SCIENTIFIC MEASUREMENT
 CHEM IH: CHAPTER 3
Stating a Measurement
In every measurement there is a
Number followed by a
 Unit from a measuring device
The number should also be as precise as the
measuring device.
Ex: Reading a Meterstick
. l2. . . . I . . . . I3 . . . .I . . . . I4. .
First digit (known)= 2
Second digit (known)
cm
2.?? cm
= 0.7
2.7? cm
Third digit (estimated) between 0.05- 0.07
Length reported
=
2.75 cm
or
2.74 cm
or
2.76 cm
UNITS OF MEASUREMENT
Use SI units — based on the metric
system
Length
Meter, m
Mass
Kilogram, kg
Volume
Liter, L
Time
Seconds, s
Temperature
Celsius degrees, ˚C
kelvins, K
Metric Prefixes
Conversion Factors
Fractions in which the numerator and denominator are
EQUAL quantities expressed in different units
Example:
1 hr. = 60 min
Factors: 1 hr.
and
60 min
60 min
1 hr.
How many minutes are in 2.5 hours?
Conversion factor
2.5 hr x 60 min
1 hr
= 150 min
cancel
By using dimensional analysis / factor-label method, the
UNITS ensure that you have the conversion right side up,
and the UNITS are calculated as well as the numbers!
Learning Check
How many seconds are in 1.4 days?
Unit plan: days
hr
min
seconds
1.4 days x 24 hr x ___min x ____ s =
1 day
hr
min
ANSWER: 120,960 s.
Significant Figures (Honors only)
The numbers reported in a measurement
are limited by the measuring tool
Significant figures in a measurement
include the known digits plus one
estimated digit
Counting Significant Figures:
Non-Zero Digits (Honors Only)
RULE 1. All non-zero digits in a measured number ARE
significant.
#of Significant Figures
38.15 cm
5.6 ft
65.6 lb
122.55 m
4
2
___
___
Counting Significant Figures:
Leading Zeros (Honors Only)
RULE 2. Leading zeros in decimal numbers are NOT
significant.
#of Significant Figures
0.008 mm
1
0.0156 oz
3
0.0042 lb
____
0.000262 mL
____
Counting Significant Figures:
Sandwiched Zeros (Honors Only)
RULE 3. Zeros between nonzero numbers ARE
significant. (They can not be rounded unless they
are on an end of a number.)
# of Significant Figures
50.8 mm
3
2001 min
4
0.702 lb
____
0.00405 m
____
Counting Significant Figures:
Zeros @ the End of a # & to the Right of a
Decimal
(Honors Only)
RULE 4. Trailing zeros at the end of a number and to
the right of a decimal numbers ARE significant.
# of Significant Figures
43.00 m.
4
200.00 yr
5
1.10 gal
____
0.04500 g
____
Counting Significant Figures:
Trailing Zeros (Honors Only)
RULE 5. Trailing zeros in numbers without decimals
are NOT significant. They are only serving as place
holders.
# of Significant Figures
25,000 in.
2
200. yr
3
48,600 gal
____
25,005,000 g
____
Counting Significant Figures:
Unlimited Sig Figs (Honors Only)
RULE 6. 2 instances in which there are an unlimited #
of sig figs.
a) Counting. Ex: 23 people in our classroom.
b) Exactly defined quantities. Ex: 1hr = 60 min.
 Both are exact values. There is no uncertainty.
 Neither of these types of values affect the process
of rounding an answer.
Learning Check
(Honors Only)
A. Which answers contain 3 significant figures?
1) 0.4760 2) 0.00476 3) 4760
B. All the zeros are significant in
1) 0.00307
2) 25.300 3) 2.050 x 103
C. 534,675 rounded to 3 significant figures is
1) 535
2) 535,000
3) 5.35 x 105
Learning Check (Honors Only)
In which set(s) do both numbers contain
the same number of significant figures?
1) 22.0 and 22.00
2) 400.0 and 40
3) 0.000015 and 150,000
Significant Numbers in Calculations
(Honors Only)
 A calculated answer cannot be more precise than the
measuring tool.
 A calculated answer must match the least precise
measurement.
 Significant figures are needed for final answers from
1) adding or subtracting
2) multiplying or dividing
 If you must round to obtain the right # of sig figs, do
so after all calcs are complete
Adding and Subtracting (Honors Only)
The answer has the same number of decimal places
as the measurement with the fewest decimal places.
25.2 one decimal place
+ 1.34 two decimal places
26.54
answer 26.5 one decimal place
Learning Check (Honors Only)
In each calculation, round the answer to the correct
number of significant figures.
A. 235.05 + 19.6 + 2.1 =
1) 256.75
2) 256.8
3) 257
B. 58.925 - 18.2
=
1) 40.725
2) 40.73
3) 40.7
Multiplying and Dividing
(Honors Only)
Round (or add zeros) to the calculated
answer until you have the same number of
significant figures as the measurement with
the fewest significant figures.
Learning Check
A. 2.19 X 4.2 =
1) 9
(Honors Only)
2) 9.2
B. 4.311 ÷ 0.07 =
1) 61.58
2) 62
C.
2.54 X 0.0028 =
0.0105 X 0.060
1) 11.3
2) 11 3) 0.041
3) 9.198
3) 60
What is Scientific Notation?
 Scientific notation is a way of expressing
really big numbers or really small numbers.
 For very large and very small numbers,
scientific notation is more concise.
Scientific notation consists of
two parts:
 A number between 1 and 10
 A power of 10
x
N x 10
Examples
 Given: 289,800,000
 Use: 2.898 (moved 8 places)
 Answer: 2.898 x 108 (how many sig figs?
Honors only)
 Given: 0.000567
 Use: 5.67 (moved 4 places)
 Answer: 5.67 x 10-4 (How many sig figs?
Honors only)
CHEMICAL QUANTITIES:
THE MOLE
CHEM IH: CHAPTER 3 & 10
MEASURING MASS
 A mole is a quantity of things, just as…
1 dozen = 12 things
1 gross
= 144 things
1 mole
= 6.02 x 1023 things
 “Things” usually measured in moles are
atoms, molecules, ions, and formula
units
 You can measure
mass, or volume, or
you can count pieces
 We measure mass in
grams
 We measure volume
in liters
 We count pieces in
MOLES
A MOLE…
 is an amount, defined as the number of
carbon atoms in exactly 12 grams of carbon12
 1 mole = 6.02 x 1023 of the representative
particles
 Treat it like a very large dozen
6.02 x 1023 is called: Avogadro’s number
 Similar Words for an amount:
 Pair: 1 pair of shoelaces = 2 shoelaces
 Dozen: 1 dozen oranges = 12 oranges
 Gross: 1 gross of pencils= 144 pencils
 Ream: 1 ream of paper= 500 sheets of
paper
What are Representative
Particles (“RP”)?
 The smallest pieces of a substance:
1. For a molecular compound: it is the
molecule.
2. For an ionic compound: it is the formula
unit (made of ions)
3. For an element: it is the atom
 Remember the 7 diatomic elements? (made
of molecules)
Practice Counting Particles
 How many oxygen atoms in the following?
CaCO3 3 atoms of oxygen
2. Al2(SO4)3 12 (4 x 3) atoms of oxygen
1.
 How many ions in the following?
1.

