Download Unit 3

Unit III: Quantitative
Composition of Compounds
A. Atoms and Isotopes
B. Atomic Composition of Chemical
Compounds
C. Formula and Molecular Mass
D. Calculations using Moles of Atoms
E. Calculations using Moles of Compounds
F. Mass Percent Composition of Compounds
G. Empirical and Molecular Formulas
The Structure of an Atom
e
p
pn n
n
p np
e
e
e
3-A
What is an atom?
An atom is a fundamental piece of matter,
composed of subatomic particles.
An single atom is the smallest amount of an
element that maintains properties of that
element.
Atoms are not created, changed, or
destroyed during a chemical reaction
3-A
The Structure of an Atom
Proton
e
e
pn p
n
pn
e
(+1)
1.673 x 10-24 g
p
Neutron
n
1.674 x 10-24 g
Electron
e
(0)
(-1)
9.109 x 10-28 g
3-A
Atomic Mass
Protons and neutrons are regarded as
having one atomic mass unit (1 AMU)
Electrons have a mass of zero AMU
The total number of protons and neutrons in
an atom is called the atomic mass
3-A
Isotopes
While all atoms of the same element behave
similarly in chemical reactions, not all
atoms are identical
Atoms always have the same number of
protons, but may have different masses….
3-A
Isotopes of hydrogen
e
NP
1
2
3
N
H
1
Atomic mass = 3
1
2
Deuterium
Hydrogen
Tritium
3-A
Isotopes
Mass number
A
Atomic number
Z
30
Zn
zinc
65.38
E
Element symbol
The isotope zinc-65
65
30
Zn
3-A
Practice with isotopes
Name the following isotopes
13
6
C
carbon-13
15
7
N
nitrogen-15
How many protons are in each of the following isotopes?
25
12
Mg
30
15
P
3-A
Fill in the gaps in the following table
assuming each column represents a neutral
atom:
Isotopic Symbol
Number of
Protons
34
Number of
Neutrons
28
Number of Electrons
95
Mass Number
241
42
23
3-A
Kinds of isotopes
Stable and radio isotopes
Oxygen for example has isotopes with
between 4 and 16 neutrons.
Only those atoms with 8, 9, or 10 neutrons
are stable.
3-A
Atomic mass vs. Atomic weight
Atomic Mass refers to the mass of a
particular isotope of a particular element
Atomic weight refers to the weighted
average mass of all the isotopes of an
element.
(you might also see this as relative atomic
mass)
Atomic Composition of Chemical
Compounds
To calculate formula weights we need to
know the number of atoms of each
element in a compound.
Example: Hexanoic acid
CH3(CH2)3CH2COOH
6 carbon atoms
2 oxygen atoms
12 hydrogen atoms
3-B
UNIT 3-C: Formula and Molecular
weights
Molecular weight refers to covalent
compounds
Formula weight refers to the weight of an
ionic compound’s formula unit.
3-C
UNIT 3-C: Lecture Problem 1 (pg 95)
A. Find the formula weight of barium nitrate.
B. Find the molecular weight of diphosphorus
pentaoxide
3-C
Unit 3-D: Calculations Using Moles
of Atoms
When we talk about practical amounts of
elements or compound we are talking
about huge numbers.
We need a convenient term that describes a
large group of things (similar to a dozen
or gross). We use the term:
‘mole’
3-D
Moles
A ‘mole’ of anything always contains the
same number, 6.022 x 1023
This special number is called Avogadro’s
number.
It just happens that there are 6.022 x 10 23
atoms of carbon-12 in exactly 12 grams of
carbon-12.
3-D
Moles and masses
The atomic mass of C-12 is 12 amu
If the atomic mass of He-4 (for example) is
4 amu.. (1/3 the mass of C-12)
….and there are 12 g of C-12 in one mole of C12…..
Then there will be 4 g of He-4 in one mole of He-4
3-D
Lecture Problem 3-2
A. What is the atomic weight of zinc (in
amu)?
B.
How many grams are in a mol of zinc?
C. How many atoms are present in a mol
of zinc?
Page 97
3-D
Lecture Problem 3-3
A. How many grams are in 2.20 mol of
iron?
B. Calculate the number of mol in 2.45 x
10-5 g of calcium.
Pg. 97
3-D
Unit 3 Part E:
Calculations using Moles of
Compounds
Calculating molecular weights and
molar masses
If we add up the atomic weights of all the
atoms in a compound we get the
molecular weight. (units of amu)
If we want to talk about the mass of 1 mol
of a compound we replace the units of
amu with the units g/mol.
