The Mole

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

• mass of hydrogen atom 1.673 x 10-24 g

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• This is an
• infinitesimal
• mass

1.673 x 10-24 g
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• Chemists require a unit for counting which can
  express large numbers of atoms using simple
  numbers.

• Chemists have chosen a unit for counting atoms.
• That unit is the Mole

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1 mole 6.022 x 1023 objects
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6.022 x 1023 is a very
LARGE
number
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6.022 x 1023 is
Avogadros Number
number
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If 10,000 people started to count Avogadros
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.
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• 1 mole of any element contains
• 6.022 x 1023
• particles of that substance.

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• The atomic mass in grams
• of any element
• contains 1 mole of atoms.

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• This is the same number of particles6.022 x 1023
  as there are in exactly 12 grams of

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Examples
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Species Quantity Number
of H atoms
H
1 mole
6.022 x 1023
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Species Quantity Number
of H2 molecules
H2
1 mole
6.022 x 1023
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Species Quantity Number
of Na atoms
Na
1 mole
6.022 x 1023
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Species Quantity Number
of Fe atoms
Fe
1 mole
6.022 x 1023
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Species Quantity Number of
C6H6 molecules
C6H6
1 mole
6.022 x 1023
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1 mol of atoms
6.022 x 1023 atoms
6.022 x 1023 molecules
1 mol of molecules
6.022 x 1023 ions
1 mol of ions
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• The molar mass of an element is its atomic mass
  in grams.

• It contains 6.022 x 1023 atoms (Avogadros
  number) of the element.

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

• Atomic mass iron 55.85

Conversion sequence grams Fe ? moles Fe
Set up the calculation using a conversion factor
between moles and grams.
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How many iron atoms are contained in 40.0 grams
of iron?

• Atomic mass iron 55.85

Conversion sequence grams Fe ? atoms Fe
Set up the calculation using a conversion factor
between atoms and grams.
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What is the mass of 3.01 x 1023 atoms of sodium
(Na)?

• Molar mass Na 22.99 g

Conversion sequence atoms Na ? grams Na
Set up the calculation using a conversion factor
between grams and atoms.
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What is the mass of 0.365 moles of tin?

• Atomic mass tin 118.7

Conversion sequence moles Sn ? grams Sn
Set up the calculation using a conversion factor
between grams and atoms.
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How many oxygen atoms are present in 2.00 mol of
oxygen molecules?
Two conversion factors are needed
Conversion sequence
moles O2 ? molecules O ? atoms O
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Molar Mass of Compounds
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The molar mass of a compound can be determined by
adding the molar masses of all of the atoms in
its formula.
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Calculate the molar mass of C2H6O.
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
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Calculate the molar mass of LiClO4.
1 Li 1(6.94 g) 6.94 g
1 Cl 1(35.45 g) 35.45 g
4 O 4(16.00 g) 64.00 g
106.39 g
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Calculate the molar mass of (NH4)3PO4 .
3 N 3(14.01 g) 42.03 g
12 H 12(1.01 g) 12.12 g
1 P 1(30.97 g) 30.97 g
4 O 4(16.00 g) 64.00 g
149.12 g
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Molar Mass
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1 MOLE Ca
40.078 g Ca
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Avogadros Number ofH2O molecules
6 x 1023 H2O molecules
1 MOLE H2O
18.02 g H2O
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These relationships are present when hydrogen
combines with chlorine.
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In dealing with diatomic elements (H2, O2, N2,
F2, Cl2, Br2, and I2), distinguish between one
mole of atoms and one mole of molecules.
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Calculate the molar mass of 1 mole of H atoms.
1 H 1(1.01 g) 1.01 g
Calculate the molar mass of 1 mole of H2
molecules.
2 H 2(1.01 g) 2.02 g
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Problems
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How many moles of benzene, C6H6, are present in
390.0 grams of benzene?
The molar mass of C6H6 is 78.12 g.
Conversion sequence grams C6H6 ? moles C6H6
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How many grams of (NH4)3PO4 are contained in 2.52
moles of (NH4)3PO4?
The molar mass of (NH4)3PO4 is 149.12 g.
Conversion sequence moles (NH4)3PO4
? grams
(NH4)3PO4
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56.04 g of N2 contains how many N2 molecules?
The molar mass of N2 is 28.02 g.
Conversion sequence g N2 ? moles N2 ? molecules
N2
Use the conversion factors
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56.04 g of N2 contains how many N2 atoms?
The molar mass of N2 is 28.02 g.
Conversion sequence g N2 ? moles N2 ? molecules
N2 ? atoms N
Use the conversion factors
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Percent Compositionof Compounds
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Percent composition of a compound is the mass
percent of each element in the compound.
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Percent Composition From Formula
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• If the formula of a compound is known, a two-step
  process is needed to calculate the percent
  composition.

Step 1 Calculate the molar mass of the
formula. Step 2 Divide the total mass of each
element in the formula by the molar mass and
multiply by 100.
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Calculate the percent composition of
hydrosulfuric acid H2S.
Step 1 Calculate the molar mass of H2S.
2 H 2 x 1.01g 2.02 g
1 S 1 x 32.07 g 32.07 g
34.09 g
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Calculate the percent composition of
hydrosulfuric acid H2S.
Step 2 Divide the mass of each element by the
molar mass and multiply by 100.
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Percent Composition From Experimental Data
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• 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.
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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.
Step 1 Calculate the total mass of the compound
1.52 g N
3.47 g O
4.99 g
total mass of product
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Calculate the percent composition of
hydrosulfuric acid H2S.
Step 2 Divide the mass of each element by the
total mass of the compound formed.
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Empirical Formula versus Molecular Formula
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• 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.

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• 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.

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Examples
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C2H4
Molecular Formula
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C6H6
Molecular Formula
Empirical Formula
CH
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H2O2
Molecular Formula
HO
Empirical Formula
Smallest Whole Number Ratio
HO 11
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Two compounds can have identical empirical
formulas and different molecular formulas.
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CalculatingEmpirical Formulas
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• 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 elements molar
mass.
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• 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.
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• 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
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• The results of calculations may differ from a
  whole number.

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

2.9
? 3

• Deviations greater than 0.1 unit from a whole
  number usually mean that the calculated ratios
  have to be multiplied by a whole number.

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Some Common Fractions and Their Decimal
Equivalents
DecimalEquivalent
Common Fraction
0.25
0.333
Multiply the decimal equivalent by the number in
the denominator of the fraction to get a whole
number.
0.666
0.5
0.75
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Problems
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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
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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.
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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.
The simplest ratio of KCO is 213 Empirical
formula K2CO3
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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
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The percent composition of a compound is 25.94
nitrogen (N), and 74.06 oxygen (O). Calculate
the empirical formula for this substance.
Step 2 Convert the grams of each element to
moles.
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The percent composition of a compound is 25.94
nitrogen (N), and 74.06 oxygen (O). Calculate
the empirical formula for this substance.
Step 3 Divide each number of moles by the
smallest value.
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The percent composition of a compound is 25.94
nitrogen (N), and 74.06 oxygen (O). Calculate
the empirical formula for this substance.
Step 4 Multiply each of the values by 2.
N (1.000)2 2.000
O (2.500)2 5.000
Empirical formula N2O5
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Calculating the Molecular Formula from the
Empirical Formula
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• The molecular formula can be calculated from the
  empirical formula if the molar mass 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.

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What is the molecular formula of a compound which
has an empirical formula of CH2 and a molar mass
of 126.2 g?
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 14.03g
The molecular formula is (CH2)9 C9H18
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The End