Stoichiometry and the Mole SOL Review. The Mole Stoichiometry and the Mole.

Stoichiometry and the Mole

SOL Review

The Mole


The Mole Not everything that counts can

be counted, and not everything that can be counted counts.

Albert Einstein

What is a mole?

The mole is a counting unit. Like . . .

1 dozen = 12 each 1 yard = 3 feet 1 cup = 8 ounce

So then ... 1 mol = 6.022 x 1023 particles

That’s Avogadro’s Number!

Where did it come from?

Mole (n) is the SI unit for the number of particles

Amedo Avogadro determined the number of particles in a mole

The mole is the measure of the amount of a substance whose number of particles is the same as 12 grams of Carbon - 12

Calculations

Using dimensional analysis you can determine the number of particles in a mole 1 mole = 6.022 x 1023 particles, molecules, etc. 6.022 x 1023 particles = 1 mole

So, let’s count . . .

1 mol of Ag = 6.022 x 1023 atoms Ag

1 mol of CO2 = 6.022 x 1023 molecules of CO2

1 mol of pizza = 6.022 x 1023 pizzas!

The Mole Road Map

Moles to Mass Conversions

1 Moles = Molar Mass (g) =

Molar mass/Mole or (g/mol)

Now let’s apply that knowledge!

Reminder – Finding Molar Mass:The molar mass = the sum of all the atomic masses.

The Mole: Molar Mass Calculations

Example: Ca(NO3)2

Ca = 40.08 N = 2(14.01) O = 6(16.00) 164.10 grams

You try one:What is the gram formula mass (molar mass) of Mg3(PO4)2?

Mg = 3(24.305) P = 2(30.97376) O = 8(15.9994) 262.86 grams

Molar or Formula Mass

Chemical compounds are written as an empirical formula. Ex. H2SO4 is Sulfuric Acid

Calculating atomic mass, add each atom. H = 1.008 x 2 = 2.016 S = 32.07 x 1 = 32.07 O = 15.999 x 4 = 63.996

Total Atomic Mass =

2.016+32.07+63.996 = 98.08 amu

The Mole and Mole Calculations

One mole = 6.02 x 1023 representative particles

One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure

One mole = the atomic mass listed on the periodic table.

For example: one mole of Helium contains 6.02 x 1023 atoms of Helium and it has a mass of 4.00260 grams. At 0°C and

one atmosphere of pressure, it would occupy 22.4 Liters.

Sample problem: How many liters would 2.0 moles of Neon occupy?

2.0 moles Ne x 22.4 Liters Ne = 44.8 Liters Ne 1.0 moles Ne

Calculating MolesOne Step

How many moles are in 3.011 x 1023

atoms of Oxygen?

3.011 x 1023 atoms O2 1 mol Cu 6.02 X 1023 atoms

0.5 moles of Oxygen

=

More One-Step Conversions

Ex 1) Convert 4.3 grams of NaCl to moles.Mass mol

4.3 g NaCl x 1 mol NaCl = 7.4 x 10-2 mol NaCl

58.45 g NaCl

Ex 2) Convert 0.00563 mol NH3 to grams.

Mol -> mass0.00563 mol NH3 x 17 g NH3 =9.57 x 10 –2 g

NH3 1 mol NH3

Sample problem: How many moles are in 15.2 grams of Lithium?

Answer: 15.2 g Li x 1 mole Li = 2.19 mole Li

6.941 g Li

The Mole and Mole Calculations

REMINDER:

•One mole = 6.02 x 1023 representative particles

•One mole = 22.4 Liters of gas at 0°C and one atmosphere of pressure

•One mole = the atomic mass listed on the periodic table.

Sample problem: How many liters would 14 grams of Helium occupy?

Answer:14 g He x 1 mole He x 22.4 L He = 78 Liters He 4.0026 g He 1 mole He

Mole Calculations

How many atoms of Cu are present in 35.4 g of Cu?

Molar mass of Cu is 63.55 g/mol

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu 63.5 g Cu 1 mol Cu

= 3.4 X 1023 atoms Cu

And another . . .

What mass would 4.52 x 1024 molecules of water have?

