Chapter 10

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Chapter 10. THE MOLE. Avogadro’s Number. 6.022 x 10 23 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855). I didn’t discover it. Its just named after me!. Amadeo Avogadro. Calculations with Moles: Converting moles to grams. - PowerPoint PPT Presentation

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Chapter 10Chapter 10

THE MOLE

Avogadro’s NumberAvogadro’s Number6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855).

Amadeo Avogadro

I didn’t discover it. Its just named after

me!

Calculations with Moles:Calculations with Moles:Converting moles to gramsConverting moles to grams

How many grams of lithium are in 3.50 moles of lithium?

3.50 mol Li= g Li

1 mol Li

6.94 g Li 45.1

Calculations with Moles:Calculations with Moles:Converting grams to molesConverting grams to moles

How many moles of lithium are in 18.2 grams of lithium?

18.2 g Li= mol Li

6.94 g Li

1 mol Li 2.62

Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number

How many atoms of lithium are in 3.50 moles of lithium?

3.50 mol Li = atoms Li

1 mol Li

6.022 x 1023 atoms Li 2.11 x 1024

Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number

How many atoms of lithium are in 18.2 g of lithium?

18.2 g Li

= atoms Li

1 mol Li 6.022 x 1023 atoms Li

1.58 x 1024

6.94 g Li 1 mol Li

(18.2)(6.022 x 1023)/6.94

Calculating Molar MassCalculating Molar MassCalculate the formula mass of magnesium Calculate the formula mass of magnesium carbonate, MgCOcarbonate, MgCO33..

24.31 g + 12.01 g + 3(16.00 g) 24.31 g + 12.01 g + 3(16.00 g) ==

84.32 g84.32 g

Calculating Percentage Calculating Percentage CompositionComposition

Calculate the percentage composition of Calculate the percentage composition of magnesium carbonate, MgCOmagnesium carbonate, MgCO33..From previous slide:From previous slide:24.31 g + 12.01 g + 3(16.00 g) = 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g84.32 g 24.31 100 28.83%

84.32Mg

12.01 100 14.24%84.32

C 48.00 100 56.93%84.32

O 100.00

FormulasFormulas

molecular formula = (empirical formula)n [n = integer]

molecular formula = C6H6 = (CH)6

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.

FormulasFormulas (continued)(continued)

Formulas for Formulas for ionic compounds ionic compounds are are ALWAYSALWAYS empirical (lowest whole empirical (lowest whole number ratio).number ratio).Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

FormulasFormulas (continued)(continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Empirical Formula DeterminationDetermination

Base calculation on 100 grams of compound.

Determine moles of each element in 100 grams of compound.

Divide each value of moles by the smallest of the values.

Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Empirical Formula DeterminationDetermination

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32 14.107

12.01

g C mol Cmol C

g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O

Empirical Formula Empirical Formula DeterminationDetermination

(part 2)(part 2)

4.107 1.502.74

mol Cmol O

6.78 2.472.74

mol Hmol O

2.74 1.002.74

mol Omol O

Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:

Empirical Formula Empirical Formula DeterminationDetermination

(part 3)(part 3)Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2

33 55 22Empirical formula:C3H5O

2

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

1. Find the formula mass of 1. Find the formula mass of CC33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

2. Divide the molecular mass by 2. Divide the molecular mass by the mass given by the emipirical the mass given by the emipirical formula.formula.

146 273

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g146 273

3. Multiply the empirical formula by 3. Multiply the empirical formula by this number to get the molecular this number to get the molecular formula.formula.

(C(C33HH55OO22) x 2 ) x 2 ==

CC66HH1010OO44