GC-S008-Mass&Mole

Henry R. Kang (1/2010)

General Chemistry

Lecture 8

Mass and Mole

Contents

• Atomic Mass and Average Atomic Mass

• Molecular & Formula Masses

• Mole Conception

• Avogadro’s Number

• Molar Mass

• Relations between Mass, Mole, and Avogadro’s Number


Atomic Mass


Definition of Atomic Mass

• Atomic mass (or atomic weight) is the mass of an atom in atomic mass unit (amu)

One atomic mass unit is defined as a mass exactly equal to 1/12 the mass of one carbon-12 atom.

1 amu = (mass of 12C) / 12, or 12C = 12 amu

• AMU provides a relative scale for the masses of elements

On this scale: 1H = 1.008 amu and 16O = 16.00 amu

• The reason of using “amu” is because the true mass of any atom is too small a number (on the order of 10-23 g, for example, proton mass = 1.67×10-24 g) to express in normal mass unit such as gram.


Average Atomic Mass (amu): Definition

• Atomic masses given in the period table are the average atomic mass.

This is because the naturally occurring elements contain isotopes.

• The average atomic mass, matom, is the weighted average of atomic masses of isotopes, mi, weighted by their respective fractional abundances, fi.

matom = ∑ mi fi; i = 1, 2, - - - , N

N is the number of isotopes of the element.

Fractional abundance is the fraction of the isotope atoms in a given sample of the element.


Average Atomic Mass of Lithium

• 6Li = 6.015 amu at 0.0742 fractional abundance

• 7Li = 7.016 amu at 0.9258 fractional abundance

• Average atomic mass of lithium mLi = (0.0742 × 6.015 amu + 0.9258 × 7.016 amu)

= 6.942 amu

This value is within the computational uncertainty of the value listed in the period table (see next slide).


Modern Periodic Table1

1A188A

1H1.008

22A

133A

144A

155A

166A

177A

2He4.003

3Li6.941

4Be9.012

5B

10.81

6C12.01

7N

14.01

8O16.00

9F

19.00

10Ne20.18

11Na22.99

12Mg24.31

33B

44B

55B

66B

77B

88B

98B

108B

111B

1212B

13Al26.98

14Si28.09

15P

30.97

16S

32.07

17Cl35.45

18Ar39.95

19K39.10

20Ca40.08

21Sc44.96

22Ti47.88

23V50.94

24Cr52.00

25Mn54.94

26Fe55.85

27Co58.93

28Ni58.69

29Cu63.55

30Zn65.39

31Ga69.72

32Ge72.59

33As74.92

34Se78.96

35Br79.90

36Kr83.80

37Rb85.47

38Sr87.62

39Y88.91

40Zr91.22

41Nb92.91

42Mo95.94

43Tc(98)

44Ru101.1

45Rh102.9

46Pd106.4

47Ag107.9

48Cd112.4

49In114.8

50Sn118.7

51Sb121.8

52Te127.6

53I

126.9

54Xe131.3

55Cs132.9

56Ba137.3

57La138.9

72Hf178.5

73Ta180.9

74W183.9

75Re186.2

76Os190.2

77Ir192.2

78Pt195.1

79Au197.0

80Hg200.5

81Tl204.4

82Pb207.2

83Bi208.9

84Po(209)

85At(210)

86Rn(222)

87Fr(223)

88Ra(226)

89Ac(227)

104Rf(257)

105Db(260)

106Sg(263)

107Bh(262)

108Hs(265)

109Mt(266)

110Ds(271)

111Uuu(272)

112Uub(277)

114Uuq(296)

116Uuh(298)

118Uuo(?)

58Ce140.1

59Pr140.9

60Nd144.2

61Pm(147)

62Sm

(150.4)

63Eu152.0

64Gd157.3

65Tb158.9

66Dy162.5

67Ho164.9

68Er167.3

69Tm168.9

70Yb173.0

71Lu175.0

90Th232.0

91Pa(231)

92U(238)

93Np(237)

94Pu(242)

95Am(243)

96Cm(247)

97Bk(247)

98Cf(249)

99Es(254)

100Fm(253)

101Md(256)

102No(254)

103Lr(257)

Lanthanides

Actinides

Metals

Nonmetals

Metalloids

Average atomic mass (6.941)


Average Atomic Mass of Carbon

• 12C = 12.0000 amu at 0.9890 fractional abundance• 13C = 13.00335 amu at 0.0110 fractional

abundance• 14C has a very small percentage and will not affect

the average atomic mass. It is radioactive, therefore, not stable.

