Ch 2 Matter & Atoms
Natural Approach to Chemistry
Lab Aids
Condensed
2.1 Frames 2-28
2.2 Frames 29-62
2.3 Fr. 63-106
2.1 Matter & the Elements
• Standards:
• SC.912.P.8.5 relate properties of atoms & their
position in the periodic table to the
arrangement of their electrons
• SC.912.P.8.9 apply the mole concept and the
law of conservation of mass to calculate
quantities of chemicals participating in
reactions
What are things made of?
The “stuff” that we are made of is called matter.
Chemistry tells us how one kind of matter can be changed into a
completely different kind of matter.
Corn oil is a pure
substance
substance: a kind of matter that can’t be separated into other
substances by physical means such as heating, cooling,
filtering, drying, sorting, or dissolving.
Oil and vinegar
dressing is
a mixture of
substances
Corn oil is a pure
substance
mixture: matter that contains more than one substance.
Physical properties
Examples:
Mass
Density
Color
Physical properties can be
measured or seen through
direct observations.
Physical changes include changes in
shape, phase or temperature.
Chemical properties are
observed when
a substance changes into a
different substance.
Chemical change
Iron
Rust
Signs
that a
chemical change
has occurred
element: a substance that cannot be separated into simpler
substances by chemical means.
Each element is a unique type of atom.
All oxygen atoms
are identical.
An oxygen atom is different from a
silicon atom or
a potassium atom.
How small is an atom?
1 grain of rice = 0.01 gram
1 hydrogen atom = 1.678 x 10–24 grams
= 0.000000000000000000000001678 grams
The periodic table
The periodic table organizes elements according to how they combine with other
elements (based on their chemical properties).
Increasing atomic number
Increasing atomic number
Hydrogen
Lightest element
Atomic number: 1
Uranium
Heaviest naturally occurring element
Atomic number: 92
Elements that belong to the same group (column) have
similar chemical properties.
Reminder
1 atomic mass unit (amu) = 1.66 x 10-24 g
The atomic mass is
the mass of:
1) a single atom in amu.
2) a mole of atoms in grams.
What does that
mean?
One single hydrogen atom weighs 1.01 amu.
One mole of hydrogen atoms weighs 1.01 g.
The atomic mass is
the mass of:
1) a single atom in amu.
2) a mole of atoms in grams.
What does that
mean?
One single carbon atom weighs 12.0 amu.
One mole of carbon atoms weighs 12.0 g.
Avogadro’s number
One mole contains
6.02 x 1023 atoms
How many moles are in 100 g of sulfur (S)?
How many moles are in 100 g of sulfur (S)?
Asked:
Given:
Relationships:
The number of moles
The element is sulfur and there are 100 g
One mole of sulfur has a mass of 32.065 g
How many moles are in 100 g of sulfur (S)?
Asked:
Given:
Relationships:
The number of moles
The element is sulfur and there are 100 g
One mole of sulfur has a mass of 32.065 g
Solve:
 1 mole S 
100 g S  
  3.12 moles S
32.065
g
S


