Pre-University Chemistry

1. Chapter 3
Atoms: the Building Blocks of Matter

2. • The parts that make up an atom are called subatomicsubatomic
particlesparticles.
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
nucleusnucleus of an atom.
• Electrons orbit around
the nucleus.

3. 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.

4. Atomic Structure
Here are the basics; you need to know these.
1
H
1.0076
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
Atomic Number
Atomic Symbol
Atomic Mass

5. Lets practice! Find the missing
information?
Element Atomic # Atomic
Mass
Protons Electrons Neutrons
Ar 18 39.948
amu
He 2 2
O 15.999
amu
8

6. The Famous Gold Foil
Experiment
This showed us that the atom is made of mostly
empty space.

7. 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.

8. Mass Numbers of Hydrogen Isotopes
What would the masses be?

9. 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

10. The Mole (aka Avagadro’s Number):
6.02 x 1023

11. 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)

12. 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

13. Examples:
Substance Representative
Particle
Chemical
Formula
Representative
Particles in 1.00
mol
Atomic
nitrogen
Atom N 6.02 x 1023
Water Molecule H2O 6.02 x 1023
Calcium
ion
Ion Ca2+
6.02 x 1023

14. Solve
Substance Representative
Particle
Formula
Unit
Representative
Particles in
1.00 mol
Nitrogen
gas
N2 Molecule
Calcium
Fluoride
CaF2 Molecule
Sucrose C12H22O11 Molecule
Carbon C Molecule

15. 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

16. 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

17. Solve: How many atoms are in a
mole of
• 1. Carbon monoxide – CO
• 2. Glucose – C6H12O6
• 3. Propane – C3H8
• 4. Water – H2O

18. 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

19. 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

20. 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 ?

21. Molar Mass
• Gram atomic mass (gam) – atomic
mass of an element taken from the
periodic table
• Gram molecular mass (gmm) – mass
of one mole of a molecular compound
• Gram formula mass (gfm) – mass of
one mole of an ionic compound
• Can use molar mass instead of gam,
gmm, or gfm

22. 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

23. 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

24. 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

25. Answers
• 1. 718 g N2O3 (one mole is 76.0g)
• 2. 0.578 mol Fe2O3 (one mole is 159.6 g)

26. 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

27. • 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)

28. 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

29. 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

30. 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%

31. 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%

32. 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

33. Answer
• Total molar mass = 44.0 g/mol
• 36.0 g C = 81.8%
• 8.0 g H = 18.2%

34. Calculate the mass of carbon in
52.0 g of propane (C3H8)
1. 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

35. 1) Find the percent composition of
Aluminum Oxide (Al3O2)
2) How much of a 5-g piece of Iron
Bromide (FeBr3) is iron?

36. 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

37. 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

38. 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

39. • 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

40. 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?

41. Answers
• Compound Empirical Formula
• 1. H2O2 HO
• 2. CO2 CO2
• 3. N2H4 NH2
• 4. C6H12O6 CH2O
• 5. HCN

42. 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

43. 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

44. Answer
• Molar mass (efm) is 30.0 g
• 60.0 g divided by 30.0 g = 2
• Answer: C2H8N2

45. 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.

46. 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