1. 2 The mass of a single atom is too small to measure on a balance. mass of hydrogen atom = 1.673 x 10 -24 g.

1

2

The mass of a single atom is too small to measure on a balance.

mass of hydrogen atom = 1.673 x 10-24 g

3

This is an

infinitesimal

mass

1.673 x 10-24 g

4

• Chemists have chosen a unit for counting atoms.

• That unit is the

• Chemists require a unit for counting which can express large numbers of atoms using simple numbers.

MOLE

5

1 mole = 6.02 x 1023 objects

6

LARGE6.02 x 1023

is a very

number

7

6.02 x 1023

is

number

Avogadro’s Number

8

Amadeo (Amedeo) Avogadro

9

If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.

http://youtu.be/1R7NiIum2TI

10

1 mole of any element contains

6.02 x 1023

particles of that substance.

11

The atomic weight in grams

of any element23

contains 1 mole of atoms.

12

This is the same number of particles6.02 x 1023

as there are in exactly 12 grams of

C126

13

ExamplesExamplesExamplesExamples

14

Species

Quantity

Number of H atoms

H

1 mole

6.02 x 1023

15

Species

Quantity

Number of H2 molecules

H2

1 mole

6.02 x 1023

16

Species

Quantity

Number of Na atoms

Na

1 mole

6.02 x 1023

17

Species

Quantity

Number of Fe atoms

Fe

1 mole

6.02 x 1023

18

Species

Quantity

Number of C6H6 molecules

C6H6

1 mole

6.02 x 1023

19

1 mol of atoms = 6.02 x 1023 atoms

6.02 x 1023 molecules

6.02 x 1023 ions

1 mol of molecules =

1 mol of ions =

20

• The mole weight of an element is its atomic weight in grams.

• It contains 6.02 x 1023 atoms (Avogadro’s number) of the element.

21

ElementAtomic mass

Mole weightNumber of

atoms

H 1.008 amu 1.008 g 6.02 x 1023

Mg 24.31 amu 24.31 g 6.02 x 1023

Na 22.99 amu 22.99 g 6.02 x 1023

22

ProblemsProblemsProblemsProblems

23

Convert ToMoles!

24

Weight (mass) Numbers

Moles

Periodic Table Avogadro’s Number6.02 × 1023

25

Atomic weight iron = 55.85

How many moles of iron does 25.0 g of iron represent?

Conversion sequence: grams Fe → moles Fe

1 mol Fe(grams Fe)

55.85 g Fe

1 mol Fe(25.0 g Fe)

55.85 g Fe

0.448 mol Fe

Set up the calculation using a conversion factor between moles and grams.

(40.0 g Fe)1 mole Fe55.85 g Fe

236.02 x 10 atoms Fe1 mole Fe

26

Atomic weight iron = 55.85Conversion sequence is:

grams Fe → moles Fe → atoms Fe

How many iron atoms are contained in 40.0 grams of iron?

234.31 x 10 atoms Fe

23(3.01 x 10 atoms Na) 23

1 mole Na6.02 x 10 atoms Na

22.99 g Na1 mole Na

27

Mole weight Na = 22.99 g

Conversion sequence:

atoms Na → moles Na → grams Na

What is the mass of 3.01 x 1023 atoms of sodium (Na)?

11.5 g Na

28

2(2.00 mole O )23

2

2

6.02 x 10 molecules O1 mole O

2

2 atoms O1 molecule O

Conversion sequence:

moles O2 → molecules O2 → atoms O

How many oxygen atoms are present in 2.00 moles of oxygen molecules?

24= 2.41 x10 atoms O

29

Mole Weight of Mole Weight of CompoundsCompounds

30

The mole weight of a compound can be determined by adding the mole weights of all of the atoms in its formula.

31

2 C = 2(12.01 g) = 24.02 g

6 H = 6(1.01 g) = 6.06 g

1 O = 1(16.00 g) = 16.00 g

46.08 g

Calculate the mole weight of C2H6O.

32

1 Li = 1(6.94 g) = 6.94 g1 Cl = 1(35.45 g) = 35.45 g4 O = 4(16.00 g) = 64.00 g

106.39 g

Calculate the mole weight of lithium perchlorate.

LiClO4

33

Calculate the mole weight of ammonium phosphate.