Answer:
100 g of sulfur contains 3.12 moles of sulfur atoms.
How many grams of calcium (Ca) do you need to
have 2.50 moles of calcium?
Asked:
Given:
Relationships:
The number of grams
The element is calcium and there are 2.50 moles
One mole of calcium has a mass of 40.078 g
How many grams of calcium (Ca) do you need to
have 2.50 moles of calcium?
Asked:
Given:
Relationships:
The number of grams
The element is calcium and there are 2.50 moles
One mole of calcium has a mass of 40.078 g
Solve:
 40.078 g Ca 
2.50 moles Ca  
  100.2 g Ca
 1 mole Ca 
2.50 moles of calcium has a mass of 100. g.
Answer:
Physical properties can be seen and
measured
Chemical properties are observed when
one substance is changed into another
Elements are
organized in
a periodic table
This allows to convert
grams to moles and vice versa.
2.2 Molecules and
Compounds
Sc.912.P.8.7 interpret formula
representations of molecules and
compounds in terms of composition and
structure.
Sc.912.P.8.9 mole concept and law of
conservation of mass
One O2 molecule
One H2 molecule
molecule: a group of atoms chemically bonded together.
One O2 molecule
One H2 molecule
One H2O molecule
molecule: a group of atoms chemically bonded together.
compound: a substance containing more than one element in
which atoms of different elements are chemically bonded
together.
What do all these have in common?
What do all these have in common?
They are made up of only 3 types of atoms:
carbon, oxygen and hydrogen.
The alphabet of chemistry
More than 200,000 words
in the English language
Trillions of substances that
make up the universe
The chemical formula
The chemical formula
Write a chemical formula for a compound that has three hydrogen (H)
atoms for each atom of nitrogen (N).
Write a chemical formula for a compound that has three hydrogen (H)
atoms for each atom of nitrogen (N).
Asked:
Given:
Relationships:
Chemical formula
3 hydrogen (H) and 1 nitrogen (N)
The subscript tells the number of each element in the
compound.
Write a chemical formula for a compound that has three hydrogen (H)
atoms for each atom of nitrogen (N).
Asked:
Given:
Relationships:
Answer:
Chemical formula
3 hydrogen (H) and 1 nitrogen (N)
The subscript tells the number of each element in the
compound.
NH3
The properties of a compound depend more on
the exact structure of the molecule
than on the individual elements from which it is made.
F
A
E
F
U
N
E
R
R
N
A
U
L
L
R
E
A
The arrangement of letters matters!
L
F
U
N
The arrangement of atoms matters!
Representation
There are many ways of representing the same thing.
Representation
Ionic compounds
Salt (NaCl) is not a molecule!
ionic compound: a compound such as a salt in which positive
and negative ions attract each other to keep matter together.
ion: an atom or small molecule with an overall positive or
negative chare as a result of an imbalance of protons and
electrons.
Formula mass
The formula mass of water (H2O) is 18 g.
What is the mass of 1 mole of methane (CH4)?
What is the mass of 1 mole of methane (CH4)?
Asked:
The mass of 1 mole of methane
Given:
Methane (CH4) contains 1 carbon (C) and 4 hydrogen (H) atoms
Relationships:
The formula mass is the sum of the atomic masses for each atom
in the compound
What is the mass of 1 mole of methane (CH4)?
Asked:
The mass of 1 mole of methane
Given:
Methane (CH4) contains 1 carbon (C) and 4 hydrogen (H) atoms
Relationships:
The formula mass is the sum of the atomic masses for each atom
in the compound
Solve:
Answer:
g
4  H   C  4 1.0079   12.011  16.04
CH 4
moleg.
One mole of methane (CH4) has a mass of 16.04
How many moles are in 100 grams of water (H2O)?
How many moles are in 100 grams of water (H2O)?
Asked:
The moles in 100 g of water
Given:
Water (H2O) contains 2 hydrogen (H) atoms and 1 oxygen (O)
atom.
Relationships:
The formula mass is the sum of the atomic masses for each atom
in the compound.
How many moles are in 100 grams of water (H2O)?
Asked:
The moles in 100 g of water
Given:
Water (H2O) contains 2 hydrogen (H) atoms and 1 oxygen (O)
atom.
Relationships:
The formula mass is the sum of the atomic masses for each atom
in the compound.
Solve:
2  H   O  2 1.0079   15.999  18.015
 1 mole 
100 g  
  5.55 moles
18.01
5
g


Answer:
100 g of water (H2O) contains 5.55 moles.
g
H2O
mole
How many grams are in 2.300 moles of butane (C4H10)?
Butane is used as a lighter fluid in disposal lighter.
How many grams are in 2.300 moles of butane (C4H10)?
Butane is used as a lighter fluid in disposal lighter.
Asked:
The mass in grams of 2.300 moles of butane
Given:
2.300 moles of C4H10
Relationships:
The formula mass is the sum of the atomic masses for each atom
in the compound.
How many grams are in 2.300 moles of butane (C4H10)?
Butane is used as a lighter fluid in disposal lighter.
Asked:
The mass in grams of 2.300 moles of butane
Given:
2.300 moles of C4H10
Relationships:
Solve:
The formula mass is the sum of the atomic masses for each atom
in the compound.
g
4  H   10 O   4 12.011  10 1.0079   58.12 3
H 2O
mole
 58.123 g C4H10 
g
2.300