3-E
Lecture Problem III-4 (pg. 99)
What is the molar mass of ammonium
sulfide?
(NH4)2S
Molar mass of (NH4)2S is 68.17 g/mol
3-E
Unit factors
The concept of the ‘mol’ is very useful in
constructing unit factors..
Using the example of glucose (C6H12O6)
With a molar mass of 180.18g/mol
Write as many unit factors as possible
utilizing the mol concept.
3-E
Lecture Problem 3-5 (pg. 101)
A. How many mol of hydrogen atoms are
in 83.5 g of C6H12O6?
B. How many water molecules are in
0.946 mol of water?
Extra practice: How many total atoms are in
55.3 g of sulfur dioxide?
3-E
Using mols to describe
composition of compounds
Lecture problem III-6 (pg. 102)
A sample of ammonium dichromate contains 3.07 mol
of nitrogen atoms. What is the mass of the sample
of ammonium dichromate?
(NH4)2Cr2O7 has a molar mass of 252.10 g/mol
386.97 g of (NH4)2Cr2O7 rounded to 387 g
3-E
Lecture problem III-7 (pg. 103)
How many grams of bromine are present in a sample
of Ca(BrO3)2 that is found to contain 10.0 g of
calcium?
Ca has a molar mass of 40.08 g/mol and
Br has a molar mass of 79.90 g/mol
39.87 g of Br rounded to 39.9 g
3-E
Unit 3-F: Mass Percent
Composition
1. Calculate the formula or molecular
weight of the compound.
2. Determine the total weight of each
element in the compound.
% composition of an element = Total weight of that element x 100
formula or molecular weight
Unit 3-F: Mass Percent
Composition
Lecture problem III-8 (pg. 105).
Calculate the percent by mass of calcium in calcium
phosphate.
Ca3(PO4)2
molar mass of 310.2 g/mol
38.76% Ca by mass in CaPO4
Unit 3-G: Empirical and Molecular
Formulas
The empirical formula of a compound
represents the smallest ration of atoms that are
present in a compound.
The formula for an ionic compound is in essence an
empirical formula
The molecular formula is the true formula,
representing the number of atoms of each
element that are present in a single molecule of
a compound
Unit 3-G: Empirical and Molecular
Formulas
Consider dinitrogen tetraoxide
Based on the compound name we can write its
molecular formula: N2O4
The empirical formula describes the smallest ratio
of atoms that are present in a compound.
There is 1 nitrogen atom for every 2 oxygen atoms
So the smallest ratio, or empirical formula) is…
NO2
Unit 3-G: Empirical and Molecular
Formulas
Given the percent composition of an
unknown compound, how can we
determine its identity?
(sometimes actual masses for each element may be given)
The first step is to turn the mass percentage
into a ratio of atoms in the compound,
called an empirical formula.
Solving Empirical Formula Problems
1. If starting with % composition assume exactly
100 grams of sample. Then percentages
become mass quantities. (If starting with mass, skip to
step 2)
2. Calculate the number of mol of each element
present using our unit analysis method.
3. Find the simplest mol ratio by dividing all of the
mol quantities by the smallest number of mol of
an element present.
Lecture problems III-9 (page 108-109)
A.
Calculate the empirical formula of a
compound that contains 56.4% phosphorus and
43.6% oxygen
B. A compound contains 1.79g of phosphorus and
2.29g of oxygen. What is the empirical formula
of the compound?
C. It is found that 2.231 grams of selenium react
with fluorine to form a selenium fluoride with a
mass of 5.452 grams. What is the empirical
formula of this compound?
Finding the Molecular Formula
Find the empirical formula for the compound.
– Calculate the empirical molar mass.
Divide the molecular molar mass by the empirical
molar mass. This is our multiplying factor
Multiply the quantity of each element in the
empirical formula by this factor.
The result is the molecular formula.
Molecular Formula Determination
Lecture problem (III-10 pg. 111):
A. What is the molecular formula for galactose if the
empirical formula is CH2O and the molar mass is
180.2 g/mol?
B. A molecular compound contains 40.0% carbon,
6.7% hydrogen and 53.3% oxygen by mass.
A 0.100 mol sample of this compound weighs 6.00
grams. What is the molecular formula of the
compound?
Extra practice with ‘mystery metal’
problems
0.125 mol of chlorine gas react with 0.0625 mol of
a metal to form 16.3 grams of a binary ionic
compound. What is the metal, and what is the
formula for the compound?
5.11 grams of chlorine gas reacts with 0.0480 mol
of a metal to form 14.56 grams of a new solid
compound with the formula MCl3. What is the
metal (symbol)?