Chapter 4
The Mole and
Stoichiometry
Chemistry: The Molecular Nature
of Matter, 6E
Brady/Jespersen/Hyslop
• The molecular scale versus the
laboratory scale:
• Defining the mole:
• A number equal to the number of atoms
in exactly 12 gram of 𝟏𝟐𝑪 atoms.
• 1 mole of element X= gram atomic mass of X
• Example:
• 1 mole of sulfur = 32.6 g
atomic mass of sulfur = 32.6 u
• The mole concept applied to compound:
• 1 mole of molecules X = gram molecular
mass of X
• Example:
• The molecular mass of water = 18.02 u
• The sum of atomic mass of two H atoms
and one O atmo.
• 1 mole of 𝑯𝟐 O= 18.02 g
• 1 mole of ionic compound X = gram
formula mass of X
• Example:
• 1 mole of 𝑨𝑳𝟐 𝑶𝟑 =101.96 g
• 𝑨𝑳𝟐 𝑶𝟑 has 2AL with an atomic masses
26.98u and three oxygen with mass of
16 u.
• 1 mole of X = gram molar mass of X
• Converting from gram to moles:
• Example:
• In experiment to prepare of titanium(IV)
oxide we start with 23.5g sample of
titanium. How many moles of titanium do
we have?
• Solution:
• 1 mole of element X = gram atomic mass of
X
• 1 mole of Ti = 47.867g Ti
• 23 .5g Ti = ? Mole Ti
• Conversion factor:
𝟏 𝒎𝒐𝒍𝒆 𝑻𝒊
𝟒𝟕. 𝟖𝟔𝟕 𝒈𝑻𝒊
• 23.5 gTi ×
𝟏 𝒎𝒐𝒍𝒆 𝑻𝒊
𝟒𝟕,𝟖𝟔𝟕 𝒈𝑻𝒊
= 0.491 mole Ti
• Conversion from mole to grams:
• Example:
• We need 0.254 moles of 𝑭𝒆𝑪𝑳𝟑 for
certain experiment. How many grams
would you need weight?
• Solution:
• 1 mole of 𝑭𝒆𝑪𝑳𝟑 = 162.204 g
• Molar mass 𝑭𝒆𝑪𝑳𝟑 = 55.845 g/mole +
( 3× 35.453) g/mole = 162.204 g/mole
• 0.254 mole 𝑭𝒆𝑪𝑳𝟑 = ? g 𝑭𝒆𝑪𝑳𝟑
• Conversion factor:
𝟏𝟔𝟐.𝟐𝟎𝟒 𝒈𝑭𝒆𝑪𝑳𝟑
𝟏 𝒎𝒐𝒍𝒆 𝑭𝒆𝑪𝑳𝟑
0.254 ×
𝟏𝟔𝟐.𝟐𝟎𝟒 𝒈𝑭𝒆𝑪𝑳𝟑
𝟏 𝒎𝒐𝒍𝒆 𝑭𝒆𝑪𝑳𝟑
= 41.2 g 𝑭𝒆𝑪𝑳𝟑
Learning check
* How many moles of aluminum are
there iN 3.47 gram sheet of aluminum foil.
• Avogadros number:
• The relationship between the atomic scale
and laboratory scale as
• 1 mole of X = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 unit of X
• The unit can be atoms , molecules , formula
unit.
• Example :
• 1 mole of Xe = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 atoms of Xe
• 1 mole of 𝑵𝑶𝟐 = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 molecules
of 𝑵𝑶𝟐
• Converting from the laboratory scale to
the atomic scale.
• Example:
• In lightbulb the tungsten weight 0.653 g.
how many atoms of tungsten are there
in such sample.
• Solution:
• 0.632 g W = ? Atoms of W
• gram W to mole W to atom W
• 1 mole of W = 183.84 g W
• 1 mole W = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 atoms W
• Conversion factor :
• 1.
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟏𝟖𝟑.𝟖𝟒 𝒈 𝑾
• 2.
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑾
𝟏 𝒎𝒐𝒍𝒆 𝑾
• 0.635 ×
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟏𝟖𝟑.𝟖𝟒 𝒈 𝑾
×
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑾
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟐𝟏
2.08 × 𝟏𝟎
=
atoms W
• Calculating the mass of molecules:
• Example :
• What is the average mass of one molecule of
carbon tetrachloride.
• Soluation:
• 1 molecule 𝑪𝑪𝑳𝟒 = ? g 𝑪𝑪𝑳𝟒
• 1 mole of 𝑪𝑪𝑳𝟒 = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 molecules of
𝑪𝑪𝑳𝟒
• 1 mole of 𝑪𝑪𝑳𝟒 = 153.823 g 𝑪𝑪𝑳𝟒
• Molar mass of 𝑪𝑪𝑳𝟒 = 12 + (4×
35.45)=153.823 g\mole
• Conversion factor :
• 1.
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝑪𝑪𝑳𝟒
• 2.
𝟏𝟓𝟑.𝟖𝟐𝟑 𝒈𝑪𝑪𝑳𝟒
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
• 1 molecule 𝑪𝑪𝑳𝟒 ×
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
𝟏𝟓𝟑.𝟖𝟐𝟑 𝒈𝑪𝑪𝑳𝟒
×
𝟐𝟑
𝟔.𝟎𝟐𝟐×𝟏𝟎 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝑪𝑪𝑳𝟒
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
𝟐. 𝟓𝟓𝟓 × 𝟏𝟎−𝟐𝟐 g 𝑪𝑪𝑳𝟒 .
