StoichiometryStoichiometryTopic 1.4.1 – 1.4.3Topic 1.4.1 – 1.4.3

Chocolate Chip Cookies!!Chocolate Chip Cookies!!1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

How many eggs are needed to make 3 dozen cookies?

How much butter is needed for the amount of chocolate chips used?

How many eggs would we need to make 9 dozen cookies?

How much brown sugar would I need if I had 1 ½ cups white sugar?

Cookies and Chemistry…!?!?Cookies and Chemistry…!?!?

Just like chocolate chip Just like chocolate chip cookies have recipes, cookies have recipes, chemists have recipes as wellchemists have recipes as well

Instead of calling them Instead of calling them recipes, we call them reaction recipes, we call them reaction equationsequations

Furthermore, instead of using Furthermore, instead of using cups and teaspoons, we use cups and teaspoons, we use molesmoles

Lastly, instead of eggs, butter, Lastly, instead of eggs, butter, sugar, etc. we use chemical sugar, etc. we use chemical compounds as ingredientscompounds as ingredients

Interpreting balanced, chemical equationsInterpreting balanced, chemical equations

massmass and and atomsatoms are always conserved are always conservedmolecules/compounds, moles, and molecules/compounds, moles, and

volumes may or may not be conservedvolumes may or may not be conserved theoretical yields theoretical yields are what we should get are what we should get

based on the balanced equation and the based on the balanced equation and the periodic tableperiodic tableexample: example: 2 Na + Cl2 Na + Cl2 2 2 NaCl 2 NaCl

by mixing 2 moles of sodium with 1 mole of chlorine, by mixing 2 moles of sodium with 1 mole of chlorine, we will get 2 moles of sodium chloridewe will get 2 moles of sodium chloride

what if we wanted 10 moles of NaCl? what if we wanted 10 moles of NaCl? would need 10 mol of Na and 5 mol of Clwould need 10 mol of Na and 5 mol of Cl22

Mole to MoleMole to Mole

write the balanced reaction for hydrogen gas write the balanced reaction for hydrogen gas reacting with oxygen gas to form waterreacting with oxygen gas to form water

2 H2 H22 + O + O22 2 H 2 H22OO How many moles of reactants are needed?How many moles of reactants are needed?

What if we wanted 4 moles of water?What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much What if we had 3 moles of oxygen, how much hydrogenhydrogen

would we need to react and how much would we need to react and how much waterwater would we would we get?get?

What if we had 50 moles of hydrogen, how much What if we had 50 moles of hydrogen, how much oxygenoxygen would we need and how much would we need and how much waterwater produced? produced?

3

6

6 of H2 and 6 of H2O

25 of O2 and 50 of H2O

mole ratios can be used to calculate the mole ratios can be used to calculate the moles of one chemical from the given moles of one chemical from the given amount of a amount of a different different chemical chemical

Example: How many moles of chlorine are Example: How many moles of chlorine are needed to react with 5 moles of sodium?needed to react with 5 moles of sodium?

2 Na + Cl2 Na + Cl22 2 NaCl 2 NaCl

5 mol Na 1 mol Cl2

2 mol Na= 2.5 moles Cl2

Mole-Mole ConversionsMole-Mole Conversions

How many moles of sodium chloride will How many moles of sodium chloride will be produced if you react 2.6 moles of be produced if you react 2.6 moles of chlorine gas with an excess (more than chlorine gas with an excess (more than you need) of sodium metal?you need) of sodium metal?

2 Na + Cl2 Na + Cl22 2 NaCl 2 NaCl2.6 mol Cl2.6 mol Cl22 x x 2 mol NaCl 2 mol NaCl = 5.2 mol NaCl = 5.2 mol NaCl

1 mol Cl1 mol Cl22

Mole to Mass ConversionsMole to Mass Conversions

amounts are given in moles but often need to be amounts are given in moles but often need to be in grams insteadin grams insteadHow many grams of chlorine are required to How many grams of chlorine are required to

react completely with 5.00 moles of sodium to react completely with 5.00 moles of sodium to produce sodium chloride?produce sodium chloride?

