Chemistry Review
Part 4: Molar Relationships
•
The mole and molar calculations
•
Stoichiometry
•
Solution Concentrations
•
Chemical Equilibrium
You will need a calculator and periodic table
to complete this section.
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Chemistry Review— Molar Relationships
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?
Answer:
2.0 moles Ne x 22.4 Liters Ne = 44.8 Liters Ne
1.0 moles Ne
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
Chemistry Review— Molar Relationships
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.
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
Chemistry Review— Molar Relationships
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.
You try one:
What is the mass of 9.0 Liters of Argon gas at 0°C and one atmosphere of
pressure?
Chemistry Review— Molar Relationships
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.
You try one:
What is the mass of 9.0 Liters of Argon gas at 0°C and one atmosphere of
pressure?
9.0 L Ar x 1 mol Ar x
22.4 L Ar
39.948 g Ar = 16 g Ar
1 mole Ar
Chemistry Review— Molar Relationships
The Mole and Mole Calculations
The molar mass = the sum of all the atomic masses.
Example Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 grams
You try one:
What is the gram formula mass (molar mass) of Mg3(PO4)2?
Chemistry Review— Molar Relationships
The Mole and Mole Calculations
The molar mass = the sum of all the atomic masses.
Example Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 grams
You try one:
What is the gram formula mass (molar mass) of Mg3(PO4)2?
3(24.305) + 2(30.97376) + 8(15.9994) = 262.86 grams
Chemistry Review— Molar Relationships
The Mole and Mole Calculations
The molar mass = the sum of all the atomic masses.
Example Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 grams
You try one:
What is the gram formula mass (molar mass) of Mg3(PO4)2?
3(24.305) + 2(30.97376) + 8(15.9994) = 262.86 grams
What is the percent Magnesium in Mg3(PO4)2?
Answer: 3(24.305) x 100 = 27.7%
262.86
You try one:
What is the percent Lithium in Li2SiO3?
molar mass = 2(6.941) + 28.0855 + 3(15.9994) = 89.9657 g
% Li = 2(6.941) x 100 = 15.4%
89.9657
Chemistry Review— Molar Relationships
A Brief Return to Empirical Formulas
Empirical Formulas are the reduced form of Molecular formulas.
For example: The empirical formula for C5H10 is CH2.
What is the empirical formula of a compound that contains 30% Nitrogen and
70% Oxygen?
a) N2O
b) NO2
c) N2O5
d) NO
This is really a percent
composition problem. Figure
out which compound contains
30% nitrogen.
Chemistry Review— Molar Relationships
The Mole and Mole Calculations
At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L
You can use this to calculate the density of a gas in g/Liter at STP.
Example: What is the density of CO2 gas at STP?
The molar mass of CO2 = 12.0111 + 2(15.9994) = 44.0099 g
Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L
Chemistry Review— Molar Relationships
The Mole and Mole Calculations
At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L
You can use this to calculate the density of a gas in g/Liter at STP.
Example: What is the density of CO2 gas at STP?
The molar mass of CO2 = 12.0111 + 2(15.9994) = 44.0099 g
Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L
You try one:
What is the density of Cl2 gas at STP?
Chemistry Review— Molar Relationships
The Mole and Mole Calculations
At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L
You can use this to calculate the density of a gas in g/Liter at STP.
Example: What is the density of CO2 gas at STP?
The molar mass of CO2 = 12.0111 + 2(15.9994) = 44.0099 g
Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L
You try one:
What is the density of Cl2 gas at STP?
Answer: molar mass = 2(35.453) = 70.906 g
70.906 g/22.4 L = 3.165 g/L
Chemistry Review— Molar Relationships
Stoichiometry
For reaction calculations, the molar ratio is used.
Example:
How many moles of nitrogen will react with 9 moles of hydrogen to
produce ammonia according to this equation?
2N2(g) +3 H2(g) → 3NH3(g)
Given: 9 moles H2, Find moles N2
9 mol H2 x 2 mol N2 = 6 mol N2
3 mol H2
Mole ratio
Chemistry Review— Molar Relationships
Stoichiometry
For reaction calculations, the molar ratio is used.
Example 2:
How many grams of nitrogen are needed to react with 2.0 grams of
hydrogen using this equation?
2N2(g) +3 H2(g) → 3NH3(g)
Given: 2.0 grams H2, Find grams N2
2.0 g H2 x 1 mol H2
2.016g H2
x 2 mol N2 x 28.014 g N2
3 mol H2
1 mol N2
= 18.53 g N2
Chemistry Review— Molar Relationships
Solution Concentrations
Calculating molarity:
Memorize this equation: Molarity = moles/liters or M = mol/L
Memorize conversion factor: 1000 mL = 1 L
Some example of using this equation:
Example 1: the molarity of 2.0 moles of HCl in a 0.50 L solution of water is:
molarity = 2.0 mole HCl/0.50 L = 4.0 Molar or 4 M
Example 2: The molarity of 0.40 moles of HCl in a 300. mL L solution of
water is:
molarity = 0.40 moles HCl/0.300. L = = 1.3 M
Chemistry Review— Molar Relationships
Solution Concentrations
Example 3:
The molarity of 72.9 g of HCl in 5.0 liters of aqueous solution is:
Answer: first calculate the moles of HCl
72.9 g HCl
x 1 mol HCl
= 2.00 mol HCl
36.46 g HCl
Then calculate molarity of solution:
2.00 mol HCl/5.0 L = 0.40 M HCl
Chemistry Review— Molar Relationships
Solution Concentrations
You try one:
What is the molarity of 1.2 grams LiF in a 50. mL aqeous solution?
Answer: first calculate the moles of LiF
1.2 g LiF
x
1 mol LiF
= 0.046 mol LiF
25.94 g LiF
Then calculate molarity of solution (remember convert mL to Liters):
0.046 mol LiF/0.050 L = 0.95 M LiF
Chemistry Review--Molar Relationships
Chemical Equilibrium
Exothermic and Endothermic Reactions
Catalysts lower the Activation energy barrier, making reactions faster.
100
100
Joules >
Endothermic reactions absorb heat
Joules >
Exothermic reactions release heat
0
0
rxn progress >
A + B = AB + heat
rxn progress >
A + B + heat = AB
Chemistry Review--Molar Relationships
Chemical Equilibrium
Reversible Reactions
Some reactions are REVERSIBLE, which means that they can go
backwards (from product to reactant)
Example: The reaction between nitrogen and hydrogen, where a “”
indicates a reversible reaction
N2(g) + 3H2(g)  2 NH3(g) + heat
The forward reaction takes place at the same rate as the reverse
reaction. The equilibrium position of products and reactants depends on
the conditions of the reaction. If we change the reaction conditions, the
equilibrium changes.
Chemistry Review--Molar Relationships
Chemical Equilibrium
Methods to Speed up Reactions:
•Use a catalyst
•Reduce the particle size
•Increase the heat
•Increase reactant concentration
Chemistry Review--Molar Relationships
References
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