1. Introduction to Chemical Principles
Chapter 9: The Mole Concept and Chemical Formulas or
Quantitative Relationships Between Elements and Compounds or
“Chemical Arithmetic”

2. The Law of Definite Proportions
The law of definite proportions states:
In a pure compound the elements are always present in the same definite proportion by mass.
Compounds have chemical and physical properties based on this law of definite proportions.
No mater how much water is in a container, it will always boil at 100°C. It’s composition doesn’t change with mass.

3. Composition of Water
Water always contains the same two elements: hydrogen and oxygen.
Water is always 11.2% hydrogen by mass.
Water is always 88.8% oxygen by mass.
Water always has these percentages. If the percentages were different the compound would not be water.

4. Determine which compounds have the same composition.
1.926 g R
3.000 g comp. 1)
* 100% = % R
3.440 g R
5.000 g comp. 2)
* 100% = % R
4.494 g R
7.000 g comp. 3)
* 100% = % R
Since experiments 1 and 3 produced compounds with the same composition, they more than likely produced the same compound.

5. Percent Composition of Compounds from the Formula or from Experiment

6. Percent Composition From the Formula

7. If the formula of a compound is known, a two-step process is needed to calculate the percent composition.
Step 1 Calculate the formula mass or molecular mass.
Step 2 Divide the mass of each element in the formula by the total mass and multiply by 100.

8. Formula Mass and Molecular Mass
Formula mass: the sum of the atomic masses of the atoms in one formula unit.
Molecular mass: the sum of the atomic masses of the atoms in one molecule.
May also be called formula weight or molecular weight.
Formula Mass for NaCl
Formula Mass for MgCl2
Molecular Mass for H2O
22.99 amu (Na)
24.30 amu (Mg)
[2 x 1.01 amu (H)]
amu (Cl)
+ [2 x amu (Cl)]
amu (O)
58.44 amu NaCl
95.20 amu MgCl2
18.02 amu H2O

10. Step 1 Calculate the molecular mass of H2S.
Calculate the percent composition of hydrosulfuric acid: H2S.
Step 1 Calculate the molecular mass of H2S.
2 H = 2 x 1.01amu = amu
1 S = 1 x amu = amu
34.09 amu

11. Calculate the percent composition of hydrosulfuric acid: H2S.
Step 2 Divide the mass of each element by the molecular mass and multiply by 100.
%H = amu / amu x 100 = %
%S = amu / amu x 100 = 94.07%
H 5.93%
S %

12. Percent Composition From Experimental Information

13. Percent composition can be calculated from experimental data without knowing the composition of the compound.
Step 1 Calculate the mass of the compound formed.
Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.

14. Step 1 Calculate the total mass of the compound
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.
Step 1 Calculate the total mass of the compound
1.52 g N
3.47 g O
4.99 g
= total mass of product

15. A compound containing nitrogen and oxygen is found to contain 1
A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.
Step 2 Divide the mass of each element by the total mass of the compound formed.
N 30.5%
O 69.5%

16. Counting Atoms and Molecules
Atoms and molecules are so tiny, that you generally need billions of them to have a measurable amount.
Chemists required a unit for counting which could express large numbers of atoms using simple numbers. Thus, the unit would be like other counting units we employ…
This is an
infinitesimal
mass
mass of hydrogen atom = x g
A mole is just a number, like a pair is 2, a dozen is 12, and a gross is 144. 16

17. MOLE Well, chemists have chosen a unit for counting atoms.
That unit is the…

18. 1 mole = x 1023 objects

19. 1 mole of atoms =
6.022 x 1023 atoms
1 mole of molecules =
6.022 x 1023 molecules
1 mole of ions =
6.022 x 1023 ions

20. 6.022 x 1023
is a very
LARGE
number

21. 6.022 x 1023 Avogadro’s Number This number is called:
Named after Lorenzo Romano Amedeo Carlo Avogadro, an Italian physicist.

22. Species Quantity Number of Na atoms
1 mole of atoms
6.022 x 1023

23. Species Quantity Number of Fe atoms
1 mole of atoms
6.022 x 1023

24. Species Quantity Number of C6H6 molecules
1 mole of molecules
6.022 x 1023

25. Curiously enough, chemists decided on the following abbreviation for the mole unit:

26. The Mass of a Mole Moles of Substance Grams of Substance
The molar mass of a compound is a mass in grams that is numerically equal to the formula mass or molecular mass of the compound.
Molecular mass of H2O = amu
Mass of 1 mole H2O molecules = g
So the units of molar mass are given as g/mol.
Moles of Substance
Molar
mass
Grams of Substance

27. Avogadro’s Number of Particles
6.022 x Particles
1 MOLE
Molar Mass

28. Avogadro’s Number of Ca atoms
6.022 x Ca atoms
1 MOLE Ca
40.08 g Ca

29. Avogadro’s Number of H2O molecules
6.022 x H2O molecules
1 MOLE H2O
18.02 g H2O

30. Avogadro’s Number of H2SO4 molecules
6.022 x H2SO4 molecules
1 MOLE H2SO4
98.08g H2SO4

31. How many moles of iron does 25.0 g of iron represent?
Molar mass iron = g/mol
Conversion sequence: grams Fe → moles Fe
Set up the calculation using a conversion factor between moles and grams.

