Chemistry SOL Review
by Anne Mooring (Jamestown High School, Williamsburg VA, 2006)
Part 4: Molar Relationships
•
The mole and molar calculations
•
Stoichiometry
•
Gas Laws (Boyle, Charles, Combined, Ideal, Dalton, Graham)
•
Solution Concentrations
•
Chemical Equilibrium
•
Acid/Base Theory
Use the SOL periodic table. Click here for link
You will need a calculator and periodic table
to complete this section.
This section represents 8/50 of the SOL questions
Chemistry SOL 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 SOL 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 SOL 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 SOL 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 SOL 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 SOL 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 SOL 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 SOL 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.
A favorite SOL type question:
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 SOL 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 SOL 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 SOL 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 SOL 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) → 2NH3(g)
Given: 9 moles H2, Find moles N2
9 mol H2 x 2 mol N2 = 6 mol N2
3 mol H2
Mole ratio
Chemistry SOL 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 SOL Review— Molar Relationships
Gas Laws
1.
General Properties of Gases
There is a lot of “free” space in a gas.
Gases can be expanded infinitely.
Gases fill containers uniformly and completely.
Gases diffuse and mix rapidly.
The Gas Laws
The Combined Gas Law
Boyle’s Law
Inverse relationship
Charles Law
P1V1 = P2V2
T1
T2
P1V1 = P2V2
V1
T1
=
V2
T2
Always use degrees
Kelvin
°C + 273 = K
Chemistry SOL Review— Molar Relationships
Gas Laws
P1V1
Some Problems
A balloon contains 8.0 liters of gas at 100 K. What is the
balloon’s volume at 200K?
8
Answer:
=
100
= 16 Liters
200
A balloon contains 10. Liters at 3 atmospheres and 275 K. What is
the volume of the balloon at 0.50 atmospheres and 200K?
Answer:
(3.0)(10) = (0.50)V2
275
T1
200
= 45 Liters
P2V2
T2
P1V =
P2V2
1
V1
V2
=
T1
=
V2
T2
Chemistry SOL Review— Molar Relationships
Gas Laws
The Ideal Gas Law
Memorize: PV = nRT
•P= pressure in kPa
•V= liters
•N= moles
•T= temperature in Kelvin
•R = universal gas law constant = 8.31
kPa x L
Moles x K
•The SOL test uses
Chemistry SOL Review— Molar Relationships
Gas Laws
The Ideal Gas Law
R = 8.31 kPa x L
Memorize: PV = nRT
Example 1:A 15 liter tank contains 2.0 moles of nitrogen gas
at 27 °C. What is the pressure of nitrogen inside the tank?
Answer:
P=?, V=15 L, n=2.0, T=300K (remember to convert)
P(15)=2.0(8.31)(300) so P = 332.4 kPa
You try: How many moles of Hydrogen gas are in a 20. L tank
pressurized to 1000. kPa at 300K?
Answer:
P=1000., V=20. L, n=? T=300K
(1000.)(20) = n(8.31)(300) so n = 8.0 moles Hydrogen
Moles x K
Chemistry SOL Review— Molar Relationships
Gas Laws
Dalton’s Law of Partial Pressures
Memorize: Ptotal = P1 + P2 + P3 + …
Example A tank containing nitrogen, hydrogen and ammonia gas has a total
pressure of 12 atmospheres. The partial pressure of the hydrogen is 6
atmospheres, the partial pressure of the ammonia is 4 atmospheres. What is
the partial pressure of the nitrogen?