= 135 g H2O

4.52 x 1024 mlcs of H20 1 mol 18.02 g

6.02 X 1023 mlcs 1 mol

StoichiometryStoichiometry


Stoichiometry Stoichiometry means that if you know

one piece of information about ONE compound in an equation, you can determine EVERYTHING else!

If you have 3L of Nitrogen, how many liters of ammonia will you produce?

N2 + 3H2 2NH3

Let’s look at that last reaction again.N2(g) + 3H2(g) 2NH3(g)

If you start out with 1 mole of Nitrogen gas and 3 moles of Hydrogen gas, you will make 2 moles of Ammonia gas.

It is important in industry to know the exact proportions of your ingredients so that you will not have excess waste in your product.

Stoichiometry

How many grams of silver chloride can be produced from the reaction of 17.0 g silver nitrate with excess sodium chloride solution?

1. Write the balanced equation

AgNO3 + NaCl AgCl + NaNO3

2. Given and asked for

17.0g ?g

3. Moles of given

17.0g AgNO3 x 1 mol = 170 g

0.100 mol AgNO3

Stoichiometry

StoichiometryMass-Mass Problem

StoichiometryMass-Mass Problem

4. Moles asked for

0.100 mols AgNO3 x 1 mol AgCl =

1 mol AgNO3

0.100 mol AgCl

5. Convert your answer

0.100 mol AgCl x 144 g AgCl = 1 mol AgCl

14.4 g AgCl

AgNO3 + NaCl AgCl + NaNO3

1 mol of a gas=22.4 Lat STP

C. Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

Limiting Reactant Limiting Reactant ProblemsProblems


Limiting Reactant Limiting Reactant Problems:Problems:

Given the following reaction:Given the following reaction:2Cu + S 2Cu + S Cu Cu22SS

• What is the limiting reactant when What is the limiting reactant when 82.0 g of Cu reacts with 25.0 g S?82.0 g of Cu reacts with 25.0 g S?

• What is the maximum amount of What is the maximum amount of CuCu22S that can be formed? S that can be formed?

• How much of the other reactant is How much of the other reactant is wasted?wasted?

Our 1Our 1stst goal is to calculate how goal is to calculate how much S would react if all of the Cu much S would react if all of the Cu was reacted.was reacted.

From that we can determine the From that we can determine the limiting reactant (LR).limiting reactant (LR).

Then we can use the Limiting Then we can use the Limiting Reactant to calculate the amount Reactant to calculate the amount of product formed and the amount of product formed and the amount of excess reactant left over.of excess reactant left over.82g Cu82g Cu mol Cumol Cu mol Smol S g Sg S


2Cu + S Cu2S2Cu + S Cu2S

82.0gCu82.0gCu1molCu1molCu

63.5gCu63.5gCu

1mol S1mol S

2molCu2molCu

32.1g S32.1g S

1mol S1mol S

=20.7 g S So if all of our 82.0g of Copper were So if all of our 82.0g of Copper were reacted completely it would require reacted completely it would require only only 20.7 grams20.7 grams of Sulfur. of Sulfur.

Since we initially had 25g of S, we are Since we initially had 25g of S, we are going to run out of the Cu, the limiting going to run out of the Cu, the limiting reactant) & end up with reactant) & end up with 4.3 grams of S4.3 grams of S

Copper being our Limiting Reactant is then used to determine how much product is produced.

The amount of Copper we initially start with limits the amount of product we can make.

82.0gCu82.0gCu

1molCu1molCu

63.5gCu63.5gCu 1molCu1molCu22SS

2molCu2molCu22SS________________159gCu159gCu22SS

1molCu1molCu22SS

= 103 g = 103 g CuCu22SS


So the reaction between 82.0g of Cu and 25.0g of S can only produce 103g of Cu2S.

The Cu runs out before the S and we will end up wasting 4.7 g of the S.


Percent YieldPercent Yield


Calculating Percent Calculating Percent YieldYield

In theory, when a teacher gives an In theory, when a teacher gives an exam to the class, every student exam to the class, every student should get a grade of 100%. Sadly, should get a grade of 100%. Sadly, this is not always true. The this is not always true. The calculation for percent yield is calculation for percent yield is similar.similar.

