3.1 - Measurement & Calculations
Measurement - An Observed Value.
Science requires a measurement to be
observable and repeatable.
•
Calculation - A Derived Value
• Many concepts in science cannot be directly
measured, they are derived mathematically from other
measured values.
• Examples: Density
Concentration
3.2 - Density
Density = the amount of matter (mass) that
occupies a given volume
Density =
LOW
DENSITY
Mass
Volume
HIGH
DENSITY
3.2 - Density
Density = the amount of mass that occupies
a given volume
Density =
Mass
Volume
Mass units
-> grams
Volume units -> mL or cm3
Density units -> g/mL or g/cm3
3.2 - Density
Example: You find a gold colored rock in your yard. You
measure its volume at 34.5 mL and its mass at 309.2
grams. Use the table of known densities to determine if
the rock is gold.
Given:
V = 34.5 mL
m = 309.2 g
D=?
Substance
Density
3)
(g/cm
Gold
Lead
Copper
Brass
Iron
Chromium
Aluminum
19.3
11.4
8.96
8.83
7.87
7.19
2.70
Density =
Mass
Volume
309.2 g
=
34.5 mL
= 8.96 g/mL
Ans. = No, its copper
3.2 - Density
Density cannot be measured directly. Instead, the mass
and volume are measured and density must be calculated.
How is this done?
Mass is easy to measure.
We use a mass scale.
Volume is not so easy. Two methods exist:
1) Geometrical calculation (if object geometry is simple)
2) Water displacement method
3.2 - Density
Measuring volume
1) Geometrical Method - If the
object has a simple geometric
shape. Measure the dimensions
and calculate the volume.
r
Example:
What is the
volume of a
sphere with
radius = 4 cm
2) Water Displacement Method Few objects are simple
geometrics. Water Displacement
finds the volume of any shaped
objects.
67.5
mL
77.5
mL
Volumesphere = 4/3πr3
= 4/3π43
= 268 cm3
Volumeheart = 77.5 - 67.5 = 10.0 mL
3.2 - Density
Example: You want to know the mass of the gold ring given by your
fiance. You measure its volume using water displacement. The
original volume of water is 77.7 mL and after dropping in the ring,
the water volume measures 78.5 mL. What is the ring’s mass?
Given:
D = 19.3 g/mL
V = 78.5 - 77.7 = 0.8 mL
m=?
Substance
Density
3)
(g/cm
Gold
Lead
Copper
Brass
Iron
Chromium
Aluminum
19.3
11.4
8.96
8.83
7.87
7.19
2.70
D= m
V
Vx D = m
V
xV
m=VxD
m = 0.8 mL x 19.3 g/mL
m = 15.44 = 20 g
3.3 - Concentration
Concentration = the amount of matter (mass)
dissolved in a given volume of solution
+
=
Solute = Substance
being dissolved.
Solvent = Substance
doing the dissolving.
Solution = Mixture of
solute and solvent.
Example: NaCl
Example: H2O
Example: Saline solution
3.3 - Concentration
Remember - Density is matter per volume
LOW
DENSITY
HIGH
DENSITY
3.3 - Concentration
Concentration = the amount of matter (mass)
dissolved in a given volume of solution
Low
Concentration
High
Concentration
3.3 - Concentration
Concentration = the amount of matter (mass)
dissolved in a given volume of solution
Concentration is measured in
numerous different methods:
Molarity = Amount solute (mol)
Volume solution (L)
Molality =
Amount solute (mol)
Mass solvent (kg)
pH = -log(H+ molarity) = -log[H+]
Units = Molars = M
Units = Molals = m
Units = none
3.3 - Concentration
Concentration = the amount of matter (mass)
dissolved in a given volume of solution
Concentration is measured in
numerous different methods:
Molarity
Mole fractions
Molality
Percent by volume
pH
ppm
ppb
Mass fraction
3.3 - Concentration
Example: What is the molar concentration of a solution
made by mixing 4.55 mols of glucose (C6H12O6) into water
to make a 735.45 mL solution?
Given:
M=?