Answer:
Ptotal = 12 atm, PN2=?, PH2=6, PNH3=4
12 = PN2 + 6 + 4 so PN2 = 2
Chemistry SOL 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 SOL 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 SOL 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 SOL Review— Molar Relationships
Solution Concentrations
Diluting concentrated solutions
Memorize: M1V1
= M2V2
•M1 and V1 are the beginning molarities and volumes
•M2 and V2 are the ending molarities and volumes
•V1 and V2 can be in Liters or mLs, but must be the same units for both
Example:
What is the molarity of a 10. mL sample of 2.0 M aqueous HCl diluted to
40. mL
Answer:
(2.0)(10.) = (M2)(40.) so M2 = 0.5 Molar HCl
Chemistry SOL Review— Molar Relationships
Solution Concentrations
Diluting concentrated solutions
Memorize: M1V1
= M2V2
•M1 and V1 are the beginning molarities and volumes
•M2 and V2 are the ending molarities and volumes
•V1 and V2 can be in Liters or mLs, but must be the same units for both
You try one:
How many milliliters of 6.0 Molar HCl are required to prepare 240 mL of
2.0 Molar HCl?
Answer:
(6.0)(V1) = (2.0)(240) so V1 = 80. mL HCl
Chemistry SOL 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 SOL 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 SOL Review--Molar Relationships
Chemical Equilibrium
Reversible Reactions
Le Chatelier’s Principle: If a system at equilibrium is stressed, the
equilibrium will shift in a direction that relieves that stress.
Equilibrium will shift AWAY from what is added. Here, N2 is added.
N2
More “product” made
N2(g) + 3H2(g)  2 NH3(g) + heat
Chemistry SOL Review--Molar Relationships
Chemical Equilibrium
Reversible Reactions
Le Chatelier’s Principle: If a system at equilibrium is stressed, the
equilibrium will shift in a direction that relieves that stress.
Equilibrium will shift AWAY from what is added. Here, NH3 is added.
More “reactants” made
NH3
N2(g) + 3H2(g)  2 NH3(g) + heat
Chemistry SOL Review--Molar Relationships
Chemical Equilibrium
Reversible Reactions
Le Chatelier’s Principle: If a system at equilibrium is stressed, the
equilibrium will shift in a direction that relieves that stress.
Equilibrium will shift TOWARDS what is removed. Here H2 is removed.
H2
N2(g) + 3H2(g)  2 NH3(g) + heat
More “reactants”
made
Chemistry SOL Review--Molar Relationships
Chemical Equilibrium
Reversible Reactions
Le Chatelier’s Principle: If a system at equilibrium is stressed, the
equilibrium will shift in a direction that relieves that stress.
Equilibrium will shift TOWARDS what is removed. Here heat is
removed.
heat
N2(g) + 3H2(g)  2 NH3(g) + heat
More “product” made
Chemistry SOL Review--Molar Relationships
Chemical Equilibrium
Methods to Speed up Reactions:
•Use a catalyst
•Reduce the particle size
•Increase the heat
•Increase reactant concentration
Chemistry SOL Review--Molar Relationships
Acid/Base Theory
Acids and Bases
Generic formula for acids = HX (HCl, HNO3, H2SO4)
Generic formula for bases = MOH where M is any metal (NaOH, KOH,
Ca(OH)2 Ammonia, NH3, is also a base.
Acid solutions have a pH less than 7
Basic solutions have a pH more than 7
Arrhenius acids:
sour
Taste _______
turn litmus paper red.
SAFETY NOTES
Arrhenius bases
Taste _______
bitter
slippery
Feel __________
Turn litmus paper blue.
If you spill acid or base on
yourself, rinse with lots of water.
Always add acid to water when
diluting
Chemistry SOL Review--Molar Relationships
Acid/Base Theory
What is pH?
pH indicates the hydrogen ion molarity [H+] in a solution
pH = -log[H+]
pOH indicates the hydroxide ion molarity [OH-] in a solution.
pOH = -log[OH-]
Example: A 1.0 x 10-3 molar solution of HCl would have a pH of ___
3
4
Example: A 1.0 x 10-4 molar solution of KOH would have a pOH of ___
Memorize: pH + pOH = 14.
6
Example: A solution with a pH of 8 will have a pOH of: ____.
Chemistry SOL Review--Molar Relationships
References
http://www.markrosengarten.com/ for New York Regent’s exam
powerpoint.