2 Na + Cl2 Na + Cl22 2 NaCl 2 NaCl

5.00 moles Na 1 mol Cl2 70.90g Cl2

2 mol Na 1 mol Cl2

= 177g Cl2

Calculate the mass in grams of iodine required to Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum.react completely with 0.50 moles of aluminum.

2Al + 3I2Al + 3I22 2AlI 2AlI33

0.50 mol Al 3 mol I2 253.8 g I2

2 mol Al 1 mol I2

= 95.2g l2

Mass to MoleMass to Molecan also start with mass and convert to moles can also start with mass and convert to moles

of product or another reactantof product or another reactant in a in a combustion reactioncombustion reaction, calculate the number , calculate the number

of moles of ethane (Cof moles of ethane (C22HH66) needed to produce ) needed to produce

10.0 g of water10.0 g of water 2 C2 C22HH66 + 7 O + 7 O22 4 CO 4 CO22 + 6 H + 6 H220 0

10.0 g H2O 1 mol H2O 2 mol C2H6

18.0 g H2O 6 mol H20

= 0.185 mol C2H6

PracticePractice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxideCalculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

4 Al + 3 O4 Al + 3 O22 2 Al 2 Al22OO33

10.0 g Al10.0 g Al22OO33 x x 1 mol Al1 mol Al22OO33 x x 3 mol O3 mol O22

102 g Al102 g Al22OO33 2 mol Al 2 mol Al22OO33

= .147 mol O= .147 mol O22

Mass-Mass ConversionsMass-Mass Conversionsmost often given a starting mass of one of

the reactants and want to find out the mass of a product we should get (called theoretical yield)

or how much of another reactant we need to completely react with it (no leftover ingredients)

therefore, go from grams to moles, find the mole ratio, and back to grams of compound we are interested in

Mass-Mass ConversionMass-Mass Conversion

Ex. Calculate how many grams of ammonia Ex. Calculate how many grams of ammonia (NH(NH33) are produced when you react 2.00g of ) are produced when you react 2.00g of

nitrogen with excess hydrogen.nitrogen with excess hydrogen.NN2 2 + 3 H+ 3 H2 2 2 NH 2 NH33

2.00g N2 1 mol N2 2 mol NH3 17.06g NH3

28.02g N2 1 mol N2 1 mol NH3

= 2.4 g NH3

Percent YieldPercent Yield

simply calculating how much you got in an simply calculating how much you got in an experiment to what you should have experiment to what you should have gottengotten% yield = experimental yield/theoretical yield % yield = experimental yield/theoretical yield

(100)(100)What if in the previous example the What if in the previous example the

experiment only yielded 1.9 g of NHexperiment only yielded 1.9 g of NH33 instead instead

of the theoretical yield of 2.4g?of the theoretical yield of 2.4g?1.9g / 2.4 g (100) = 79%1.9g / 2.4 g (100) = 79%

PracticePractice

How many grams of calcium nitride are How many grams of calcium nitride are produced when 2.00 g of calcium reacts produced when 2.00 g of calcium reacts with an excess of nitrogen?with an excess of nitrogen?

2.39 g Ca2.39 g Ca33NN22

Other Stoichiometric CalculationsOther Stoichiometric Calculations

Ex. Calculate the number of Ex. Calculate the number of grams of Ligrams of Li33N N

that must be added to an excess of water that must be added to an excess of water to produce to produce 15 L of NH15 L of NH33 at STP.at STP.

LiLi3 3 NN(s) (s) + 3H+ 3H22OO(l) (l) NH NH3 3 (g) (g) 3LiOH3LiOH(aq)(aq)

15L NH3 1 mol NH3 1 mol Li3 N 34.7 g Li3 N

22.4L NH3 1 mol NH3 1 mol Li3 N

= 23.2 g Li3N

Limiting Reactant: CookiesLimiting Reactant: Cookies1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

•If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?

•What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?

•What if we only had one egg, could we make 3 dozen cookies?