=
• Chemical formulas and stoichiometry:
• Mole to mole conversion factors:
• Calculating the amount of a compound
by analyzing one element:
• The formula
𝑷𝟒 𝑶𝟏𝟎 𝐢𝐦𝐩𝐥𝐢𝐞𝐬 𝐨𝐭𝐡𝐞𝐫 𝐞𝐪𝐮𝐢valance
each with two conversion factors:
• 1 mole 𝑷𝟒 𝑶𝟏𝟎 ↔ 4 mole P
1 mole 𝑷𝟒 𝑶𝟏𝟎
𝟒 𝒎𝒐𝒍𝒆 𝑷
• C.v:
or
𝟒 𝒎𝒐𝒍𝒆 𝑷
1 mole 𝑷𝟒 𝑶𝟏𝟎
• 1 mole 𝑷𝟒 𝑶𝟏𝟎 ↔ 10 mole O
• C.V:
1 mole 𝑷𝟒 𝑶𝟏𝟎
𝟏𝟎 𝒎𝒐𝒍𝒆 𝑶
•
or
𝟏𝟎 𝒎𝒐𝒍𝒆 𝑶
1 mole 𝑷𝟒 𝑶𝟏𝟎
• Example:
• Sample of calcium phosphate is found to
contain 0.864 moles of phosphours.
How many moles of 𝑪𝒂𝟑 (𝑷𝑶𝟒 )𝟐 .
• Solution:
• 0.864 mole P = ? mole 𝑪𝒂𝟑 (𝑷𝑶𝟒 )𝟐
• 1 mole 𝑪𝒂𝟑 (𝑷𝑶𝟒 )𝟐 = 2 mole P
• Conversion factor:
•
1 mole of 𝑪𝒂𝟑 (𝑷𝑶𝟒 )𝟐
𝟐 𝒎𝒐𝒍𝒆 𝑷
• 0.864 mole P ×
1 mole of 𝑪𝒂𝟑 (𝑷𝑶𝟒 )𝟐
𝟐 𝒎𝒐𝒍𝒆 𝑷
=
0.432 mole 𝑪𝒂𝟑 (𝑷𝑶𝟒 )𝟐
Learning check
• Aluminum sulfate is analysize and the sample
contain 0.774 moles of sulfate ions. How
many moles of aluminum are in sample?
• 0.774 mole 𝑨𝑳𝟐 (𝑺𝑶𝟒 )3 = ? mole AL
• Mass to mass calculation:
• Calculating the amount of one element
from the another of another in
compound:
• Example:
• Sample of green pigment has formula
𝑪𝟐𝟐 𝑯𝟕𝟐 𝑴𝒈 𝑵𝟒 𝑶𝟓 if 0.0011g Mg is
avalable how many grams of carbon will
be required to completely use up the
magnisum?
• Solution :
• 0.0011g Mg ↔ 𝒈 C
•
•
•
•
1 mole of Mg = 55 mole of C
1 mole Mg = 24.3050 g Mg
1mole C = 12 g C
0.0011 g Mg → 𝒎𝒐𝒍 𝑴𝒈 → 𝒎𝒐𝒍 𝑪 →
𝒈𝑪
• Conversion factor:
• 1.
𝟏𝑴𝒐𝒍 𝑴𝒈
𝟐𝟒.𝟑 𝒈 𝑴𝒈
•
•
𝟓𝟓 𝒎𝒐𝒍 𝑪
2.
𝟏 𝒎𝒐𝒍 𝑪
𝟏𝟐.𝟎 𝒈 𝑪
3.
𝟏 𝒎𝒐𝒍 𝑪
• 0.0011 g Mg ×
𝟏𝑴𝒐𝒍 𝑴𝒈
𝟐𝟒.𝟑 𝒈 𝑴𝒈
𝟏𝟐.𝟎 𝒈 𝑪
𝟏 𝒎𝒐𝒍 𝑪
= 0.03 g C
×
𝟓𝟓 𝒎𝒐𝒍 𝑪
𝟏 𝒎𝒐𝒍 𝑪
×
• Percentage composition:• Called percentage by mass of element:
is the number of grams of the element
present in 100g of the compound.
• Percentage by mass of element =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒘𝒉𝒐𝒍𝒆 𝒔𝒂𝒎𝒑𝒍𝒆
× 𝟏𝟎𝟎%
• Example:
• A sample of liquid with a mass of 8.657 g
was decomposed into its element and gave
5.217g of carbon, 0.9620 g of hydrogen, and
2.478 g of oxyge. What is the percentage
composition of this compound?
• solution:
• For C :
• For H :
𝟓.𝟐𝟏𝟕 𝒈
× 𝟏𝟎𝟎% = 𝟔𝟎. 𝟐𝟔%
𝟖.𝟔𝟓𝟕 𝒈
𝟎.𝟗𝟔𝟐 𝒈
× 𝟏𝟎𝟎% = 𝟏𝟏. 𝟏𝟏 %
𝟖.𝟔𝟓𝟕 𝒈
C
• For O :
𝟐.𝟒𝟕𝟖 𝒈
𝟖.𝟔𝟓𝟕 𝒈
× 𝟏𝟎𝟎 % = 𝟐𝟖. 𝟔𝟐 %
• Sum of percentage: 99.99%
Learning check
* An organic compound weighing 0.6672 g
is decomposed , giving 0.3481 g of carbon
, 0,087 g hydrogen. What is the
percentage of hydrogen and carbon in this
compound?