Download Lecture 8

10/5/2013
Today:

◦ Mass Percent Composition:
 Empirical Formulas
 Molecular Formulas
 Understand the differences
between these two types of
formulas
Course Survey 2:
◦ Please provide constructive
feedback on this course and
the instruction
 Using experimental data to calculate:

UPDATE:
◦ Suggested homework
problems are include for
each reading
◦ Introduction to Chemical
Reactions:

 Balancing Equations
Next Meeting
◦ Concept Check:
 Law of Conservation of Matter
 General Types of Chemical Reactions
 Covers today’s material
SCORE DISTRIBUTIONS FOR MIDTERM EXAM 1
60
Percentage of Class

50
40
30
20
10
0
< 60 %
60-70 %
70-80 %
80-90 %
90-100 %
Exam Score
1
10/5/2013
Graded Concept Check:
Gram to Mole Conversions
How many moles of hydrogen atoms are present in 39.0 g
of C6H6?
A. ½ mole of H atoms
B. 1.5 mole of H atoms
Information that MIGHT be important:
C = 12.01 amu
H = 1.00794 amu
Density of C6H6 = 0.876 g/mL
C. 3 moles of H atoms
D. 6 moles of H atoms
E. 44.5 moles of H atoms
How is the formula of a compound determined?

We cannot count individual atoms to determine the formula.

BUT: We can measure the composition of a substance
in MASS PERCENT for each element in a compound.
For Example: 100. grams of Natural Gas consists of:
 74.8 grams of Carbon
 25.2 grams of Hydrogen

THEN: The elemental composition by mass can be
related to the composition in moles of each element.
◦ Knowing the mole to mole ratio of each element in a
compound gives us the Empirical formula for that substance.
◦ The empirical formula provides the LOWEST whole number
ratio of the atoms in the substance.
2
10/5/2013
Calculating the Empirical Formula from Mass %
100. grams of Natural Gas consists of:
74.8 grams of Carbon & 25.2 grams of Hydrogen
1. Convert the masses of each element to the number of moles of
each element (using the atomic mass):
2. THEN: Divide the number of moles of each element by the
smallest value present. This gives a mole to mole ratio (or atom to
atom ratio) for each element. THIS IS THE EMPIRICAL FORMULA.
Elemental Composition in Mass Percent
• The mass percent composition of a substance can be
calculated from the chemical formula
The MASS PERCENT tells us the portion of each element by MASS:
% Mass Element
in compound
=
(atomic mass of element )·(# atoms) x 100%
molar mass of the compound
3
10/5/2013
100 g of Benzene


100 g of Acetylene
Liquid
Solvent

 Gas
Used in
welding
Carbon
Hydrogen
Carbon
Hydrogen
92.26 g
7.74 g
92.26 g
7.74 g
100 g of Benzene (C6H6)
Carbon
C1H1
Hydrogen
100 g of Acetylene (C2H2)
Carbon
C1H1 Hydrogen
• Benzene & Acetylene have the same composition of Carbon and
Hydrogen BY MASS. They have the same EMPRICAL FORMULA.
BUT: How are they different? They have different…
MOLECULAR FORMULAS:
• The molecular formula indicates the TOTAL number of atoms
in each individual molecule
4
10/5/2013
iClicker Question:
Converting between units of VOLUME
Glucose is known to have a molar mass of approximately
180.16 g/mol. Which of the following is a possible
EMPIRICAL FORMULA for glucose?
A. C12H24O12
B. C6H12O6
C. CHO
D. CH2O
E. None of the above
Empirical and Molecular Formulas

Empirical Formula - the smallest whole number ratio
of atoms (or moles of atoms) in a compound.
 Calculated from Percent Composition by MASS.