• Average atomic mass of carbon mC = 0.9890 × 12 amu + 0.0110 × 13.00335 amu

= 12.01 amu This number is exactly the same as the one given by the period

table.Henry R. Kang (1/2010)

Example 1 of Average Atomic Mass• Chromium, Cr, has the following isotopic masses and fractional

abundances: Mass number Isotope mass (amu) Fractional abundance

50 49.9461 0.0435

52 51.9405 0.8379

53 52.9407 0.0950

54 53.9389 0.0236

What is the atomic mass of chromium?

• Answer: Average atomic mass of chromium

= 0.0435 × 49.9461 amu + 0.8379 × 51.9405 amu

+ 0.0950 × 52.9407 amu + 0.0236 × 53.9389 amu

= 51.99 amu


Example 2 of Average Atomic Mass

• The average atomic mass of copper is 63.546 amu; it has two stable isotopes 63Cu = 62.930 amu and 65Cu = 64.9278 amu. Determine the fractional abundances for the isotopes.

• Answer: Use the equation: matom = ∑ mi fi

63.546 amu = 62.930 amu × fCu-63 + 64.9278 amu × fCu-65

63.546 amu = 62.930 amu × (1 – fCu-65) + 64.9278 amu × fCu-65

0.616 amu = 1.9978 amu × fCu-65

fCu-65 = 0.308

fCu-65 = 0.692


Example 3 of Average Atomic Mass

• The average atomic mass of boron is 10.811 amu; it has two stable isotopes 11B = 11.0093 amu at 80.22% and 10B. Determine the atomic mass of 10B.

• Answer: Use the equation: matom = ∑ mi fi

The fractional abundance of 11B is 0.8022; thus, the fractional abundance of 10B is

1 – 0.8022 = 0.1978

10.811 amu = 11.0093 amu × 0.8022 + mB-10 × 0.1978

1.9793 amu = mB-10 × 0.1978

mB-10 = 10.00677 amu = 10.007 amuHenry R. Kang (1/2010)

Mass and

Moles of Substance


Molecular Mass• Molecular mass ( or molecular weight) is the sum of the

atomic masses of all the atoms in a molecule.• Examples:

CHCl3

Molecular mass = 12.01 amu + 1.008 amu + 3 × 35.45 amu

= 119.368 amu = 119.37 amu

←round to two places after decimal point

(NH4)2CO3

Molecular mass = 2× (14.01 amu + 4 × 1.008 amu)

+ 12.01 amu + 3 × 16.00 amu

= 2× (18.04 amu) + 12.01 amu + 48.00 amu

= 96.09 amuHenry R. Kang (1/2010)

Formula Mass• Formula mass (or formula weight) is the sum of

the atomic masses of all the atoms in a formula unit of an ionic compound. Formula mass is used for ionic compounds that have

no definite molecular structure.

• Example: Fe2(SO4)3

Formula mass = 2 × 55.845 amu

+ (32.065 amu + 4 × 15.9994 amu) × 3

= 111.690 amu + 96.063 amu × 3

= 399.879 amu Henry R. Kang (1/2010)

Mole Concept

• Mole Definition A mole (symbol mol) is defined as the quantity

of a given substance that contains the same number of elementary entities (molecules, formula units, atoms, ions, etc.) as the number of atoms in exactly 12 grams of carbon-12.

• Now the question is “How many atoms in 12 g of carbon-12?”


Avogadro’s Number

• Avogadro’s number (NA) is the actual number of atoms in 12 grams of 12C. NA = 6.0221367×1023 (round off to 6.022×1023 )

• Mole (mol) is the amount of a substance that contains as many elementary entities as the Avogadro’s number. 1 mole of H atoms contains 6.022×1023 H atoms

1 mole of Na atoms contains 6.022×1023 Na atoms

1 mole of e- contains 6.022×1023 electrons

1 mole of H2O contains 6.022×1023 water molecules 2 × 6.022×1023 of H atoms = 1.204×1022 H atoms

6.022×1023 of O atoms

1 mole of Na2CO3 contains 6.022×1023 formula units of Na2CO3 units

2 × 6.022×1023 Na+ cations = 1.204×1022 Na+ cations

6.022×1023 CO32- anions


Size of Avogadro’s Number• The volume occupied by an Avogadro’s number (6.022×1023) of 2-

liter soda bottle is about the size of the Earth (radius = 6400 km). Volume of the earth