3 N = 3(14.01 g) = 42.03 g12 H = 12(1.01 g) = 12.12 g

1 P = 1(30.97 g) = 30.97 g4 O = 4(16.00 g) = 64.00 g

149.12 g

(NH4)3PO4

34

Water in a hydrate is known as water of hydration or water of crystallization.

Solids that contain water as part of their crystalline structure are known as hydrates.

35

Formulas of hydrates are written by first writing the formula for the anhydrous compound and then adding a dot followed by the number of water molecules present.

6H2OCoCl2

36

Calculate the mole weight of NaC2H3O2 · 3 H2O

1 Na = 1(22.99 g) = 22.99 g 2 C = 2(12.01 g) = 24.02 g

9 H = 9(1.01 g) = 9.09 g 5 O = 5(16.00 g) = 80.00 g

136.10 g

Sodium acetate trihydrate

37

Avogadro’s Number of Particles

6.02 x 1023 Particles

Mole Weight

1 MOLE

38

1 MOLE Ca

Avogadro’s Number ofCa atoms

6.02 x 1023 Ca atoms

40.078 g Ca

39

1 MOLE H2O

Avogadro’s Number of

H2O molecules

6.02 x 1023 H2O

molecules

18.02 g H2O

40

In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.

41

Calculate the mole weight of 1 mole of H atoms.

1 H = 1(1.01 g) = 1.01 g

Calculate the mole weight of 1 mole of H2 molecules.

2 H = 2(1.01 g) = 2.02 g

42

ProblemsProblemsProblemsProblems

43

How many moles of benzene, C6H6, are present in 390.0 grams of benzene?

Conversion sequence: grams C6H6 → moles C6H6

6 6

6 6

78.12 grams C HUse the conversion factor:

1 mole C H

6 6

6 6

1 mole C H 78.12 g C H

6 6(390.0 g C H ) 6 6= 4.992 moles C H

The mole weight of C6H6 is 78.12 g.

44

How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?

Conversion sequence:

moles (NH4)3PO4 → grams (NH4)3PO4

4 3 4

4 3 4

149.12 grams (NH ) POUse the conversion factor:

1 mole (NH ) PO

4 3 4(2.52 mol (NH ) PO )) 4 3 4

4 3 4

149.12 g (NH ) PO1 mol (NH ) PO

4 3 4= 376g (NH ) PO

The mole weight of (NH4)3PO4 is 149.12 g.

45

2(56.04 g N ) 2

2

1 mol N28.02 g N

232

2

6.02 x 10 molecules N

1 mol N

56.04 g of N2 contains how many N2 molecules?

The mole weight of N2 is 28.02 g.

Conversion sequence: g N2 → moles N2 → molecules N2

Use the conversion factors

2

2

1 mol N 28.02 g N

232

2

6.02 x 10 molecules N

1 mol N

242= 1.20 x 10 molecules N

46

2

2

1 mol N28.02 g N

2(56.04 g N )23

2

2

6.02 x 10 molecules N

1 mol N

56.04 g of N2 contains how many N atoms?

The mole weight of N2 is 28.02 g.

Conversion sequence: g N2 → moles N2 → molecules N2

→ atoms NUse the conversion factors

2

2

1 mol N 28.02 g N

232

2

6.02 x 10 molecules N

1 mol N 2

2 atoms N 1 molecule N

24= 2.41 x 10 atoms N2

2 atoms N1 molecule N

47

Percent CompositionPercent Compositionof Compoundsof Compounds

48

Percent composition of a compound is the mass percent of each element in the compound.

H2O11.19% H by mass 88.79% O by mass

49

Percent Composition Percent Composition From FormulaFrom Formula

50

If the formula of a compound is known, a two-step process is needed to calculate the percent composition.

Step 1 Calculate the mole weight of the formula.

Step 2 Divide the total mass of each element in the formula by the mole weight and multiply by 100.

51

total mass of the element x 100 = percent of the element

mole weight

52

Step 1 Calculate the mole weight of H2S.2 H = 2 (1.01 g) = 2.02 g

1 S = 1 (32.07 g) = 32.07 g34.09 g

Calculate the percent composition of hydrosulfuric acid H2S(aq) .

53

Calculate the percent composition of hydrosulfuric acid H2S.

