 # AP Chemistry Chapter 3 Chemical Quantities (The Mole)

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AP Chemistry Chapter 3 Chemical Quantities (The Mole)

Amedeo Avogadro Hypothesis: Equal volumes of different gases at the same temperature and pressure contain equal numbers of particles. H2 hydrogen CH4 methane Amadeo Avogadro

The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.02 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.

I didn’t discover it. Its just named after me!
Avogadro’s Number 6.02 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amedeo Avogadro ( ). I didn’t discover it. Its just named after me! Amadeo Avogadro

The Mole (Quantities) Mass: 1 mole = Molar mass (periodic table)
Volume: 1 mole = 22.4 L for a Representative Particles: 1 mole = 6.02 x 1023 atoms molecules formula units (elements) (nonmetals) (cation-anion)

1. Convert 3.00 moles of ammonia, NH3, to grams of ammonia.

You have to find molar mass.
1. Convert 3.00 moles of ammonia, NH3, to grams of ammonia. Conversion factor: 1 mole NH3 = ?? g NH3 g NH3 3.00 mol NH3 x = g NH3 mol NH3 You have to find molar mass. (Use Periodic Table)

NH3 Molar mass of ammonia. N H 1 mole NH3 = 17.04 g NH3 14.01 g 1.01 g
x 1 x 3 14.01 g g = 17.04 g NH3 1 mole NH3 = g NH3

1. Convert 3.00 moles of ammonia, NH3, to grams of ammonia.
Conversion factor: 1 mole NH3 = g NH3 1 17.04 g NH3 3.00 mol NH3 51.1 g NH3 x = g NH3 mol NH3 NH3 On calculator: 3.00 x = 51.12 x x 3 14.01 g g = g/mol Round to 3 sig. figs.

Ch. 7 (Calculating Chemical Quantities)
17.04 g NH3 1 mol NH3 1. = 51.1 g NH3 x 1 x 3 14.01 g g = g/mol 2. 3.

2. How many molecules of carbon dioxide are in 2.00 moles of CO2?

1 mole = 6.02 x 1023 representative particles
How many molecules of carbon dioxide are in 2.00 moles of CO2? Conversion factor: 1 mole CO2 = 6.02 x 1023 molecules CO2 molec. CO2 2.00 mol CO2 x = molec. CO2 mol CO2 1 mole = 6.02 x 1023 representative particles CO2 is made of all nonmetals so rep. particles are “molecules”

How many molecules of carbon dioxide are in 2.00 moles of CO2?
Conversion factor: 1 1 mole CO2 = 6.02 x 1023 molecules CO2 6.02 x 1023 molec. CO2 2.00 mol CO2 x = 1.20 x 1024 molec. CO2 molec. CO2 mol CO2 On calculator: 2.00 x (6.02 x 1023) = x 1024 Round to 3 sig. figs.

Ch. 7 (Calculating Chemical Quantities)
17.04 g NH3 1 mol NH3 1. = 51.1 g NH3 x 1 x 3 14.01 g g = g/mol 2. 1.20 x 1024 molec. CO2 3.

1 mole = 22.4 L (for any gas @ STP)
3. How many moles of oxygen are in 44.8 L of oxygen, O2? Conversion factor: 1 1 mole O2 = L O2 22.4 mol O2 44.8 L O2 x = 2.00 moles O2 moles O2 L O2 On calculator: 44.8 ÷ 22.4 = 2 Show to 3 sig. figs. 1 mole = 22.4 L (for any STP)

Ch. 7 (Calculating Chemical Quantities)
17.04 g NH3 1 mol NH3 1. = 51.1 g NH3 x 1 x 3 14.01 g g = g/mol 2. 1.20 x 1024 molec. CO2 3. 2.00 mol O2

24.3 g Li 4. How many grams of lithium are in 3.50 moles of lithium?
Conversion factor: 1 1 mole Li = g Li 6.94 g Li 3.50 mol Li x = 24.3 g Li g Li mol Li On calculator: 3.50 x 6.94 = 24.29 Round to 3 sig. figs.

