1. 4 Calculations and the Chemical Equation GENERAL CHEMISTRY
General, Organic and Biochemistry 7th Edition
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 4.1 The Mole Concept and Atoms
Atoms are exceedingly small
Unit of measurement for mass of an atom is atomic mass unit (amu)
carbon-12 assigned the mass of exactly 12 amu
1 amu = 1.66  g
Periodic table gives atomic weights in amu

3. 4.1 The Mole Concept and Atoms
Mass of Atoms
What is the atomic weight of one atom of fluorine? Answer: amu
What would be the mass of this one atom in grams?
Chemists usually work with much larger quantities
It is more convenient to work with grams than amu when using larger quantities
4.1 The Mole Concept and Atoms

4. The Mole and Avogadro’s Number
A practical unit for defining a collection of atoms is the mole
1 mole of atoms =  1023 atoms
This is called Avogadro’s number
This has provided the basis for the concept of the mole
4.1 The Mole Concept and Atoms

5. 4.1 The Mole Concept and Atoms
To make this connection we must define the mole as a counting unit
The mole is abbreviated mol
A mole is simply a unit that defines an amount of something
Dozen defines 12
Gross defines 144 2
4.1 The Mole Concept and Atoms

6. 4.1 The Mole Concept and Atoms
Atomic Mass
The atomic mass of one atom of an element corresponds to:
The average mass of a single atom in amu
The mass of a mole of atoms in grams
1 atom of F is amu amu/atom F
1 mole of F is g g/mole F
4.1 The Mole Concept and Atoms
=19.00 g/mol F

7. Calculating Atoms, Moles, and Mass
We use the following conversion factors:
Density converts grams  milliliters
Atomic mass unit converts amu grams
Avogadro’s number converts moles  number of atoms
Molar mass converts grams  moles 2
4.1 The Mole Concept and Atoms

8. Strategy for Calculations
Map out a pattern for the required conversion
Given a number of grams and asked for number of atoms
Two conversions are required
Convert grams to moles
1 mol S/32.06 g S OR g S/1 mol S
Convert moles to atoms
mol S  (6.022  1023 atoms S) / 1 mol S
4.1 The Mole Concept and Atoms

9. Practice Calculations
Calculate the number of atoms in 1.7 moles of boron.
Find the mass in grams of 2.5 mol Na (sodium).
Calculate the number of atoms in 5.0 g aluminum.
Calculate the mass of 5,000,000 atoms of Au (gold)
4.1 The Mole Concept and Atoms

10. Interconversion Between Moles, Particles, and Grams
4.1 The Mole Concept and Atoms

11. 4.2 The Chemical Formula, Formula Weight, and Molar Mass3
Chemical formula - a combination of symbols of the various elements that make up the compound
Formula unit - the smallest collection of atoms that provide two important pieces of information
The identity of the atoms
The relative number of each type of atom

12. 4.2 The Chemical Formula, Formula Weight and Molar Mass
Consider the following formulas:
H2 – 2 atoms of hydrogen are chemically bonded forming diatomic hydrogen, subscript 2
H2O – 2 atoms of hydrogen and 1 atom of oxygen, lack of subscript means one atom
NaCl – 1 atom each of sodium and chlorine
Ca(OH)2 – 1 atom of calcium and 2 atoms each of oxygen and hydrogen, subscript outside parentheses applies to all atoms inside
4.2 The Chemical Formula, Formula Weight and Molar Mass

13. 4.2 The Chemical Formula, Formula Weight and Molar Mass
Consider the following formulas:
(NH4)3SO4 – 3 ammonium ions and 1 sulfate ion
Ammonium ion contains 1 nitrogen and 4 hydrogen
Sulfate ion contains 1 sulfur and 4 oxygen
Compound contains 3 N, 12 H, 1 S, and 4 O
CuSO4.5H2O
This is an example of a hydrate - compounds containing one or more water molecules as an integral part of their structure
5 units of water with 1 CuSO4
4.2 The Chemical Formula, Formula Weight and Molar Mass

14. Comparison of Hydrated and Anhydrous Copper Sulfate
4.2 The Chemical Formula, Formula Weight and Molar Mass
Hydrated copper sulfate
Anhydrous copper sulfate
Marked color difference illustrates the fact
that these are different compounds

15. Formula Weight and Molar Mass
Formula weight - the sum of the atomic weights of all atoms in the compound as represented by its correct formula
expressed in amu
What is the formula weight of H2O?
16.00 amu + 2(1.008 amu) = amu
Molar mass – mass of a mole of compound in grams / mole
Numerically equal to the formula weight in amu
What is the molar mass of H2O?
18.02 g/mol H2O 4
4.2 The Chemical Formula, Formula Weight and Molar Mass

16. Formula Unit 4.2 The Chemical Formula, Formula Weight and Molar Mass
Formula unit – smallest collection of atoms from which the formula of a compound can be established
4.2 The Chemical Formula, Formula Weight and Molar Mass
What is the molar mass of (NH4)3PO4?
3(N amu) + 12(H amu) + P amu + 4(O amu)=
3(14.01) + 12(1.008) (16.00)= g/mol (NH4)3PO4

17. 4.2 The Chemical Formula, Formula Weight and Molar Mass
Molar mass - The mass in grams of 1 mole of atoms
What is the molar mass of carbon?
12.01 g/mol C
This means counting out a mole of Carbon atoms (i.e.,  1023) they would have a mass of g
One mole of any element contains the same number of atoms,  1023, Avogadro’s number
4.2 The Chemical Formula, Formula Weight and Molar Mass

