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The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons, etc. The Mole (mol) Avogadro’s number 1 mole of Carbon= 6.02 x 10 23 carbon atoms 1 mole water= 6.02 x 10 23 water molecules 1 mole of H + = 6.02 x 10 23 H + ions 1 mole NaCl= 6.02 x 10 23 NaCl formula units IN GENERAL: 1 mole of a substance = 6.02 x 10 23 particles

The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

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Page 1: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 1023 particles

These particles can be atoms, ions, formula units,molecules, electrons, etc.

The Mole (mol)Avogadro’s number

1 mole of Carbon= 6.02 x 1023 carbon atoms

1 mole water= 6.02 x 1023water molecules

1 mole of H+ = 6.02 x 1023 H+ ions

1 mole NaCl= 6.02 x 1023 NaCl formula units

IN GENERAL: 1 mole of a substance = 6.02 x 1023particles

Page 2: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Examples (# of particles)How many atoms of Lithium are in 0.43 moles of Li atoms?

2.00 x 1032 ions of Na+ mol of Na+1

6.02 x 1023

0.43 mol of Li 1

6.02 x 1023

mol of Lix

Atoms of Li= 2.6 x 1023 atoms of Li

How many moles of sodium ions are if we have 2.00 X 1032 sodium ions

xions of Na+

= 3.32 x 108 mol of Na+

Page 3: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

More examples (# of particles)

How many moles of carbon dioxide are if we have 6.00 x 1024

molecules of carbon dioxide?

6.00 x 1024 molec of CO2

6.02 x 1023

1

molec of CO2

x mole of CO2=

9.97 moles of CO2

How many formula units of sodium chloride are in 0.25 moles of sodium chloride?

0.25 mol of NaCl 1

6.02 x 1023

mol of NaClx Formula units of NaCl =1.5 x 1023 form. Units of NaCl

Page 4: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Molar Mass and Atomic Mass

It is equal to the atomic mass from the periodic table , expressed in grams.

Atomic mass: the average mass of all the isotopes of an element considering their percentage of abundance.

The atomic mass can be expressed in amu´s (mass of atoms) or in g (mass of mole of atoms).

An amu (atomic mass unit) is the mass of 1/12 the mass of one atom of 12C.

Molar Mass: The mass of one mole of substance ( 6.02 x 1023 particles).

Page 5: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

The molar mass can be: atomic mass (for atoms), molecular mass (for molecules) or formula mass (for salts), but the term molar mass can be applied to all of them.

Examples: What is are the molar masses of Carbon, Sodium Chloride, Carbon Dioxide, Oxygen Atoms and Oxygen Molecules?

Molar Mass

Remember: 1 Mole = Molar Mass (in grams)

Page 6: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

The mole and the mass of substances

0.250 mol Namol Na

g Na22.99

1

What is the mass of 0.250 moles of sodium?

= x 5.75 g Na

How many moles of methane (CH4) are in 45.0g of methane?

From the Periodic Table, the molar mass of Na is 22.99g

45.0 g CH4

g CH4

mol CH4 1

16.05

Molar mass of CH4: 1C= 12.01

4H= 4 x 1.01= 4.04

16.05

+

=x 2.80 mol CH4

Page 7: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Relative atomic and molecular mass

The relative molecular mass Mr is the mass of the molecule relative to carbon 12. It has no units.

The relative atomic mass Ar is the mass of an atom relative to carbon 12. It has no units.

It is important to distinguish between the atomic mass (A or Ar ) of elements and the molecular mass (M or Mr ) of elements. For example, the atomic mass of chlorine is 35.45 and its molecular mass is 70.90 because it exists as Cl2

Page 8: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Mass of elements

Molecules of a compound or formula units of a salt.

Mass of a compound

Atoms of an element

12g Fe x

g Fe

mol Fe1

55.85

0.50 mol Na xmol Na

g Na22.99

1

12g NaCl x

gNaCl

mol NaCl1

58.44

0.64 mol NaOH x

mol NaOH

g NaOH

1

40.00

3.1 x 1040molec.NaF Xmolec NaF

mol1

6.02x1023

0.53 mol CaCl2 xmol CaCl2

molec CaCl2

16.02 x 1023

3.1 x 1040atoms of B Xatoms of B

mol of B1

6.02 x 1023

0.72 mol of Ni xmol of Ni

atoms of Ni

1

6.02 x 1023

mol

Page 9: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Mass of elements

Molecules of a compound or formula units of a salt.Mass of a compound

Atoms of an element

..mol

Divide by the Molar Mass (of the element)

Multiply by the Molar Mass (of the element)

Divide by the Molar Mass (of the compound)

Multiply by the Molar Mass (of the compound)

Divide by 6.02 x 1023

Multiply by 6.02 x 1023

Multiply by 6.02 x 1023

Divide by 6.02 x 1023

Page 10: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Formula: % composition of X = mass of X in the compound x 100%

mass of the compound

Percent composition: The percent by mass of each element in a compound. It is also called percent by mass.

Percent composition

Page 11: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Example 1: Calculate the percent composition of each element in sulfuric acid.

Note: If the formula of the compound is not known, it can be determined experimentally.

Example 2: In the laboratory 32.0 g of a compound were analyzed. The compound contained 24.0g of Carbon and 8.00g of hydrogen. What is the percent composition of the of each element in the compound

Example 3: Sodium Hydrogen Carbonate, NaHCO3, also called baking soda is an active ingredient in some antacids used for the relief of indigestion. Determine the percent composition of sodium hydrogen carbonate.

Examples

Page 12: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Molecular formula: The formula for molecular compounds, indicating total atoms of each element in a compound. (For example H2O2).

Empirical formula: The formula that represents the smallest whole number ratio of types of atoms in a compound (or formula unit). (For example HO)

Empirical and Molecular Formulas

The molecular formula can be equal to or a multiple of the empirical formula.

Page 13: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

If the data is in percentage assume 100 g sample and use the percentages as grams of each element.

•Calculate the moles of each element

•Divide by the smallest number of moles obtained

•If the numbers are whole numbers they are the subscripts in the formula.

• If the numbers are not whole numbers, multiply them by the smaller number that makes them all whole numbers. The obtained values are the subscripts in the formula.

To determine the empirical and molecular formulas:

Page 14: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Example –   Methyl acetate, is a compound commonly used in some paints, inks, and adhesives. Determine the empirical formula of methyl acetate, with the following chemical analysis: Percent by mass: 48.65% C, 8.16% H 43.20% O

Page 15: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

• To obtain the molecular formula divide the molar mass of the compound (given) by the molar mass of the empirical formula.

• Multiply the subscripts of the empirical formula by the number obtained in the previous step.

To obtain the Molecular formula

Page 16: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Example –   Methyl acetate, is a compound commonly used in some paints, inks, and adhesives. Determine the empirical formula of methyl acetate, with the following chemical analysis: Percent by mass: 48.65% C, 8.16% H 43.20% O

Page 17: The mole (abbreviation: mol) is the amount of substance equal to 6.02 x 10 23 particles These particles can be atoms, ions, formula units,molecules, electrons,

Example: Succinic acid is a substance produce by lichens. Chemical analysis indicates it is composed of 40.68% oarbon, 5.08 % hydrogen and 54.24% oxygen and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.