4πr3/3 = 4 × 3.1416 × (6400 km)3 = 1.1×1012 km3

Convert 2-L bottle to km3

Volume of bottle = 2.00 L × (1000 mL/1 L) × (1 cm3/1 mL) × (1 km / 105 cm)3

= 2.00×10-12 km3

1 Avogadro’s number of 2-L bottle Volume = (6.022×1023)× 2.00×10-12 km3 = 1.2×1012 km3

• If an Avogadro’s number of hydrogen atoms (diameter = 74 picometers or 7.4×10-12 meters) is laid side by side, the length is enough to encircle the Earth about one hundred-thousand times. (6.022×1023) × (7.4×10-12 m) = 4.5×1012 m = 4.5×109 km

length of equator = 2πr = 4.02×104 km

(4.5×109 km) / (4.02×104 km) = 1.1×105Henry R. Kang (1/2010)

Molar Mass

• Molar mass (M) is the mass (in grams or kilograms) of 1 mole of a substance. Molar mass (in grams) of an atomic substance is

numerically equal to its atomic mass in amu.

(Molar Mass of an atom) / g = (Atomic Mass) / amu

• Examples: Molar mass of 12C is 12.00 g

Atomic mass is 12.00 amu

Molar mass of Na is 22.99 g

Atomic mass is 22.99 amuHenry R. Kang (1/2010)

Example of Mass of a Single Atom

• What is the mass in grams of a chlorine atom, Cl? 1 mole Cl atoms = 6.022 × 1023 Cl atoms

1 mole Cl atoms = 35.45 g

Using these relationships, we can compute the mass in grams of a single Cl atom

• Answer

Mass of a Cl atom = = 5.887×10-23 g35.45 g

6.022 × 1023


Example of Mass of a Single Molecule

• What is the mass in grams of a hydrogen bromide molecule, HBr? 1 mole HBr molecules = 6.022 × 1023 molecules

1 mole HBr molecules = 1.008 g + 79.90 g

= 80.91 g

• Answer:

Mass of an HBr = = 1.343×10-22 g80.91 g

6.022 × 1023


Relationships between Mass, Mole, and Avogadro’s Number

Mass of element (m)

# moles of element (n)

# atoms of element (N)

n = m/M

m = n M

N = nNA

n = N/NA

n: number of molesm: massM: molar massN: number of atomsNA: Avogadro’s number, 6.022 × 1023

N = (m /M)NA

m = (N/NA)M


Examples of Mole to Mass

• How many grams of zinc iodide (ZnI2) are in 0.354 moles of ZnI2?

1 mole ZnI2 = 65.39 g + 2 × 126.9 g = 319.2 g

• Calculate the number of grams of lead (Pb) in 10.8 moles of Pb? 1 mole Pb = 207.2 g Pb

319.2 g ZnI2

1 mole ZnI2

= 113 g ZnI20.354 mole ZnI2 ×

10.8 mole Pb ×207.2 g Pb

1 mole Pb= 2.24×103 g Pb


Examples of Mass to Mole

• How many moles of lead(II) chromate (PbCrO4) are in 65.8 g of PbCrO4?

1 mole PbCrO4 = 207.2 g + 51.996 g + 4 × 16.00 g = 323.2 g

• How many moles of magnesium (Mg) atoms are in 72.3 g of Mg? 1 mole Mg = 24.31 g Mg

65.8 g PbCrO4 ×1 mol PbCrO4

323.2 g PbCrO4

= 0.204 mol PbCrO4

72.3 g Mg × 1 mole Mg24.31 g Mg = 2.97 mole Mg


Example of Mass to #Atoms

• How many atoms are in 26.3 g of iron (Fe)?• Answer:

Grams of Fe → moles of Fe → number of Fe atoms 1 mole Fe = 55.85 g Fe

1 mole = 6.022×1023 atoms

26.3 g Fe ×

= 2.84×1023 atoms

# mole Fe = 1 mole Fe26.3 g Fe ×55.85 g Fe

# Fe atoms = mole Fe × 1 mole Fe6.022×1023 atoms

1 mole Fe55.85 g Fe

6.022×1023 atoms1 mole Fe

×


Example of Mass to Molecules

• How many molecules in a 6.43-g sample of hydrogen chloride, HCl?

• Answer: 1 mole HCl = 1.008 g + 35.45 g = 36.46 g

6.43 g HCl ×1 mol HCl36.46 g HCl

= 1.06×1023 molecules

×1 mol HCl

6.022 × 1023 molecules


Example of Mass to Particle Number

• How many hydrogen atoms are present in 52.8 g of isopropanol (C3H7OH)?

