Composition Stoichiometry
and Mole Concepts
Atomic Mass
Atomic mass is the mass of an atom in atomic mass units (amu)
By definition:
1 atom
C “weighs” 12 amu
12
The average atomic mass is the weighted average of all of the
naturally occurring isotopes of the element.
The MOLE
• A mole is Avogadro’s number of items
• defined as the amount of substance that contains as many entities (atoms,
molecules, or other particles) as there are atoms in exactly 0.012 kg of
pure carbon-12 atoms.
Avogadro’s number, NA
1 mole = 6.022 x 1023 particles (atoms, molecules or ions)
Formula Weight (FW) of a substance
• sum of the atomic weights (AW) of the elements in the formula, each taken
the number of times the element occurs. Hence a formula weight gives the
mass of one formula unit in atomic mass units.
Molar mass
• in grams/mole, numerically equal to the sum of the masses (in amu) of
the atoms in the formula
3
Examples
Determine the Formula Weight (formula mass) of the
following:
a. H2O Answer: 18.02 amu
Answer: 399.86 amu
b. Fe2(SO4)3
Answer: 249.71 amu
c. CuSO4 . 5H2O
4
Converting Between Mass and
Atoms
1 1.66 10– g or
amu 24
1 g 6.022 amu
1023
M = molar mass in
NA = Avogadro’s number
5
Examples
1. Calculate the number of moles, n, contained in 35.8 g of
water.
ANSWER : 1.98 moles
Examples
2. Calculate the mass contained in 69.8 moles of water.
ANSWER : 1.26 x 103 g
Examples
3. (a) Calculate the number of molecules and (b) number of
hydrogen atoms present in 69.8 moles of water.
ANSWER : a) 4.20 x 1025 molecules
b) 8.41 x 1025 H atoms
Examples
4. Calculate the mass of hydrogen present in 69.8 moles of
water.
ANSWER : 1.41 x 102 g
Percent Composition and Formulas of Compounds
Consider a compound having a chemical formula:
AaBb
Example:
Calculate the percent composition by mass of water.
ANSWER : 11.21% H; 88.79% O
10
Chemical Formula
• Empirical Formula
• Simplest formula of a compound
• Molecular Formula
• True formula of a compound
• Whole-number multiple of the Empirical Formula
Note:
Empirical formula can be the molecular formula but Molecular may not be the
empirical formula
Example
11
*
12
*COMMON FRACTIONS IN DECIMAL FORM in the
whole number ratio of elements in the compound*
Decimal Fraction Multiply by
equivalent
0.50 1/2 2
0.33 1/3 3
0.67 2/3 3
0.25 ¼ 4
0.75 ¾ 4
0.20 1/5 5
*Page69 Whittens (9th ed) 13
Determining Empirical Formula
Experimentally
g CO2 mol CO2 mol C g C
g H2O mol H2O mol H g H
g of O g of sample – g of C g of H
© McGraw-Hill Education. 3-14
1. A compound is composed of 56.4% phosphorus and the remainder
oxygen. Determine the empirical formula of the compound.
ANSWER : EF = P2O3
COMMON FRACTIONS IN
DECIMAL FORM*
Decimal Fraction Multiply
equivalent by
0.50 1/2 2
0.33 1/3 3
0.67 2/3 3
0.25 ¼ 4
0.75 ¾ 4
0.20 1/5 5
th
*Page69 Whittens (9 ed)
2. A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What
is its empirical formula?
KNOW COMMON
Decimal FRACTIONS
Fraction MultiplyIN
equivalentFORM*
DECIMAL by
0.50 1/2 2
0.33 1/3 3
0.67 2/3 3
0.25 ¼ 4
0.75 ¾ 4
0.20 1/5 5
*Page69 Whittens (9th ed)
**Empirical Formula via Combustion Reaction
3. A 0.1014-g sample of purified glucose was burned in a C-H combustion
train to produce 0.1486 g of CO2 and 0.0609 g of H2O. An elemental
analysis showed that glucose contains only carbon, hydrogen, and oxygen.
Other experiments show that its molecular weight is approximately 180
amu. Determine the simplest formula and the molecular formula of
glucose.
Strategy:
g CO2 mol CO2 mol C g C
g H2O mol H2O mol H g H
g of O g of sample – g of C g of H
Empirical Formula Weight= 30.03 amu
Empirical Formula
Molecular Formula
ANSWER : EF = CH2O ; MF = C6H12O6
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