  133.7 g C4H10
mole  1 mole C4H10 
Answer:
133.7 g are in 2.300 moles of butane (C4H10).
How many oxygen atoms are in 200.0 g of glucose (C6H12O6)?
Asked:
Number of oxygen atoms
Given:
200.0 g of C6H12O6
Relationships:
Formula mass of glucose:
6 C   12 O   6  H 
 6 12.011  12 1.0079   6 15.999   180.15
g
mole
Remember:
Avogadro’s number indicates that one mole contains 6.02 x 1023 atoms
Solve:
First we find how many moles are in 200.0 g of glucose:
200.0 g C6H12O6 
1 mole C6H12O6
 1.110 moles C6H12O6
180.15 g C6H12O6
Solve:
First we find how many moles are in 200.0 g of glucose:
200.0 g C6H12O6 
1 mole C6H12O6
 1.110 moles C6H12O6
180.15 g C6H12O6
Next we find how many molecules are contained in 1.11 moles of glucose:
6.022  1023 molecules C6H12O6
1.110 moles C6H12O6 
1 mole C6H12O6
 6.684  1023 molecules C6H12O6
Solve:
First we find how many moles are in 200.0 g of glucose:
200.0 g C6H12O6 
1 mole C6H12O6
 1.110 moles C6H12O6
180.15 g C6H12O6
Next we find how many molecules are contained in 1.11 moles of glucose:
6.022  1023 molecules C6H12O6
1.110 moles C6H12O6 
1 mole C6H12O6
 6.684  1023 molecules C6H12O6
Then we find how many O atoms are contained in 6.684 x 1023 molecules of
glucose:
6.684  1023 molecules C6H12O6 
6 O atoms
 4.010  1024 O atoms
1 molecule C6H12O6
Solve:
First we find how many moles are in 200.0 g of glucose:
200.0 g C6H12O6 
1 mole C6H12O6
 1.110 moles C6H12O6
180.15 g C6H12O6
Next we find how many molecules are contained in 1.11 moles of glucose:
6.022  1023 molecules C6H12O6
1.110 moles C6H12O6 
1 mole C6H12O6
 6.684  1023 molecules C6H12O6
Then we find how many O atoms are contained in 6.684 x 1023 molecules of
glucose:
6.684  1023 molecules C6H12O6 
Answer:
6 O atoms
 4.010  1024 O atoms
1 molecule C6H12O6
There are 4.010 x 1024 atoms of O in 200.0 g of glucose (C6H12O6).
We can build compounds
using atoms
- The type of atom matters
- The arrangement of atoms
also matters
We can calculate the formula
mass of compounds:
2.3 Mixtures and Solutions
Sc.912.P.8.9 mole concept and law of
conservation of mass
Sc.912.P.12.12 explain how various
factors such as concentration,
temperature, and presence of a catalyst
affect the rate of a chemical reaction
CHAPTER 2
Matter
and Atoms
2.3 Mixtures and
Solutions
A recipe calls for you to “mix until homogeneous.”
What does that mean?
A mixture that is uniform throughout.
A mixture that is uniform throughout.
Different samples may have
different compositions.
What happens when sugar is
added to water?
What happens when sugar is
added to water?
A solution is obtained.
Is orange juice a solution?
Hint: Are things dissolved in orange juice?
Is orange juice a solution?
It is partially a solution because some chemicals are
dissolved in water, but some bits (like pulp) are not.
These four solutions contain the same solvent and solute.
Which one is the most concentrated solution?
Which one is the most dilute solution?
concentration: the amount of each solute compared to the
total solution.
Variables
• Can you dissolve salt in water?
Amount of salt
Amount of water
Variables
• Can you dissolve salt in water?
• Can you continue adding more and more
salt, and still get a solution?
Amount of salt
Amount of water
Variables
• Can you dissolve salt in water?
• Can you continue adding more and more
salt, and still get a solution?
• Can you dissolve more salt when you
increase the temperature?
Amount of salt
Amount of water
Amount of salt
Amount of water
Temperature
Solubility of common substances in water at 25oC
solubility: the amount of a solute that will dissolve in
a particular solvent at a particular temperature and pressure.
If the concentration of a sugar solution is 75 g/L, how much solution do you
need if you want 10 g of sugar?
If the concentration of a sugar solution is 75 g/L, how much solution do you
need if you want 10 g of sugar?
Asked:
Volume of solution
Given:
10 g of solute and concentration of 75 g/L
Relationships:
Liters of solution 
mass of solute
concentration in g L
If the concentration of a sugar solution is 75 g/L, how much solution do you
need if you want 10 g of sugar?
Asked:
Volume of solution
Given:
10 g of solute and concentration of 75 g/L
Relationships:
Liters of solution 
Solve:
10 g
 0.133 L or 133 mL
75 g L
mass of solute
concentration in g L
How much menthol do you need to make 10 kg of mouthwash if
the concentration of menthol must be 0.05%?
How much menthol do you need to make 10 kg of mouthwash if
the concentration of menthol must be 0.05%?
Asked:
Mass of solute
Given:
10 kg of solution, solute concentration of 0.05%
Relationships:
 concentration in % 
mass of solute  mass of solution  