Molar mass of C3H8O

= 3 × 12.01 g + 8 × 1.008 g + 16.00 g = 60.09 g 1 mole isopropanol = 60.09 g isopropanol 1 mole = 6.022×1023 molecules 1 molecule = 8 H atoms

• Answer:52.8 g × 6.022×1023 molecules

60.09 g8 H atoms1 molecule

= 4.23×1024 H atoms

×


Molecular Mass & Molar Mass

• Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in the molecule. Molecular mass of water (H2O)

= 2 (atomic mass of H) + atomic mass of O

= 2 (1.008 amu) + 16.00 amu = 18.02 amu

• Molar mass of a compound in grams is numerically equal to its molecular mass in amu. Molar mass of water = 18.02 g

• One mole of a compound contains one Avogadro’s number of molecules.


Example 1 of Molar Mass

• Calculate the molecular mass (in amu) and molar mass (in g) of dinitrogen pentoxide (N2O5).

• Answer:

Molecular mass of N2O5

= 2 (atomic mass of N) + 5 (atomic mass of O)

= 2 × 14.01 amu + 5 × 16.00 amu

= 108.02 amu

Molar mass of N2O5 = 108.02 g Henry R. Kang (1/2010)


• Calculate the molecular mass (in amu) and molar mass (in g) of caffeine (C8H10N4O2).

• Answer: Molecular mass of C8H10N4O2

= 8 (atomic mass of C) + 10 (atomic mass of H)

+ 4 (atomic mass of N) + 2 (atomic mass of O)

= 8 × 12.01 amu + 10 × 1.008 amu

+ 4 × 14.01 amu + 2 × 16.00 amu

= 194.20 amu

Molar mass of C8H10N4O2 = 194.20 gHenry R. Kang (1/2010)


• Calculate the molecular mass (in amu) and molar mass (in g) of Ethanol (C2H5OH).

• Answer:

Molecular mass of C2H5OH

= 2 (atomic mass of C) + 6 (atomic mass of H)

+ atomic mass of O

= 2 × 12.01 amu + 6 × 1.008 amu + 16.00 amu

= 46.07 amu

Molar mass of C2H5OH = 46.07 g Henry R. Kang (1/2010)

Example 4 of Mass to Mole

• How many moles of propane (C3H8) are in 16.07 g of C3H8?

• Answer:

Mass of C3H8 → moles of C3H8

Molar mass of C3H8

= 3 (molar mass of C) + 8 (molar mass of H)

= 3 × 12.01 g + 8 × 1.008 g = 44.09 g

1 mole CH4 = 44.09 g CH4

(16.07 g C3H8) × (1 mole C3H8 / 44.09 g C3H8)

= 0.3645 mole C3H8


Example 5 of Mole to Mass

• Calculate the mass of Ca3(PO4)2 in 0.198 moles?

• Answer:

Molar mass of Ca3(PO4)2

= 3 (molar mass of Ca)

+ 2 {(molar mass of P) + 4 (molar mass of O)}

= 3 × 40.08 g + 2 × (30.97 g + 4 × 16.00 g)

= 310.18 g

1 mole Ca3(PO4)2 = 310.18 g Ca3(PO4)2

0.198 moles × (310.18 g / 1 mole)

= 61.4 g Ca3(PO4)2


Example of Mass to #Atoms• How many hydrogen atoms are present in 25.6 g of urea

[(NH2)2CO]? The molar mass of urea is 60.06 g.

• Answer: Grams of urea → moles of urea → molecules of urea

→ atoms of urea 1 mole urea = 60.06 g urea 1 mole = 6.022×1023 atoms 1 molecule = 4 H atoms

25.6 g urea ×60.06 g urea

6.022×1023 urea molecules1 urea molecule

4 H atoms×

= 1.03×1023 H atoms


Example of #Atoms to Mass

• What is the mass butanol (C4H10O)? If there are 5.81×1024 hydrogen atoms in the sample of butanol.

• Answer: Molar mass of C4H10O

= 4 × 12.01 g + 10 × 1.008 g + 16.00 g = 74.12 g 1 mole butanol = 74.12 g butanol 1 mole = 6.022×1023 molecules 1 molecule = 10 H atoms

= 71.5 g 6.022×1023 molecules

74.12 g10 H atoms1 molecule5.81×1024 H atoms ×


×

GC-S008-Mass&Mole

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