Transcript 11.4: Empirical and Molecular Formulas

Table of Contents
Chapter 11: More on the Mole
11.4: Empirical and Molecular Formulas
11.5: Hydrates
11.4: Empirical and Molecular Formulas
• Percent composition: the percent by mass of each
element in a compound
% by mass=
(moles element) x (mass of element)
Mass of compound
• The sum of the mass percents = 100%
x100
11.4: Percent composition
• What is the percent composition of H2O?
H:
(2mol H) x (1.01g/mol H)
x 100 = 11.2% H
18.02 g H2O
O: (1mol O) x (16.00g/mol O)
x 100 = 88.80% O
18.02 g H2O
• The percent composition of H2O is 11.2% H
and 88.80% O.
11.4: Percent composition
• What is the percent composition of NaHCO3?
Na: (1mol Na) x (22.99g/mol Na)
84.01 g NaHCO3
H:
(1mol H) x (1.008g/mol H)
84.01 g NaHCO3
C:
(1mol C) x (12.01g/mol C)
84.01 g NaHCO3
O:
(3mol O) x (16.00g/mol O)
84.01 g NaHCO3
x 100 = 27.37% Na
x 100 = 1.2% H
x 100 = 14.3% C
x 100 = 57.14% O
11.4: Empirical and Molecular Formulas
• Why? This process can be used to determine the
composition of an unknown compound
• Empirical Formula: the formula with the simplest
whole number ratios
11.4: Empirical and Molecular Formulas
Finding Empirical Formulas
• You have an substance made of 40.05 % Sulfur and
59.95% Oxygen. What is its empirical formula?
1) Assume percentages are masses
• S = 40.05 g
• O = 59.95 g
11.4: Empirical and Molecular Formulas
You have an substance made of 40.05 % Sulfur and
59.95% Oxygen. What is its empirical formula?
2) Convert grams to moles → ratio
• S = 40.05 g 1 mol S
32.07 g S
• O = 59.95 g 1 mol O
16.00 g O
= 1.249 mol S
= 3.747 mol O
• Ratio S : O = 1.249 : 3.747
11.4: Empirical and Molecular Formulas
You have an substance made of 40.05 % Sulfur and
59.95% Oxygen. What is its empirical formula?
3) Simplify ratio: divide each by smallest moles
• S = 1.249 mol S
1.25
= 1 mol S
• O = 3.747 mol O = 3 mol O
1.25
• Empirical formula: SO3
11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a
compound with 48.64 % C, 8.16% H, and 43.20% O.
1) Assume percentages are masses
• C = 48.64 g
• H = 8.16 g
• O = 43.20 g
11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a
compound with 48.64 % C, 8.16% H, and 43.20% O.
2) Convert grams to moles → ratio
• C = 48.64 g 1 mol C
12.01 g C
• H = 8.16 g
1 mol H
1.008 g H
• O = 43.20 g 1 mol O
16.00 g O
= 4.050 mol C
= 8.10 mol H
= 2.700 mol O
11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a
compound with 48.64 % C, 8.16% H, and 43.20% O.
3) Simplify ratio: divide each by smallest moles
• C = 4.050 mol C = 1.5 mol C
2.700
• H = 8.10 mol H
2.700
= 3 mol H
• O = 2.700 mol O = 1 mol O
2.700
11.4: Empirical and Molecular Formulas
EXAMPLE 1: Determine the empirical formula for a
compound with 48.64 % C, 8.16% H, and 43.20% O.
3) Simplify ratio: whole numbers
• C = 1.5 mol x 2 = 3 mol C
• H = 3 mol x 2 = 6 mol H
• O = 1 mol x 2 = 2 mol O
• Empirical formula: C3H6O2
11.4: Empirical and Molecular Formulas
Practice:
Ex 2) A blue solid has 36.84% nitrogen and 63.16%
Oxygen. What is its empirical formula?
N2O3
Ex 3) Aspirin contains 60.00% carbon, 4.44%
hydrogen, and 35.56% oxygen. What is its empirical
formula?
C9H8O4
11.4: Empirical and Molecular Formulas
• Sometimes substances with different properties have
the same percent composition and empirical
formula. WHAT?
• Molecular Formula: the actual number of atoms in
a molecule/formula unit of a substance
11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
• The molar mass of acetylene is 26.04 g/mol. The
mass of its empirical formula (CH) is 13.02 g/mol
1) Divide:
correct molar mass
mass of empirical formula
=
26.04 g/mol
13.02 g/mol
= 2
So… the molecular formula is 2 times the empirical
formula
11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
• The molar mass of acetylene is 26.04 g/mol. The
mass of its empirical formula (CH) is 13.02 g/mol
2) Multiply:
CH x 2 = C2H2
11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
EXAMPLE 4: Succinic acid has a molar mass of
118.1 g/mol and an empirical formula of C2H3O2
1) Divide:
correct molar mass
mass of empirical formula
=
118.1 g/mol
59.04 g/mol
= 2
So… the molecular formula is 2 times the empirical
formula
11.4: Empirical and Molecular Formulas
Finding Molecular Formulas
EXAMPLE 4: Succinic acid has a molar mass of
118.1 g/mol and an empirical formula of C2H3O2
2) Multiply:
C2H3O2 x 2 = C4H6O4
11.4: Empirical and Molecular Formulas
Practice:
Ex 5) A photographic developing fluid has a chemical
composition of 65.45% C, 5.45% H, and 29.09% O.
Its molar mass is 110.0 g/mol. What is its molecular
formula? (Find its empirical formula first)
C6H6O6
Ex 6) A colorless liquid is composed of 46.68%
nitrogen and 53.32% oxygen has a molar mass of
60.01 g/mol. What is the molecular formula?
N2O2