Transcript
Page 1: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Chapter 3:Stoichiometry

3.1 & 3.2 Atomic Masses

3.3 The Mole

3.4 Molar Mass

3.5 Percent Composition

3.6 Chemical Formulas

Page 2: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

AtomicMass

Page 3: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Atomic Mass• Atoms are so small, it is difficult to discuss

how much they weigh in common units.

• The decimal numbers on the table are atomic masses in amu.

• Atomic masses are not decimals because they are based on averages of atoms & of isotopes.

• The average atomic mass is from the mass of the isotopes and their relative abundance.

Page 4: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Atomic Mass Examples• There are two isotopes of carbon 12C with a

mass of 12.00000 amu(98.892%), and 13C with a mass of 13.00335 amu (1.108%).

• There are two isotopes of nitrogen, one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each?

Page 5: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Moles

Page 6: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Avogadro’s Number

• The mole is a number.

• A very large number, but still, just a number.

• 6.022 ´ 1023 of anything is a mole

• A large dozen.

Page 7: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Avogadro’s Number

1 mole of 12C has 6.02 ´ 1023 atoms and a mass of 12 g

Page 8: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Molar Mass

• By definition, these are the mass of 1 mol of a substance (i.e., g/mol).

–The molar mass of an element is the mass number for the element that we find on the periodic table

–Often called molecular weight or formula weight (in g/mol)

Page 9: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

Page 10: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.

• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

Page 11: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Percent Composition

Page 12: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Percent Composition

One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

% element = ´ 100mass of element in sample

total mass of sample

Page 13: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Percent Composition

So the percentage of carbon in ethane, C2H6, is…

%C =(2)(12.0 amu)

(30.0 amu)

24.0 amu

30.0 amu= x 100

= 80.0%

Page 14: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Empirical Formulas

Page 15: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

One can calculate the empirical formula from the percent composition.

Calculating Empirical Formulas

Page 16: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Calculating Empirical Formulas

The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

Page 17: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g × = 5.105 mol C

H: 5.14 g × = 5.09 mol H

N: 10.21 g × = 0.7288 mol N

O: 23.33 g × = 1.456 mol O

1 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g

Calculating Empirical Formulas

Page 18: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Calculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol

Calculating Empirical Formulas

Page 19: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Calculating Empirical Formulas

These are the subscripts for the empirical formula:

C7H7NO2

Page 20: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Combustion Analysis

• Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like this– C is determined from the mass of CO2 produced.

– H is determined from the mass of H2O produced.

– O is determined by difference after the C and H have been determined.

Page 21: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Elemental Analyses

Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.

Page 22: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Molecular Formulas

Page 23: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Empirical To Molecular Formulas

• Empirical is lowest ratio.• Molecular is actual molecule.• Need Molar mass.• Ratio of empirical to molar mass will tell

you the molecular formula.• Must be a whole number because...

Page 24: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Example

• A compound is made of only sulfur and oxygen. It is 50.0% S by mass. Its molar mass is 192 g/mol. What is its formula?

Page 25: Chapter 3: Stoichiometry 3.1 & 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass

Chapter 3:Stoichiometry

3.1 & 3.2 Atomic Masses

3.3 The Mole

3.4 Molar Mass

3.5 Percent Composition

3.6 Chemical Formulas


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