Mole Notes

1. Atomic Mass Unitamu – atomic mass unit, used to describe the

mass of an atom

1 amu = 1.66 x 10-24 g

Example:How many amu are in 27.0 grams of mercury?

x ________________amu Hg

1.66 x 10-24

1

g Hg

27.0 g Hg= 1.63 x 1025 amu Hg

Learning Check

• How many grams are in 1.73 x 1025 atomic mass units of silver?

2. The Mole

mole (mol) – indicates a quantity of a substance that has a mass in grams numerically equal to its atomic mass.

****Round the atomic mass on the periodic table to the HUNDREDTHS PLACE.****

Example:1 mol of copper = ______________ g1 mol of calcium = ______________ g1 mol of chromium = ______________ g

63.55

40.0852.00

Learning Check:

• 1 mol of sodium = ______________ g

3. Molar Massmolar mass (g/mol) – indicates the mass of one

mole of a compound

Example:

Calculate the molar mass of sodium chloride

Calculate the molar mass of silver phosphate

Calculate the molar mass of barium hydroxide

NaCl

“make a little chart”

Na = 1 x 22.99 = 22.99Cl = 1 x 35.45 = 35.45

58.44 g/mol

Ag3PO4

Ag = 3 x 107.87 = 323.61 P = 1 x 30.97 = 30.97 418.58 g/mol O = 4 x 16.00 = 64.00

Ba = 1 x 137.33 = 137.33 O = 2 x 16.00 = 32.00 H = 2 x 1.01 = 2.02

Ba(OH)2 171.35 g/mol

Learning Check:

• Calculate the molar mass of dihydrogen monoxide

4. Avogadro’s Number

Avogadro’s number – indicates the number of atoms, molecules or particles in a mole.

1 mol = 6.02 x 1023 units of a substance

(atoms, molecules, particles, formula units, ions)

Examples:

Mole Mass

What is the mass of 5.0 mol of sulfur?

Mass Mole

How many moles are in 17.0 g of bromine,Br2?

x ______________g S

1

32.06

mol S5.0 mol S = 160 g S

x ____________mol Br2

159.80

1

g Br2

17.0 g Br2 = 0.106 mol Br2

Br = 2 x 79.90 = 159.80 g/mol

Mole Atoms (molecules or particles)

How many atoms are in 2.3 moles of copper?

Atoms (molecules or particles) Mole

How many moles are in 1.24 x 1024 molecules of carbon dioxide?

x __________________atoms Cu

1

6.02 x 1023

mol Cu2.3 mol Cu = 1.4 x 1024 atoms Cu

x ____________________mol CO2

6.02 x 1023

1

molecules CO2

1.24 x 1024 molecules CO2 =

2.06 mol CO2

Atoms (molecules or particles) Grams

How many grams are in 2.4 x 1025 particles of KCl?

Grams Atoms (molecules or particles)

How many atoms are in 514 g of Pb?

x ______________g KCl

mol KCl

74.55

1

x ____________________mol KCl

6.02 x 1023

1

particles KCl

2.4 x 1025 particles KCl =

3.0 x 103 g KCl

K 1 x 39.10 = 39.10

Cl 1 x 35.45 = 35.45

74.55 g/mol

x __________________atoms Pb

mol Pb

6.02 x 1023

1

x ______________mol Pb

207.2

1

g Pb

514 g Pb=

1.49 x 1024 atoms Pb

Learning Check:

• How many particles are in 8.75 g of silver nitrate?

Mole Lab

5. Percent Composition of Compounds

Mass Percent for = mass of the element present in 1 mole of the compound x100%

a given element mass of 1 mol of the compound

Steps for Calculating Percent Composition

1. Calculate the molar mass of the compound.

2. Divide the mass of each element in the compound by the mass of the compound.

3. Multiply each by 100%.

4. Double check. The sum of the mass percents should be 100.

Part

Whole______ x 100%

Examples:

CCl4

NaOH

C = 1 x 12.01 = 12.01Cl = 4 x 35.45 = 141.80

153.81 g/mol

12.01 g

153.81 g% C = ______ x 100%

141.80g

153.81g% Cl = ______ x 100%

= 7.808 %

= 92.192 %

O = 1 x 16.00 = 16.00 H = 1 x 1.01 = 1.01

40.00 g/mol

22.99 g

40.00 g% Na = ______ x 100%

16.00 g

40.00 g% O = ______ x 100%

= 57.48 %

= 40.00 %

Na = 1 x 22.99 = 22.99

1.01 g

40.00 g% H = ______ x 100% = 2.53 %

Learning Check:

• Determine the percent composition for each element in dinitrogen pentoxide.

6. Formulas of Compounds

Empirical formula – the formula of a compound that expresses the smallest whole-number ratio of the atoms present.

Molecular formula – the actual formula of a compound, the formula that tells the actual composition of the molecules that are present.

7. Calculation of Empirical Formulas

Steps for Calculating the Empirical Formula1. Obtain the mass of each element, generally given, but

may involve a subtraction step. For percentages assume a 100 gram sample.

2. Convert grams to moles.3. Find the ratio of elements: Divide the number of moles

of each element by the smallest number of moles. If all calculated values are whole numbers, these are the subscripts in the empirical formula. If NOT whole numbers go to step four.

4. Multiply all the numbers from step three by the smallest whole number that will convert all of them to whole numbers.

Calculate the empirical formula of a compound for a sample that contains 7.808 g of C and 92.192 g Cl.

0.3545 g V reacts with oxygen to achieve a final mass of 0.6330 g. Calculate the empirical formula of the compound.

Calculate the empirical formula of a compound that contains 65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl.

7.808 g C

92.192 g Cl

x _________

x _________g Cl

mol Clg C

mol C1

12.011

35.45

=

=

.6501 mol C

2.601 mol Cl

/ 0.6501

/ 0.6501

= 1

= 4

CCl4

0.3545 g V

0.2785 g O

x _________

x _________g O

mol Og V

mol V1

50.941

16.00

=

=

0.006959 mol V

0.01741 mol O

/ 0.006959

/ 0.006959

= 1

= 2.5

x 2 = 2

x 2 = 5

V2O5

65.02 g Pt

9.34 g N

x ________

x _________g N

mol Ng Pt

mol Pt1

195.081

14.01

=

=

0.3333 mol Pt

0.667 mol N

/ 0.3333

/ 0.3333

= 1

= 2

2.02 g H x _______g H

mol H1

1.01= 2.00 mol H / 0.3333 = 6

23.63 g Cl x ________g Cl

mol Cl1

35.45= 0.6666 mol Cl

/ 0.3333 = 2

PtN2H6Cl2

0.6330 g

- 0.3545 g

0.2785 g

8. Calculation of Molecular Formulas

To calculate the molecular formula, the empirical formula and molecular molar mass are needed.

Molecular Formula = (empirical formula)n

n = Molecular molar mass Empirical molar mass

Steps for Calculating the Molecular Formula1. Calculate the empirical formula, if necessary.2. Find the molar mass of the empirical formula.3. Divide the molecular molar mass by the empirical molar mass.4. multiply the subscripts in the empirical formula by the result of #3.

Example:Calculate the molecular formula of a compound that has a molar mass

of 283.88 g and an empirical formula of P2O5. P 2 x 30.97 = 61.94

O 5 x 16.00 = 80.00

141.94 g/mol

n = 283.88 g

141.94 g

_________ = 2 P4O10

Learning Check:

Determine the empirical formula of a molecule with the percent composition of 25.93 % nitrogen and 74.06 % oxygen. Calculate the molecular formula if the molar mass is 216.04 g/mol.