Atoms: the Building Blocks of
Matter

The parts that make up an atom are called subatomic
particles.
1) Protons (p+) positively charged particle
2) Neutron (no) neutral particle (uncharged)
3) Electrons (e-) negatively charged particle
 Neutrons and Protons are
located in the nucleus of
an atom.
 Electrons orbit around
the nucleus.
Q- How are atoms of different elements distinguished from
one another? In other words, how do we distinguish a
helium atom from a carbon atom?
A- Their number of protons, indicated by the atomic number
Let’s look at helium, He.
It has an atomic number of 2,
which means that is has 2
protons in it’s nucleus.
Atomic Structure
Here are the basics; you need to know these.
1
Atomic Number
H
Atomic Symbol
1.0076
Atomic Mass
Hydrogen
Atomic Number (Z): the number of protons (p+)
Atomic Mass: the number of protons (p+) + the number of neutrons (n0)
▪ measured in atomic mass units (amu) which is one twelfth
the mass of a carbon-12 atom.
▪ the mass of electrons (1/1860 p+) is negligible.
Number of Neutrons: the atomic mass - the atomic number
Lets practice! Find the missing
information?
Element
Atomic # Atomic
Mass
Ar
18
Electrons Neutrons
39.948
amu
He
O
Protons
2
15.999
amu
2
8
The Famous Gold Foil Experiment
This showed us that the atom is made of mostly
empty space.
Isotopes
Atoms of the same element with different
number of neutrons
Because they have the same number of protons, all
isotopes of an element have the same chemical
properties.
Mass Numbers of Hydrogen Isotopes
What would the masses be?
The Mole: A
Measurement of
Matter
At the end of this section, you should
be able to:
Describe
how Avogadro’s number is
related to a mole of any substance
Calculate the mass of a mole of
any substance
The Mole
(aka Avagadro’s
Number):
6.02 x
23
10
The Mole and Avogadro’s Number
SI unit that measures the amount of substance
 1 mole = 6.02 x 1023 representative particles
 Representative particles are usually atoms,
molecules, or formula units (ions)

But Why the Mole?
Just as 12 = 1 dozen, or 63,360 inches = 1 mile,
the mole allows us to count microscopic items
(atoms, ion, molecules) on a macroscopic scale.
So, 1 mole of any substance is a set number of
Items, namely: 6.02 x 1023.
Chemistry = awesome
Examples:
Substanc
e
Representative
Particle
Atomic
nitrogen
Atom
N
6.02 x 1023
Water
Molecule
H2O
6.02 x 1023
Ca2+
6.02 x 1023
Calcium ion Ion
Chemical
Formula
Representative
Particles in
1.00 mol
Solve
Substance
Representative
Particle
Formula
Unit
Nitrogen
gas
Calcium
Fluoride
N2
Molecule
CaF2
Molecule
Sucrose
C12H22O11
Molecule
Carbon
C
Molecule
Representativ
e Particles in
1.00 mol
Answers
Nitrogen gas-molecule-N2
 Calcium fluoride-formula unit-CaF2
 Sucrose-molecule-C12H22O11
 Carbon-atom-C

All have 6.02 x 1023 representative
particles in 1.00 mol
How many atoms are in a mole?





Determined from the chemical formula
List the elements and count the atoms
Solve for CO2
C - 1 carbon atom
O - 2 oxygen atoms
Add: 1 + 2 = 3
Answer: 3 times Avogadro’s number of atoms
Solve: How many atoms are in a
mole of
1.
 2.
 3.
 4.

Carbon monoxide – CO
Glucose – C6H12O6
Propane – C3H8
Water – H2O
How many moles of magnesium is
1.25 x 1023 atoms of magnesium?
Refer to page 174 in text
 Divide the number of atoms or molecules
given in the example by 6.02 x 1023
 Divide (1.25 x 1023) by (6.02 x 1023)
 Express in scientific notation
 Answer = 2.08 x 10-1 mol Mg

Objectives
Use the molar mass to convert between
mass and moles of a substance
 Use the mole to convert among
measurements of mass, volume, and
number of particles

Molar mass
Mass (in grams) of one mole of a
substance
 Broad term (can be substituted) for gram
atomic mass, gram formula mass, and
gram molecular mass
 Can be unclear: What is the molar mass
of oxygen?
O or O2 ? - element O or molecular
compound O2 ?

Calculating the Molar Mass of
Compounds (Molecular and Ionic)
1. List the elements
 2. Count the atoms
 3. Multiply the number of atoms of the
element by the atomic mass of the
element (atomic mass is on the periodic
table)
 4. Add the masses of each element
 5. Express to tenths place

What is the molar mass (gfm) of
ammonium carbonate (NH4)2CO3?
N 2 x 14.0 g = 28.0 g
H
8 x 1.0 g =
8.0 g
C
1 x 12.0 g = 12.0 g
O
3 x 16.0 g = 48.0 g
 Add
________
 Answer
96.0 g

Practice Problems
1. How many grams are in 9.45 mol
of dinitrogen trioxide (N2O3) ?
a. Calculate the grams in one mole
b. Multiply the grams by the number
of moles
 2. Find the number of moles in 92.2
g
of iron(III) oxide (Fe2O3).
a. Calculate the grams in one mole
b. Divide the given grams by the
grams in one mole

Answers
1. 718 g N2O3 (one mole is 76.0g)
 2. 0.578 mol Fe2O3 (one mole is 159.6 g)

