โมล (Mole)

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โมล (Mole)

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Molar mass is the mass of 1 m in gramsatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)


Contains 6.02 x 1023 particles

1 mole C = 6.02 x 1023 C atoms

1 mole H2O = 6.02 x 1023 H2O

molecules1 mole NaCl = 6.02 x 1023 Na+ ions and

6.02 x 1023 Cl– ions

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Moles of elements

1 mole Mg = 6.02 x 1023 Mg atoms

1 mole Au = 6.02 x 1023 Au atoms

Moles of compounds

1 mole NH3 = 6.02 x 1023 NH3

molecules

1 mole C9H8O4 = 6.02 x 1023 aspirin

molecules

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Avogadro’s Number

6.02 x 1023

1 mole of 12C has a mass of 12 g


How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K1 mol K

39.10 g Kx x

6.022 x 1023 atoms K

1 mol K=

8.49 x 1021 atoms K


Using Moles

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


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


9The Mole

12 g of 12C has 1 mole of 12C atoms. (By definition)

12.01 g of C has 1 mole of C atoms.

Because =

Then both samples of 12C and natural C contain the same no. of components (1 mole).

This is applied on all other elements when their masses are determined with respect to the mass of the 12C atom.

12g

12.01g

12 amu

12.01amu

Relative masses of a single atom of 12C

and natural C


10The Mole

Average weight = 0.5 kg

10 pieces of orange = 5 kg

Average weight = 1.0 kg

10 pieces of grapefruit = 10 kg

A sample of a natural element with a mass equal to the element’s atomic mass expressed in grams

contains 1 mole of atoms


11The Mole

For any substance (numerically):Atomic mass (amu) = Mass for 1 mole

(g/mol)

A sample of a natural element with a mass equal to the element’s atomic mass

expressed in grams contains 1 mole of atoms


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1. Number of atoms in 0.500 mole of Al1) 500 Al atoms

2) 6.02 x 1023 Al atoms3) 3.01 x 1023 Al atoms

2.Number of moles of S in 1.8 x 1024 S atoms1) 1.0 mole S atoms2) 3.0 mole S atoms3) 1.1 x 1048 mole S atoms

Learning Check


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1. Number of atoms in 0.500 mol of Al3) 3.01 x 1023 Al atoms

0.500 mol Al x 6.02 x 1023 Al atoms1 mol Al

2. Number of moles of S if a sample of S contains 4.50 x 1024 S atoms2) 3.0 mole S atoms

4.50 x 1024 S atoms x 1 mol S

6.02 x 1023 S atoms

Solution


Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

3.3Let’s try some more……………..


How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =


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Molar Mass

The molar mass of a compound is found by adding Together the molar masses of all of its elements, taking

Into account the number of moles of each element present.


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Percent composition of an element in a compound

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

3.5


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32Percent Composition of Compounds

Mass % can be calculated by comparing the molar mass of the atom to the molar mass of the molecule.

Example is ethanol (C2H5OH):

Mass % O = × 100%

= (100%) = 34.73%

Mass % H = × 100%

= (100%) = 13.13%

1 mole of C2H5OH

2 moles of C

1 mole of O

6 moles of H

1 (MM of O)

MM of C2H5OH

16.00 g/mol

46.07 g/mol

6 (MM of H)

MM of C2H5OH

6(1.01) g/mol

46.07 g/molMass %’s must be

added up to 100%


แหิล"งอุ#างอุ�ง• Martin S. Silberberg, Chemistry: The Molecular Nature

of Matter and Change, McGraw-Hill Higher Education, 2004

• Raymond Chang, Chemistry, Williams College, McGraw-Hill Higher Education, 2002

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โมล (Mole)