many grams of SnCl4 do I really need? This will determine the LR.

Set up the plan

g A x mol A x mol B x g B = g A mol A mol B

Problem #4—Step 3Problem #4—Step 3Plug in the numbers.

0.5 g A x 1mol A x 1 mol B 169.88g A 4 mol A

x 260.50 g B = 0.192 g SnCl4 1 mol A

Problem #4—Step 4Problem #4—Step 4Analyze and determine the LR

◦0.750 g of SnCl4 are available, and only 0.192 g of it need to be used to completely react with the AgNO3

•Which substance will be in excess?•Which substance will you run out

of first?

Problem #4—Step 5Problem #4—Step 5

• Now, use the given amount of the LR to find how much product you should be able to make.• Set up your plan.

g A x mol A x mol C x g C = g A mol A mol C

Problem #4—Step 6Problem #4—Step 6

• Plug in your numbers and calculate.

0.5 g A x 1 mol A x 4 mol C x 169.88 g A 4 mol A

143.32 g C = 0.422 g AgCl should be

1 mol C formed

Problem #4—Step 7Problem #4—Step 7

• Look at the actual amount of product that was formed, and determine percent yield using the theoretical you just calculated.

0.400 g AgCl x 100 = 94.8% yield0.422 g AgCl

PracticePracticeSit down with some bread,

sauerkraut, toadstools, and arsenic sauce, and work some stoichiometry problems just for fun.