Volume of a Mole of Gas
Varies with a change in temperature or a
change in pressure
 At STP, 1 mole of any gas occupies a
volume of 22.4 L
 Standard temperature is 0°C
 Standard pressure is 101.3 kPa
(kilopascals), or 1 atmosphere (atm)
 22.4 L is known as the molar volume


22.4 L of any gas at STP contains 6.02 x
1023 representative particles of that gas

One mole of a gaseous element and one
mole of a gaseous compound both
occupy a volume of 22.4 L at STP
(Masses may differ)

Molar mass (g/mol) = Density (g/L) x
Molar Volume (L/mol)
Objectives
Define the terms
 Calculate the percent composition of a
substance from its chemical formula or
experimental data
 Derive the empirical formula and the
molecular formula of a compound from
experimental data

Terms to Know
Percent composition – relative amounts of
each element in a compound
 Empirical formula – lowest whole- number
ratio of the atoms of an element in a
compound

An 8.20 g piece of magnesium
combines completely with 5.40 g
of oxygen to form a compound.
What is the percent composition
of this compound?
1. Calculate the total mass
2. Divide each given by the total mass
and then multiply by 100%
3. Check your answer: The
percentages should total 100%
Answer
The total mass is 8.20 g + 5.40 g = 13.60
g
 Divide 8.2 g by 13.6 g and then multiply
by 100% = 60.29412 = 60.3%
 Divide 5.4 g by 13.6 g and then multiply
by 100% = 39.70588 = 39.7%
 Check your answer: 60.3% + 39.7% =
100%

Calculate the percent composition
of propane (C3H8)
1. List the elements
 2. Count the atoms
 3. Multiply the number of atoms of
the element by the atomic mass of
the element (atomic mass is on the
periodic table)
 4. Express each element as a
percentage of the total molar mass
 5. Check your answer

Answer
Total molar mass = 44.0 g/mol
 36.0 g C = 81.8%
 8.0 g H = 18.2%

Calculate the mass of carbon in
52.0 g of propane (C3H8)
Calculate the percent composition using
the formula (See previous problem)
2. Determine 81.8% of 82.0 g
Move decimal two places to the
left (.818 x 82 g)
3. Answer = 67.1 g
1.
1) Find the percent composition of
Aluminum Oxide (Al3O2)
2) How much of a 5-g piece of Iron
Bromide (FeBr3) is iron?
Calculating Empirical Formulas
Microscopic – atoms
 Macroscopic – moles of atoms
 Lowest whole-number ratio may not be
the same as the compound formula
Example: The empirical formula of
hydrogen peroxide (H2O2) is HO

Empirical Formulas
The first step is to find the mole-to-mole
ratio of the elements in the compound
 If the numbers are both whole numbers,
these will be the subscripts of the elements
in the formula
 If the whole numbers are identical,
substitute the number 1
Example: C2H2 and C8H8 have an empirical
formula of CH
 If either or both numbers are not whole
numbers, numbers in the ratio must be
multiplied by the same number to yield
whole number subscripts

What is the empirical
formula of a compound that
is 25.9% nitrogen and
74.1% oxygen?
1. Assume 100 g of the compound, so that
there are 25.9 g N and 74.1 g O
 2. Convert to mole-to-mole ratio:
Divide each by mass of one mole
25.9 g divided by 14.0 g = 1.85 mol N
74.1 g divided by 16.0 g = 4.63 mol O
 3. Divide both molar quantities by the
smaller number of moles

4. 1.85/1.85 = 1 mol N
4.63/1.85 = 2.5 mol O
 5. Multiply by a number that
converts each to a whole number (In
this case, the number is 2 because 2
x 2.5 = 5, which is the smallest whole
number )
 2 x 1 mol N = 2
 2 x 2.5 mol O = 5
 Answer: The empirical formula is
N2O5

Determine the Empirical Formulas
1. H2O2
 2. CO2
 3. N2H4
 4. C6H12O6
 5. What is the empirical formula of a
compound that is 3.7% H, 44.4% C, and
51.9% N?

Answers
Compound
 1. H2O2
 2. CO2
 3. N2H4
 4. C6H12O6


5. HCN
Empirical Formula
HO
CO2
NH2
CH2O
Calculating Molecular Formulas
The molar mass of a compound is a
simple whole-number multiple of the
molar mass of the empirical formula
 The molecular formula may or may
not be the same as the empirical
formula

Calculate the molecular formula
of the compound whose molar
mass is 60.0 g and empirical
formula is CH4N.
1. Using the empirical formula, calculate the
empirical formula mass (efm)
(Use the same procedure used to calculate
molar mass.)
 2. Divide the known molar mass by the efm
 3. Multiply the formula subscripts by this value
to get the molecular formula

Answer
Molar mass (efm) is 30.0 g
 60.0 g divided by 30.0 g = 2
 Answer: C2H8N2

Practice Problems

1) What is the empirical formula of a compounds
that is 25.9% nitrogen and 74.1% oxygen?
2) Calculate the empirical formula of a compound
that is 32.00% C, 42.66% O, 18.67% N, and 6.67%
H.
3) Calculate the empirical formula of a compound
that is 42.9% C and 57.1% O.
Practice Problems

4) What is the molecular formula for each
compound:
a) CH2O, 90 g/mol
b) HgCl, 472.2 g/mol
c) C3H5O2, 146 g/mol