Asolute = 4.55 mol
Vsolution = 735.45 mL
= 0.73545 L
Molarity = Amount solute (mol)
Volume solution (L)
Molarity = 4.55 mol
0.73545L
Molarity = 6.186688422
= 6.19 M
3.3 - Concentration
Example: How many milliliters of water must be mixed
with 2.5 moles of potassium nitrate (KNO3) to make a
3.544 molal solution?
Given:
VH2O = ?
Asolute = 2.5 mol
m = 3.544m
m= A
M Note: For water only
mass = volume
m= A
V
V x m = A xV
V
Vxm=A
Vxm=A
m m
V=A
m
=
2.5 = 0.705L = 705mL
3.544
3.3 - Concentration
Example: How many moles of sodium hydroxide (NaOH)
need to be added to water to make 2.33 Liters of a
0.500M solution?
Given:
ANaOH = ?
Vsolution = 2.33 L
M = 0.500M
M=A
V
Vx M = A
V
xV
A=VxM
A = 2.33L x 0.500M
A = 1.165 = 1.17 mol
3.4 - Dilution
Dilution = The process of adding more solvent to a
solution to decrease the concentration.
+
Solution
=
Solvent
Dilute
Solution
3.4 - Dilution
Dilution problems are one of the most common preformed by
chemists. Chemistry labs contain hundreds of solutions, but
the chemist must commonly dilute these to a concentration
appropriate for an experiment.
Stock
Solution
Needed
Solution
12M HCl
M1V1
3M HCl
=
M2V2
3.4 - Dilution
Example: How much water needs to be added to a 12.0M solution
of hydrochloric acid (HCl) to make 50.0 mL of 3.00M solution?
M1V1
12M HCl
3M HCl
Given:
Vwater = ?
M1 = 12.0M
V2 = 50.0 mL= 0.0500L
M2 = 3.00M
=
M2V2
M2V2
V1 =
M1
(3.00M)(0.0500L)
V1 =
12.0M
V1 = 0.0125L = 12.5 mL
3.4 - Dilution
Example: How much water needs to be added to a 12.0M solution
of hydrochloric acid (HCl) to make 50.0 mL of 3.00M solution?
V1 = 0.0125L = 12.5 mL
12M HCl
3M HCl
Given:
Vwater = ?
M1 = 12.0M
V2 = 50.0 mL= 0.0500L
M2 = 3.00M
But V1 is the amount of
the original 12M solution
that we need to take, not
the amount of water that
needs to be added.
The amount of water needed is:
V2 - V1 = 50.0 - 12.5 = 37.5 mL
3.5 - Conversion Problems
Conversion Problems are possibly the most common
calculation performed by scientists.
Example:
•You determine the volume of a reactant to be 0.06mL
but you need to express the value in Liters to plug it into
the dilution equation.
•You find the distance to the nearest star in light years
but for comparison purposes you need it expressed in
parsecs.
•You run an experiment and determine energy in Joules,
but the lab you are passing the data to want the values
in calories
3.5 - Conversion Problems
Conversion Problems are completed using
“conversion factors”
Conversion = To change
Factor =
A value that multiplies
Conversion Factor = a
value that multiples to
change another value
3.5 - Conversion Problems
Conversion Problems are completed using
“conversion factors”
For every equality there are two conversion factors:
Example: 12 inches = 1 foot
Conversion Factor are:
12 in
1 ft
&
1 ft
12 in
3.5 - Conversion Problems
Sample Problem: How many inches are in
234.567 feet?
Step 1: Input (with
units) over 1
Step 2: Unit Analysis
to apply conversion
factors
Step 3: Do the math
234.567 ft
1
234.567 ft
1
12 in
1 ft
6
∞
234.567 ft
1
12 in
1 ft
∞
∞
= 2814.804
= 2814.80 in
3.5 - Conversion Problems
Sample Problem: How many years are in
6.55seconds?
Step 1: Input over 1
Step 2: Unit Analysis
Step 3: Do the math
6.55 s
1
1 min
60 s
1 hr
60 min
1 day
24 hr
1 Yr
= 2.08X10-7 Yr
365 days