We already know that we do not get We already know that we do not get a 100% yield of products in an a 100% yield of products in an reaction. reaction.

Consider the Following Reactions:

Mg + 2HCl Mg + 2HCl MgCl MgCl22 + H + H22

5.0 g Mg is reacted with an excess of 5.0 g Mg is reacted with an excess of HCL. How much MgClHCL. How much MgCl2 2 will be produced.will be produced.

= 21.6 g of MgClMgCl22

5.0g 5.0g MgMg24.3 g Mg24.3 g Mg

1molMg1molMg

1mol Mg1mol Mg

1mol MgCl1mol MgCl22

1mol MgCl1mol MgCl22

105.2 g 105.2 g MgClMgCl22


You might assume that using You might assume that using stoichiometry to calculate that our stoichiometry to calculate that our reaction will produce reaction will produce 21.6 g of MgClMgCl22, but , but we will actually only recover we will actually only recover 15.2 g of MgClMgCl2 2 in the lab.in the lab.

21.6 g of MgClMgCl2 2 is the value representing is the value representing the the theoretical yieldtheoretical yield(theoretical yield is (theoretical yield is the maximum amount of product that the maximum amount of product that could be formed).could be formed).

The The 15.2 g of MgClMgCl2 2 is called the is called the actual actual yieldyield (the actual yield is less than the (the actual yield is less than the theoretical yield).theoretical yield).


The percent yield is the ratio of the actual yield to the theoretical yield as a percent

It measures the measures the efficiency of the reaction.

Percent yield=Percent yield=actual yieldactual yield

theoretical yieldtheoretical yieldx 100x 100


measured in lab

calculated on paper

Why do reactions not go to completion. Impure reactants and competing side

rxns may cause unwanted products to form.

Actual yield can also be lower than the theoretical yield due to a loss of product during filtration or transferring between containers.

If a wet precipitate is recovered it might weigh heavy due to incomplete drying, etc. .


Calcium carbonate is Calcium carbonate is synthesized by heating, as synthesized by heating, as shown in the following shown in the following equation: CaO + COequation: CaO + CO22 CaCO CaCO33• What is the theoretical yield of CaCOWhat is the theoretical yield of CaCO33 if if

24.8 g of CaO is heated with 43.0 g of 24.8 g of CaO is heated with 43.0 g of COCO22??

• What is the percent yield if 33.1 g of What is the percent yield if 33.1 g of CaCOCaCO33 is produced? is produced?

• Determine which reactant is the Determine which reactant is the limiting and then decide what the limiting and then decide what the theoretical yield is.theoretical yield is.


24.8 24.8 gg

CaOCaO

1molCa1molCaOO

56g CaO56g CaO

1mol CO1mol CO22

1mol CaO1mol CaO44 g CO44 g CO22

1molCO1molCO22

= 19.5gCO2= 19.5gCO2

24.8gCaO24.8gCaOmolCaOmolCaOmol COmol CO22gCOgCO22

24.8 g24.8 gCaOCaO

1mol 1mol CaOCaO

56g CaO56g CaO

1molCaC1molCaCOO331mol 1mol

CaOCaO

100g CaCO100g CaCO33

1molCaCO1molCaCO33

= 44.3 g CaCO3= 44.3 g CaCO3

24.8gCaO24.8gCaOmolCaOmolCaOmol CaCOmol CaCO33gCaCOgCaCO33

LRLR

CaO is our LR, so the reaction CaO is our LR, so the reaction should theoretically produce 44.3 g should theoretically produce 44.3 g of CaCOof CaCO3 3 (How efficient were we?)(How efficient were we?)

Our percent yield is:Our percent yield is:

Percent yield=Percent yield=33.1 g CaCO333.1 g CaCO3

44.3 g CaCO344.3 g CaCO3

__________________________x 100x 100

Percent yield = 74.7%Percent yield = 74.7%


GOOD LUCK!!

Mrs. Armani

Mr. Buchanan

Ms. Nichols

Mr. Smith

Stoichiometry and the Mole SOL Review. The Mole Stoichiometry and the Mole.

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