Step 2 Divide the mass of each element by the mole weight and multiply by 100.

H

5.93%

S

94.07%

32.07g SS: (100) 94.07%

34.09g

2.02 g HH: (100) = 5.93%

34.09 g

54

Percent Composition Percent Composition From Experimental DataFrom Experimental Data

55

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.

56

Step 1 Calculate the total mass of the compound1.52 g N

3.47 g O

4.99 g

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.

= total mass of product

57

Step 2 Divide the mass of each element by the total mass of the compound formed.

3.47g O(100) = 69.5%

4.99g

1.52 g N(100) = 30.5%

4.99 g

N

30.5%

O

69.5%

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.

58

Empirical Formula versus Empirical Formula versus Molecular FormulaMolecular Formula

59

• The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound.

• The empirical formula gives the relative number of atoms of each element present in the compound.

60

• The molecular formula is the true formula of a compound.

• The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

61

ExamplesExamplesExamplesExamples

62

C2H4Molecular Formula

CH2Empirical Formula

C:H 1:2Smallest Whole Number Ratio

63

C6H6Molecular Formula

CHEmpirical Formula

C:H 1:1Smallest Whole Number Ratio

64

H2O2Molecular Formula

HOEmpirical Formula

H:O 1:1Smallest Whole Number Ratio

65

66

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

67

68

CalculatingCalculatingEmpirical FormulasEmpirical Formulas

69

The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance.

C___O___

70

Step 1 Assume a definite starting quantity (usually 100.0 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 mole weight.

71

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.

72

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

73

• The results of calculations may differ from a whole number.

– If they differ ±0.1 round off to the next nearest whole number.

2.93– Deviations greater than 0.1 unit from a

whole number usually mean that the calculated ratios have to be multiplied by a whole number.

74

ProblemsProblemsProblemsProblems

75

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 = 56.58 g

C = 8.68 g

O = 34.73 g

76

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.

K: 56.58 g K1 mol K atoms

39.10 g K

1.447 mol K atoms

C: 8.68 g C1 mol C atoms

12.01 g C

0.723 mol C atoms

O: 34.73 g O1 mol O atoms

16.00 g O

2.171 mol O atoms

C has the smallest number of moles

0.723 mol C atoms

77

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.

1.447 molK = = 2.00

0.723 mol0.723 mol

C: = 1.000.723 mol

2.171 molO = = 3.00

0.723 mol

The simplest ratio of K:C:O is 2:1:3

Empirical formula K2CO3

C has the smallest number of moles

0.723 mol C atoms

78

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

Step 1 Express each element in grams. Assume 100 grams of compound.

N = 25.94 g

O = 74.06 g

79

Step 2 Convert the grams of each element to moles.

N: 25.94 g N1 mol N atoms

14.01 g N

1.852 mol N atoms

O: 74.06 g O1 mol O atoms

16.00 g O

4.629 mol O atoms

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

80

Step 3 Divide each number of moles by the smallest value.

1.852 molN = = 1.000

1.852 mol4.629 mol

O: = 2.5001.852 mol

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

This is not a ratio of whole numbers.

81

Step 4 Multiply each of the values by 2.

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

Empirical formula N2O5

N: (1.000)2 = 2.000 O: (2.500)2 = 5.000

82

Calculating the Molecular Formula Calculating the Molecular Formula from the Empirical Formulafrom the Empirical Formula

83

• The molecular formula can be calculated from the empirical formula if the mole weight 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.

mole weightn = =

empirical weight formula units number of empirical

84

What is the molecular formula of a compound which has an empirical formula of CH2 and a mole weight of 126.2 g?

The molecular formula is (CH2)9 = C9H18

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) = 2.02g

14.03g126.2 g

n 9 (empirical formula units)14.03 g

85

A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.

Old Way:

1mole30.4g 2.17mole of N

14.0g

1mole69.6g 4.35mole of O

16.0g

2.17 24.352.17 2.17

N O NO 46g/ mole

2 3 6

1383units 3(NO ) N O

46

86

A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula.

New Way:

N3O6

138gcomp'd1molecomp'd

138gcomp'd1molecomp'd

30.4gN100gcomp'd

69.6gO100gcomp'd

1moleN14.0g

1moleO16.0g

3moleN

6moleO

87