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## Stoichiometry and the Mole

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### Stoichiometryand the Mole

### Stoichiometryand the Mole

### Stoichiometryand the Mole

### Stoichiometryand 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!

Road Map

Moles to Mass Conversions

- 1 Moles = Molar Mass (g) =

Molar mass/Mole or (g/mol)

Now let’s apply that knowledge!

Molar Mass Calculations

Reminder – Finding Molar Mass:

The molar mass = the sum of all the atomic masses.

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

Example: Ca(NO3)2

Ca = 40.08

N = 2(14.01)

O = 6(16.00)

164.10 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

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 O21 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 -> mass

0.00563 mol NH3 x 17 g NH3=9.57 x 10 –2 g NH31 mol NH3

Mole Calculations

- 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

- 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 mlcs1 mol

Stoichiometry

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

Stoichiometry

- 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.

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

Stoichiometry

1. Write the balanced equation

17.0g

?g

AgNO3 + NaCl AgCl + NaNO3

2. Given and asked for

3. Moles of given

17.0g AgNO3 x 1 mol =

170 g

0.100 mol AgNO3

StoichiometryMass-Mass Problem

AgNO3 + NaCl AgCl + NaNO3

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

Limiting Reactant Problems:

Given the following reaction:

2Cu + S Cu2S

- What is the limiting reactant when 82.0 g of Cu reacts with 25.0 g S?
- What is the maximum amount of Cu2S that can be formed?
- How much of the other reactant is wasted?

Limiting Reactant Problems:

- Our 1st goal is to calculate how much S would react if all of the Cu was reacted.
- From that we can determine the limiting reactant (LR).
- Then we can use the Limiting Reactant to calculate the amount of product formed and the amount of excess reactant left over.

- 82g Cu

- mol Cu

- mol S

- g S

- So if all of our 82.0g of Copper were reacted completely it would require only 20.7 grams of Sulfur.
- Since we initially had 25g of S, we are going to run out of the Cu, the limiting reactant) & end up with 4.3 grams of S

1molCu

1mol S

32.1g S

82.0gCu

63.5gCu

2molCu

1mol S

=20.7 g S

Limiting Reactant Problems:

- 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.

1molCu

159gCu2S

2molCu2S

82.0gCu

1molCu2S

63.5gCu

1molCu2S

= 103 g Cu2S

Limiting Reactant Problems:

- 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 Yield

Calculating Percent Yield

- In theory, when a teacher gives an exam to the class, every student should get a grade of 100%. Sadly, this is not always true. The calculation for percent yield is similar.
- We already know that we do not get a 100% yield of products in an reaction.

Calculating Percent Yield

- Consider the Following Reactions:

Mg + 2HCl MgCl2 + H2

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

= 21.6 g of MgCl2

5.0g Mg

1molMg

1mol MgCl2

105.2 g MgCl2

24.3 g Mg

1mol MgCl2

1mol Mg

Calculating Percent Yield

- You might assume that using stoichiometry to calculate that our reaction will produce 21.6 g of MgCl2, but we will actually only recover 15.2 g of MgCl2 in the lab.
- 21.6 g of MgCl2 is the value representing the theoretical yield(theoretical yield is the maximum amount of product that could be formed).
- The 15.2 g of MgCl2 is called the actual yield (the actual yield is less than the theoretical yield).

Calculating Percent 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.

measured in lab

actual yield

Percent yield=

x 100

theoretical yield

calculated on paper

Calculating Percent Yield

- 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.

Calculating Percent Yield

Calcium carbonate is synthesized by heating, as shown in the following equation: CaO + CO2 CaCO3

- What is the theoretical yield of CaCO3 if 24.8 g of CaO is heated with 43.0 g of CO2?
- What is the percent yield if 33.1 g of CaCO3 is produced?
- Determine which reactant is the limiting and then decide what the theoretical yield is.

- molCaO

- mol CaCO3

- gCaCO3

- 24.8gCaO

- molCaO

- mol CO2

- gCO2

24.8 g

CaO

1molCaO

1mol CO2

44 g CO2

56g CaO

1mol CaO

1molCO2

LR

= 19.5gCO2

1mol CaO

100g CaCO3

24.8 g

CaO

1molCaCO3

56g CaO

1mol CaO

1molCaCO3

= 44.3 g CaCO3

Calculating Percent Yield

- CaO is our LR, so the reaction should theoretically produce 44.3 g of CaCO3 (How efficient were we?)
- Our percent yield is:

33.1 g CaCO3

Percent yield=

x 100

44.3 g CaCO3

Percent yield = 74.7%

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