Limiting Reactant (Reagents)Limiting Reactant (Reagents)

In a chemical reaction, an insufficient In a chemical reaction, an insufficient quantity of one of the reactants will limit quantity of one of the reactants will limit the amount of product that forms.the amount of product that forms.This is called the This is called the limiting reactant/reagent.limiting reactant/reagent.

If there is too much of a reactant If there is too much of a reactant (wasteful) it is called the (wasteful) it is called the excess reagentexcess reagent

have to calculate how much of have to calculate how much of aa product product we can get from we can get from eacheach of the reactants to of the reactants to determine which reactant is the limiting determine which reactant is the limiting oneone

thethe lowerlower amount of amount of aa product is the product is the correct answercorrect answer the reactant that makes the the reactant that makes the least amount of least amount of

product product is the is the limiting reactantlimiting reactantbe sure to pick a product

you can’t compare to see which is greater and which is lower unless the product is the same

Limiting Reactant: ExampleLimiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of 10.0g of aluminum reacts with 35.0 grams of

chlorine gas to produce aluminum chloride. Which chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how reactant is limiting, which is in excess, and how much product is produced?much product is produced?

22 Al + Al + 33 Cl Cl22 22 AlCl AlCl33 Start with Al:Start with Al:

Now ClNow Cl22::

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3

= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.9g AlCl3

limiting reactant is chlorinelimiting reactant is chlorine

LR Example ContinuedLR Example Continued

We get We get 49.4g49.4g of aluminum chloride from the given of aluminum chloride from the given amount of aluminum, but only amount of aluminum, but only 43.9g43.9g of aluminum of aluminum chloride from the given amount of chlorine. chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction the 35.0g of chlorine is used up, the reaction comes to a complete stop.comes to a complete stop.

Limiting Reactant PracticeLimiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of 15.0 g of potassium reacts with 15.0 g of

iodine. Calculate which reactant is limiting iodine. Calculate which reactant is limiting and how much product is made.and how much product is made.22 K + I K + I22 22 KI KI

15.0 g K 1 mol K 2 mol KI 166 g KI

39.1 g K 2 mol K 1 mol KI= 63.7g Kl

15.0 g I2 1 mol I2 2 mol KI 166 g KI

253.8 g I2 1 mol I2 1 mol KI= 19.6g KI

Finding the Amount of ExcessFinding the Amount of Excess

By calculating the amount of the excess By calculating the amount of the excess reactant needed to completely react with reactant needed to completely react with the limiting reactant, we can subtract that the limiting reactant, we can subtract that amount from the given amount to find the amount from the given amount to find the amount of excess.amount of excess.

Can we find the amount of excess Can we find the amount of excess potassium in the previous problem?potassium in the previous problem?

15.0 g of potassium reacts with 15.0 g of iodine. 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 K + I22 2 KI 2 KI

We found that Iodine is the limiting reactant, and We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced.19.6 g of potassium iodide are produced.

How much K will react with the 15.0 g of IHow much K will react with the 15.0 g of I22??

15.0 g I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K= 4.62 g K USED!

15.0 g K – 4.62 g K = 10.38 g K EXCESS

Given amount of excess reactant

Amount of excess reactant actually used

Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

Limiting Reactant: RecapLimiting Reactant: Recap

1.1. You can recognize a limiting reactant problem because You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.there is MORE THAN ONE GIVEN AMOUNT.

2.2. Convert ALL of the reactants to the SAME product (pick Convert ALL of the reactants to the SAME product (pick any product you choose.)any product you choose.)

3.3. The lowest answer is the correct answer.The lowest answer is the correct answer.4.4. The reactant that gave you the lowest answer is the The reactant that gave you the lowest answer is the

LIMITING REACTANT.LIMITING REACTANT.5.5. The other reactant(s) are in EXCESS.The other reactant(s) are in EXCESS.6.6. To find the amount of excess, subtract the amount used To find the amount of excess, subtract the amount used

from the given amount.from the given amount.7.7. If you have to find more than one product, be sure to If you have to find more than one product, be sure to

start with the limiting reactant. You don’t have to start with the limiting reactant. You don’t have to determine which is the LR over and over again!determine which is the LR over and over again!