The Mole (Quantities) Mass: 1 mole = Molar mass (periodic table)
Volume: 1 mole = 22.4 L for a Representative Particles: 1 mole = 6.02 x 1023 atoms molecules formula units (elements) (nonmetals) (cation-anion)

2.62 mol Li 5. How many moles of lithium are in 18.2 grams of lithium?
Conversion factor: 1 1 mole Li = g Li 6.94 mol Li 18.2 g Li x = 2.62 mol Li mol Li g Li On calculator: 18.2 ÷ 6.94 = Round to 3 sig. figs.

Use 6.02 x 1023 when looking for atoms, molecules, or formula units 6. How many atoms of lithium are in 3.50 moles of lithium? 6.02 x 1023 atoms Li 3.50 mol Li x = atoms Li 2.11 x 1024 1 mol Li

Representative Particles: Two-part problem
7. How many atoms of lithium are in 18.2 g of lithium? 6.02 x 1023 atoms Li 1 mol Li 18.2 g Li x x 6.94 g Li 1 mol Li (18.2)/6.94 x (6.02 x 1023) = atoms Li 1.58 x 1024

8. A sample of sodium chloride, (NaCl) commonly called table salt contains 7.86 g of sodium and g of chlorine. Find the % composition. Find total mass of NaCl 7.86 g g Total mass = g Take the mass of the element divided by total mass. 7.86 g %Na = X 100 = 30.23 % Na 26.00 g 18.14 g %Cl = X 100 = 69.77 % Cl 26.00 g

C3H8 C H = = 9. Find the % composition of propane, (C3H8). 12.01 g
6 C 12.01 1 H 1.01 12.01 g 1.01 g x 3 x 8 36.03 g g = 44.11 g C3H8 36.03 g %C = X 100 = 81.68 % C 44.11 g 8.08 g %H = X 100 = 18.32 % H 44.11 g

10. A 12. 8 g sample of a gas contains 6. 4 grams of sulfur and 6
10. A 12.8 g sample of a gas contains 6.4 grams of sulfur and 6.4 grams of oxygen. What is the empirical formula for this gas?

Formulas empirical formula = CH
Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound. molecular formula = (empirical formula)n where n = integer molecular formula = C6H6 = (CH)6 empirical formula = CH

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11 All can be divided by 6

Empirical Formula Determination
Convert grams values to moles for each element. Divide by lowest moles. If necessary: Multiply each number by an integer to obtain all whole numbers.

S O SO2 Empirical Formula S O x x
10. A 12.8 g sample of a gas contains 6.4 grams of sulfur and 6.4 grams of oxygen. What is the empirical formula for this gas? 1. Convert to moles. 16 S 32.07 8 O 16.00 2. Divide by lowest moles. 1 mol S x 6.4 g S = mol S = 0.20 mol S 1 32.07 g S 0.20 mol 1 mol O 6.4 g O x = 0.4 = 0.40 mol O 2 16.00 g O 0.20 mol S O SO2 Empirical Formula

H C O H2CO3 Empirical Formula H C O Assume 100 g sample = 3.16 mol H
11. An unknown clear colorless liquid with no odor is analyzed and found to contain the following. Determine the empirical formula. 3.2 % Hydrogen 19.4% Carbon 77.4% Oxygen = 3.2 g H = 19.4 g C = 77.4 g O 1. Convert to moles. 2. Divide by lowest moles. 1 H 1.01 6 C 12.01 8 O 16.00 Assume 100 g sample 1 mol H 3.2 g H = 3.16 mol H = 3.2 mol H x 2 1.01 g H 1.6 mol 1 mol C 19.4 g C x = 1.62 = 1.6 mol C 1 12.01 g C 1.6 mol 1 mol O 77.4 g O x = 4.84 = 4.8 mol O 3 16.00 g O 1.6 mol H C O H2CO3 Empirical Formula

8. A sample of sodium chloride, (NaCl) commonly called table salt contains 7.86 g of sodium and g of chlorine. Find the % composition. NaCl 11 Na 22.99 17 Cl 35.45 22.99 g 35.45 g x 1 x 1 22.99 g g = 58.44 g NaCl 22.99 g %Na = X 100 = 39.34 % Na 58.44 g 35.45 g %O = X 100 = 60.66 % Cl 58.44 g

The Mole (Quantities) Mass: 1 mole = Molar mass (periodic table)
Volume: 1 mole = 22.4 L for a Representative Particles: 1 mole = 6.02 x 1023 atoms molecules formula units (elements) (nonmetals) (cation-anion)