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Molar Relationships. Part 4: Molar Relationships The mole and molar calculations Stoichiometry Solution Concentrations Acid/Base Theory. You will need a calculator and periodic table to complete this section. The Mole and Mole Calculations.

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Part 4: Molar Relationships The mole and molar calculations Stoichiometry Solution Concentrations

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Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

Molar Relationships

  • Part 4: Molar Relationships

  • The mole and molar calculations

  • Stoichiometry

  • Solution Concentrations

  • Acid/Base Theory

You will need a calculator and periodic table to complete this section.


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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?


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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 39.948 g Ar = 16 g Ar

22.4 L Ar 1 mole Ar


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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?


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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 exam question:

What is the empirical formula of a compound that contains 30% Nitrogen and 70% Oxygen?

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

a) N2O

b) NO2

c) N2O5

d) NO


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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?


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

Mole ratio

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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

Then calculate molarity of solution:

2.00 mol HCl/5.0 L = 0.40 M HCl


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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

Then calculate molarity of solution (remember convert mL to Liters):

0.046 mol LiF/0.050 L = 0.95 M LiF


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

Chemical Equilibrium

Exothermic and Endothermic Reactions

Catalysts lower the Activation energy barrier, making reactions faster.


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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:

Taste _______

turn litmus paper red.

Arrhenius bases

Taste _______

Feel __________

Turn litmus paper blue.

SAFETY NOTES

Always add acid to water when diluting

If you spill acid or base on yourself, rinse with lots of water.

sour

bitter

slippery


Part 4 molar relationships the mole and molar calculations stoichiometry solution concentrations

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 ___

  • Example: A 1.0 x 10-4 molar solution of KOH would have a pOH of ___

  • Memorize: pH + pOH = 14.

  • Example: A solution with a pH of 8 will have a pOH of: ____.

3

4

6


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