100


How much menthol do you need to make 10 kg of mouthwash if
the concentration of menthol must be 0.05%?
Asked:
Mass of solute
Given:
10 kg of solution, solute concentration of 0.05%
Relationships:
Solve:
 concentration in % 
mass of solute  mass of solution  

100


 0.05% 
10 kg  
  0.005 kg or 5 g
10
0


molarity: the number of moles of solute per liter
of solution.
amount of solute (moles )
 molarity
volume of solvent (liters )
moles
M
L
 moles 
 L 


If 10.0 g of citric acid (C6H8O7) is added to 500 mL of water, what is the
molarity of the resulting solution?
If 10.0 g of citric acid (C6H8O7) is added to 500 mL of water, what is the
molarity of the resulting solution?
Asked:
The molarity of a solution
Given:
The amount of solute (citric acid) and the volume of solution
Relationships:
molarity 
moles solute
volume of solution  L 
If 10.0 g of citric acid (C6H8O7) is added to 500 mL of water, what is the
molarity of the resulting solution?
Asked:
The molarity of a solution
Given:
The amount of solute (citric acid) and the volume of solution
Relationships:
molarity 
Solve:
Start by calculating the formula mass of C6H8O7:
moles solute
volume of solution  L 
6 12.011  8 1.0079   7 15.999  192.12 g mole
If 10.0 g of citric acid (C6H8O7) is added to 500 mL of water, what is the
molarity of the resulting solution?
Asked:
The molarity of a solution
Given:
The amount of solute (citric acid) and the volume of solution
Relationships:
molarity 
Solve:
Start by calculating the formula mass of C6H8O7:
moles solute
volume of solution  L 
6 12.011  8 1.0079   7 15.999   192.12 g mole
Next we calculate the number of moles in 10.0 g of C6H8O7:
# moles 
1 mole
 10.0 g  0.0521 moles
192.12 g
If 10.0 g of citric acid (C6H8O7) is added to 500 mL of water, what is the
molarity of the resulting solution?
Asked:
The molarity of a solution
Given:
The amount of solute (citric acid) and the volume of solution
Relationships:
molarity 
Solve:
Start by calculating the formula mass of C6H8O7:
moles solute
volume of solution  L 
6 12.011  8 1.0079   7 15.999   192.12 g mole
Next we calculate the number of moles in 10.0 g of C6H8O7:
Answer:
# moles 
1 mole
 10.0 g  0.0521 moles
192.12 g
molarity 
0.0521 moles
 0.104 M
0.500 L
Ascorbic acid = Vitamin C
Ascorbic acid
C6H8O6
Vitamin C acts as a food
preservative by reacting with
oxygen (O2)
How much (volume) of a 1 M ascorbic acid solution will completely react
with 0.02 moles of oxygen (O2)?
How much (volume) of a 1 M ascorbic acid solution will completely react
with 0.02 moles of oxygen (O2)?
Asked:
Volume of solution
Given:
Concentration (1M) and balanced reaction
Relationships:
molarity 
moles solute
volume of solution  L 
According to the balanced reaction we need 2 moles of ascorbic acid
for every mole of O2.
How much (volume) of a 1 M ascorbic acid solution will completely react
with 0.02 moles of oxygen (O2)?
Asked:
Volume of solution
Given:
Concentration (1M) and balanced reaction
Relationships:
molarity 
moles solute
volume of solution  L 
According to the balanced reaction we need 2 moles of ascorbic acid
for every mole of O2.
Solve:
That means we need 0.04 moles of ascorbic acid:
moles solute 0.04 moles
volume  L  