32. More examples What is the mass, in grams, of 4.6 mol MgCl2?
95.20 g MgCl2
4.6 mol MgCl2
= 4.4 x 102 g MgCl2
1 mol MgCl2
How many MgCl2 formula units are in 4.6 mol MgCl2?
6.022 x 1023 formula units MgCl2
4.6 mol MgCl2
1 mol MgCl2
= 2.8 x 1024 formula units MgCl2

33. Converting from moles to grams

34. Convert from moles to gram

35. How many oxygen atoms are present in 2.00 mol of oxygen molecules?
Two conversion factors are needed:
Conversion sequence: moles O2 → molecules O2 → atoms O
= 2.41 x 1024

36. How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?
The molar mass of (NH4)3PO4 is g/mol.
Conversion sequence: moles (NH4)3PO4 →
grams (NH4)3PO4

37. How many moles of benzene, C6H6, are present in 390.0 grams of benzene?
The molar mass of C6H6 is g/mol.
Conversion sequence: grams C6H6 → moles C6H6

38. Convert from grams to moles

39. Convert from grams to moles

40. Copyright © Cengage Learning. All rights reserved
EXERCISE!
(4.48 mol Fe) × (6.022×1023 Fe atoms / 1 mol Fe) = 2.70×1024 Fe atoms.
Copyright © Cengage Learning. All rights reserved

41. Converting from moles to molecule
0.200 moles of H2O contain how many molecules
We multiple by x 1023
0.200 x 6.022x 1023 mol-1
1.20 x 1023 molecules

42. Conversion sequence: g N2 → moles N2 → molecules N2
56.04 g of N2 contains how many N2 molecules?
The molar mass of N2 is g/mol.
Conversion sequence: g N2 → moles N2 → molecules N2
Use the conversion factors

43. Conversion sequence: g N2 → moles N2 → molecules N2
56.04 g of N2 contains how many N atoms?
The molar mass of N2 is g/mol.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms N
Use the conversion factors
=2.49

44. Conversion chart
Grams to moles Moles to atom and Molecule

45. Calculate how many grams are in
4.56 moles of KCIO
5.63 moles NaOH
4.5 moles KI
2.50 moles KClO3
Calculate how many moles are in
75.57 grams of KBr
2.00 grams of H2O
100 grams of KClO4
0.750 grams of Na2CO3

46. Empirical Formula versus Molecular Formula

47. The empirical formula or simplest formula gives the smallest whole- number ratio of the atoms present in a formula unit or compound.
The empirical formula gives the relative number of atoms of each element present in the molecule of a covalent compound.

48. The molecular formula represents the total number of atoms of each element present in one molecule of a compound.
The molecular formula is the true formula of a molecular compound and can be found from the empirical formula and molar mass.

49. Examples of Formula Units and Empirical Formulas

50. Smallest Whole Number Ratio
Formula Unit
CaCl2
Empirical Formula
CaCl2
Smallest Whole Number Ratio
Ca:Cl 1:2 50

51. Examples of Molecular Formulas and Empirical Formulas51

52. Smallest Whole Number Ratio
Molecular Formula
C2H4
CH2
Empirical Formula
C:H 1:2
Smallest Whole Number Ratio

53. Smallest Whole Number Ratio
Molecular Formula
C6H6
Empirical Formula CH
C:H 1:1
Smallest Whole Number Ratio

54. Smallest Whole Number Ratio
Molecular Formula
H2O2
Empirical Formula HO
Smallest Whole Number Ratio
H:O 1:1

56. Two compounds can have identical empirical formulas and different molecular formulas.

57. Calculating Empirical Formulas

58. Step 1 Assume a definite starting quantity (usually 100
Step 1 Assume a definite starting quantity (usually g) of the compound, if not given, and express the mass of each element in grams.
Step 2 Convert the grams of each element into moles of each element using each element’s molar mass.

59. Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value
– If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula.
– If the numbers obtained are not whole numbers, go on to step 4.

60. Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers
Use these whole numbers as the subscripts in the empirical formula.
FeO1.5
Fe1 x 2O1.5 x 2
Fe2O3

61. The results of calculations may differ from a whole number.
If they differ ±0.1 round off to the next nearest whole number.
2.9
3
Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.

62. Some Common Fractions and Their Decimal Equivalents
Resulting Whole Number 1 2 3
Common Fraction
0.25
0.333…
Multiply the decimal equivalent by the number in the denominator of the fraction to get a whole number.
0.666…
0.5
0.75

63. Problems

64. The analysis of a salt shows that it contains 56. 58% potassium (K); 8
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.
Step 1 Express each element in grams. Assume 100 grams of compound.
K = g
C = g
O = g

65. C has the smallest number of moles
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.
Step 2 Convert the grams of each element to moles.
C has the smallest number of moles

66. C has the smallest number of moles
The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.
Step 3 Divide each number of moles by the smallest value.
C has the smallest number of moles
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3

67. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 1 Express each element in grams. Assume 100 grams of compound.
N = g
O = g

68. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 2 Convert the grams of each element to moles.

69. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 3 Divide each number of moles by the smallest value.
This is not a ratio of whole numbers.

70. The percent composition of a compound is 25. 94% nitrogen (N), and 74
The percent composition of a compound is 25.94% nitrogen (N), and % oxygen (O). Calculate the empirical formula for this substance.
Step 4 Multiply each of the values by 2.
N: (1.000)2 = 2.000
O: (2.500)2 = 5.000
Empirical formula N2O5

71. Calculating the Molecular Formula from the Empirical Formula

72. The molecular formula can be calculated from the empirical formula if the molar mass is known.
The molecular formula will be equal to the empirical formula or some multiple n of it.
To determine the molecular formula evaluate n.
n is the number of units of the empirical formula contained in the molecular formula.

73. What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of g?
Let n = the number of formula units of CH2.
Calculate the mass of each CH2 unit
1 C = 1(12.01 g) = 12.01g
2 H = 2(1.01 g) = g
14.03g
The molecular formula is (CH2)9 = C9H18