18. 4.5 Calculations Using the Chemical Equation
Calculating quantities of reactants and products in a chemical reaction has many applications
Need a balanced chemical equation for the reaction of interest
The coefficients represent the number of moles of each substance in the equation 9

19. 4.5 Calculations Using the Chemical Equation
General Principles
Chemical formulas of all reactants and products must be known
Equation must be balanced to obey the law of conservation of mass
Calculations of an unbalanced equation are meaningless
Calculations are performed in terms of moles
Coefficients in the balanced equation represent the relative number of moles of products and reactants
4.5 Calculations Using the Chemical Equation

20. Using the Chemical Equation
Examine the reaction:
2H2 + O2  2H2O
Coefficients tell us?
2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O
What if 4 moles of H2 reacts with 2 moles of O2?
It yields 4 moles of H2O
4.5 Calculations Using the Chemical Equation

21. Using the Chemical Equation
2H2 + O2  2H2O
The coefficients of the balanced equation are used to convert between moles of substances
How many moles of O2 are needed to react with 4.26 moles of H2?
Use the factor-label method to perform this calculation
4.5 Calculations Using the Chemical Equation

22. Use of Conversion Factors
2H2 + O2  2H2O 1 2
2.13 mol O2
4.5 Calculations Using the Chemical Equation
Digits in the conversion factor come from the balanced equation

23. Conversion Between Moles and Grams9
Requires only the formula weight
Convert 1.00 mol O2 to grams
Plan the path
Find the molar mass of oxygen
32.0 g O2 = 1 mol O2
Set up the equation
Cancel units 1.00 mol O2  32.0 g O2
1 mol O2
Solve equation  32.0 g O2 = 32.0 g O2
moles of
Oxygen
grams of
Oxygen
4.5 Calculations Using the Chemical Equation

24. Conversion of Mole Reactants to Mole Products
Use a balanced equation
C3H8(g) + 5O2(g) CO2(g) + 4H2O(g)
1 mol C3H8 results in:
5 mol O2 consumed mol C3H8 /5 mol O2
3 mol CO2 formed mol C3H8 /3 mol CO2
4 mol H2O formed mol C3H8 /4 mol H2O
This can be rewritten as conversion factors
4.5 Calculations Using the Chemical Equation

25. Calculating Reacting Quantities
Calculate grams O2 reacting with 1.00 mol C3H8
Use 2 conversion factors
Moles C3H8 to moles O2
Moles of O2 to grams O2
Set up the equation and cancel units
1.00 mol C3H8  5 mol O2  g O2 =
1 mol C3H mol O2
1.00  5  32.0 g O2 = x 102 g O2
4.5 Calculations Using the Chemical Equation
moles
C3H8
moles
Oxygen
grams
Oxygen

26. Calculating Grams of Product from Moles of Reactant
Calculate grams CO2 from combustion of 1.00 mol C3H8
Use 2 conversion factors
Moles C3H8 to moles CO2
Moles of CO2 to grams CO2
Set up the equation and cancel units
1.00 mol C3H8  3 mol CO2  g CO2 =
1 mol C3H mol CO2
1.00  3  44.0 g CO2 =  102 g CO2
4.5 Calculations Using the Chemical Equation
moles
C3H8
moles
CO2
grams
CO2

27. Relating Masses of Reactants and Products
Calculate grams C3H8 required to produce 36.0 grams of H2O
Use 3 conversion factors
Grams H2O to moles H2O
Moles H2O to moles C3H8
Moles of C3H8 to grams C3H8
Set up the equation and cancel units
36.0 g H2O  1 mol H2O  1 mol C3H8  44.0 g C3H8
18.0 g H2O mol H2O mol C3H8
36.0  [1/18.0]  [1/4]  44.0 g C3H8 = g C3H8
4.5 Calculations Using the Chemical Equation
grams
H2O
moles
H2O
moles
C3H8
grams
C3H8

28. Calculating a Quantity of Reactant
Ca(OH)2 neutralizes HCl
Calculate grams HCl neutralized by mol Ca(OH)2
Write chemical equation and balance
Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l)
Plan the path
Set up the equation and cancel units
0.500 mol Ca(OH)2  2 mol HCl  36.5 g HCl
1 mol Ca(OH)2 1 mol HCl
Solve equation  [2/1]  36.5 g HCl = 36.5 g HCl
4.5 Calculations Using the Chemical Equation
moles
Ca(OH)2
moles
HCl
grams
HCl

29. A Visual Example of the Law of Conservation of Mass
4.5 Calculations Using the Chemical Equation

30. General Problem-solving Strategy
4.5 Calculations Using the Chemical Equation

31. 4.5 Calculations Using the Chemical Equation
Sample Calculation
Na + Cl2  NaCl
Balance the equation
Calculate the moles Cl2 reacting with 5.00 mol Na
Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2
Calculate the grams Na reacting with 5.00 g Cl2
2Na + Cl2  2NaCl
4.5 Calculations Using the Chemical Equation

32. Theoretical and Percent Yield
Theoretical yield - the maximum amount of product that can be produced
Pencil and paper yield
Actual yield - the amount produced when the reaction is performed
Laboratory yield
Percent yield: 10
4.5 Calculations Using the Chemical Equation
= 125 g CO2 actual  100% = 97.4%
132 g CO2 theoretical

33. 4.5 Calculations Using the Chemical Equation
Sample Calculation
If the theoretical yield of iron was 30.0 g and actual yield was 25.0 g, calculate the percent yield:
2 Al(s) + Fe2O3(s)  Al2O3(aq) + 2Fe(aq)
[25.0 g / g]  100% = 83.3%
Calculate the % yield if 26.8 grams iron was collected in the same reaction
4.5 Calculations Using the Chemical Equation