Molecular Formula - the actual number of atoms in a
molecule (some multiple of the empirical formula).
 Calculated from two pieces of information:
1) the Empirical Formula, AND
2) the Molecular Weight (the molar mass of the
molecule)
Example: Octane

5
10/5/2013
Sample Problem: Empirical & Molecular Formulas
What is the molecular formula of the sugar ribose if it is
40.0% C, 6.7% H, and 53.3% O and it has a molecular
mass of 150.1 g/mol?
Sample Problem Solution
First, determine the empirical formula:
Step 1. Assume 100.0 g of the compound.
If you have 100 g of the material, you will have the
following masses of C, H, & O:
40.0 g C
6.7 g H
53.3 g O
6
10/5/2013
Step 2. Convert grams to moles:
Mass Ratio
40.0 g C
1 mol C
12.01 g C
Particle Ratio
= 3.33 mol C
6.7 g H
1 mol H
1.01 g H
= 6.6 mol H
53.3 g O
1 mol O
16.00 g O
= 3.33 mol O
Step 3. Find the smallest ratio of atoms.
C
H
O
3.33 / 3.33 = 1
6.6 / 3.33
3.33 / 3.33 = 1
= 1.98  2
Therefore, the empirical formula of ribose is
CH2O
7
10/5/2013
Next, determine the molecular formula
→ This requires information on the mass of an
individual molecule.
Step 4. Calculate the empirical formula mass.
C =
H =
O =
12.01 x 1 =
1.01 x 2 =
16.00 x 1 =
12.01 g/mol
2.02 g/mol
16.00 g/mol
30.03 g/mol
Step 5. Calculate the number of empirical formula units
that can fit within the molecular formula.
150.1 g/mol = 4.998  5
30.03 g/mol
THEN: Multiply the subscripts of the empirical
formula by this number to get the molecular formula.
C1x5H2x5O1x5 = C5H10O5
8
10/5/2013
Example: Empirical Formulas
Boron carbide, containing only boron and carbon, is
one of the hardest materials known. An analysis of
5.000 grams of boron carbide was found to contain
1.0868 grams of carbon. Determine the empirical
formula of boron carbide from this information.
Example: Empirical & Molecular Formulas
A 30.5-g sample of acrylic acid, used in the
manufacture of acrylic plastics, is found to contain
15.25 g C, 1.71 g H, and 13.54 g O.
In a separate mass spectrometer experiment, the
acrylic acid is found to have a molar mass of
approximately 72 g/mol.
What are the empirical and molecular formulas of
acrylic acid?
9
10/5/2013
Chemical Reactions
“The water a cow drinks turns to milk;
the water a snake drinks turns to poison.”
—Basho, Poet of Japan (1694)
"It’s amazing that the body feeds the
brain sugar and amino acids and what
comes out is poetry and pirouettes.“
--Neurologist Robert Collins
10
10/5/2013
Interpreting a Chemical Equation:
A chemical equation describes a chemical reaction much like a
sentence describes some action.
• Element Symbols  Letters
• Formulas  Words
• Equations  Sentences
(s)
=
Solid
(l)
=
Liquid
(g)
=
Gas
(aq)
=
Aqueous (dissolved in water)
+
=
"and"
=
"reacts to form" or "yields"
Balancing Chemical Equations
Balancing chemical equations is an application of both the
Modern Atomic Theory and the Law of Conservation of Matter.
BALANCING EQUATIONS: The same number of each type of
element must occur on the left (BEFORE the reaction) and on
the right (AFTER the reaction)
11
10/5/2013
Decomposition Reactions
Decomposition: Starts with one substance and breaks it into many
simpler products.
Consider the following chemical reaction:
2 H2O2 → O2(g) + 2 H2O
Sentence form:
Hydrogen peroxide breaks apart (decomposes) into oxygen
gas and water.
Combustion Reactions: Hydrocarbons & Carbohydrates
• Combustion (aka “burning”): Reacting with OXYGEN (O2)
• A hydrocarbon (something containing Hydrogen and Carbon)
burns to produce carbon dioxide (CO2) and water (H2O)
• Special balancing rule for combustion:
Balance Carbon first, Hydrogen second, and Oxygen last.
DOUBLE CHECK
12