 0.04 L or 40 mL
molarity
1M
How much (volume) of a 1 M ascorbic acid solution will completely react
with 0.02 moles of oxygen (O2)?
Asked:
Volume of solution
Given:
Concentration (1M) and balanced reaction
Relationships:
molarity 
moles solute
volume of solution  L 
According to the balanced reaction we need 2 moles of ascorbic acid
for every mole of O2.
Solve:
That means we need 0.04 moles of ascorbic acid:
moles solute 0.04 moles
volume  L  

 0.04 L or 40 mL
molarity
1M
Answer:
40 mL of the solution contains 0.04 moles of ascorbic acid, which is
enough to react with 0.02 moles of oxygen (O2).
The air you breathe is a mixture!
Air takes up less space under high pressure.
molar volume: the amount of
space occupied by a mole of gas
at STP.
It is equal to 22.4 L.
Standard
Temperature and
Pressure
0oC
1 atm
Partial pressures
The total pressure in
a mixture of gases is the sum of
the partial pressures
of each individual gas
in the mixture.
Partial pressures
Gas A
+
Gas B
Gas A
Gas B
Gas C
Total pressure
=
Partial pressure of
A
+
Partial pressure of
B
Gas C
Partial pressure of
C
If 1 L of helium (75%) and neon (25%) is at STP, what is the partial pressure
of helium?
If 1 L of helium (75%) and neon (25%) is at STP, what is the partial pressure
of helium?
Asked:
Partial pressure of helium
Given:
75% He and 25% Ne at STP conditions
Standard pressure is 101,325 Pa or 1 atm
Relationships:
The total pressure is the sum of the partial pressures of each
gas.
If 1 L of helium (75%) and neon (25%) is at STP, what is the partial pressure
of helium?
Asked:
Partial pressure of helium
Given:
75% He and 25% Ne at STP conditions
Standard pressure is 101,325 Pa or 1 atm
Relationships:
The total pressure is the sum of the partial pressures of each
gas.
Solve:
or 0.75  101,325 Pa  75,994 Pa
or 0.75  1 atm  0.75 atm
If 1 L of helium (75%) and neon (25%) is at STP, what is the partial pressure
of helium?
Asked:
Partial pressure of helium
Given:
75% He and 25% Ne at STP conditions
Standard pressure is 101,325 Pa or 1 atm
Relationships:
The total pressure is the sum of the partial pressures of each
gas.
Solve:
or 0.75  101,325 Pa  75,994 Pa
or 0.75  1 atm  0.75 atm
Answer:
The partial pressure of helium is 75,994 Pa or 0.75 atm.
Concentration of a solution can be expressed in:
mass per volume
mass percent
molarity
molarity  M  
moles of solute
liters of solvent
STP conditions:
Standard
Temperature (0oC)
Pressure (1 atm)
Air is a mixture of gases.
Based on Dalton’s law of partial pressures:
Pair  PN2  PO2  PAr  ...