Chemistry Review Part 4: Molar Relationships The mole and molar calculations Stoichiometry Solution Concentrations Chemical Equilibrium \ You will need

$Page 1: Chemistry Review Part 4: Molar Relationships The mole and molar calculations Stoichiometry Solution Concentrations Chemical Equilibrium \ You will need$
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.

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






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






You try one:

What is the mass of 9.0 Liters of Argon gas at 0°C and one atmosphere of pressure?






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 39.948 g Ar = 16 g Ar 22.4 L Ar 1 mole Ar



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?





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





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


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) N2Ob) NO2

c) N2O5

d) NO

This is really a percent composition problem. Figure out which compound contains 30% nitrogen.



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?








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


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


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 x 2 mol N2 x 28.014 g N2 = 18.53 g N2

2.016g H2 3 mol H2 1 mol N2


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



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



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

Exothermic reactions release heat

Endothermic reactions absorb heat

0

100

rxn progress >

Joul

es >

0

100

rxn progress >

Joul

es >

A + B = AB + heat A + B + heat = AB

Catalysts lower the Activation energy barrier, making reactions faster.



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.



Methods to Speed up Reactions:

•Use a catalyst

•Reduce the particle size

•Increase the heat

•Increase reactant concentration


References

http://www.markrosengarten.com/ for New York Regent’s exam powerpoint.

http://www.markrosengarten.com/


Documents

Chemistry Review Part 4: Molar Relationships The mole and molar calculations Stoichiometry Solution Concentrations Chemical Equilibrium \ You will need