2.

3.

CaCl2
3 total ions (1 Ca2+ ion and 2 Cl1- ions)
NaOH
2 total ions (1 Na1+ ion and 1 OH1- ion)
Al2(SO4)3
5 total ions (2 Al3+ + 3 SO4 ions)
CONVERSION FACTOR
 MOLES = RPs x ____1 mole___
6.02 x 1023 RPs
EXAMPLES: ATOMS  MOLES
 How many moles of B are in 3.15 x 1023 atoms
of B?

Conversion: 1 mole B = 6.02 x 1023 atoms B
(b/c the atom is the RP of boron)
3.15 x 1023 atoms of B
1 mole
B
23
6.02 x 10 atoms B
= 0.532 mole
EXAMPLES: MOLES  ATOMS
 How many atoms of Al are in 1.5 mol of Al?
 Conversion: 1 mole = 6.02 x 1023 atoms
1.5 mol of Al 6.02 x 1023 atoms Al
23 atoms
9.03
x
10
=
1 mole Al
of Al
CAUTION: Identify RPs
Carefully!
 See next slide!
EXAMPLES: MOLECULES  MOLES
How many atoms of H are there in 3 moles of
H2O? (HINT: Are atoms the RP for water?)
Conversions:
1 mole = 6.02 x 1023 molecules
(b/c molecules are the RP for H2O)
H2O molecule = 2 atoms of Hydrogen
3 moles of H2O 6.02 x 1023 molec H2O 2 atoms H
1 mole H2O
1 H2O molecule
= 3.612 x 1024 atoms H
MOLAR MASS
Def: The mass of a mole of representative particles
of a substance.
Each element & compound has a molar mass.
MOLAR MASS OF AN ELEMENT
Determined simply by looking at the periodic table
Molar mass (g) = Atomic Mass (amu)
20
Ca
40.08
* Thus,
1 mol Ca = 40 g
1 atom of Ca weighs 40.08 amu
1 mole of Ca atoms weighs 40.08 grams
MOLAR MASS FOR COMPOUNDS
 To calculate the molar mass of a compound,
find the number of grams of each element in
one mole of the compound
 Then add the masses within the compound
Example: H2O
H= 1.01
2 (1.01) + 1 (15.999)= 18.02 g/mol
O= 15.999
SOME PRACTICE PROBLEMS
 How many atoms of O are in 3.7 mol of O?
 2.2 X 1024 atoms of oxygen
 How many atoms of P are in 2.3 mol of P?
 1.4 x 1024 atoms of phosphorus
 How many atoms of Ca are there in 2.5 moles
of CaCl2?
 1.5 x 1024 atoms Ca
 How many atoms of O are there in 1.7 moles of
SO4?
 4.1 x 1024 atoms of oxygen
Remember!!!!
 The molar mass of any substance (in grams)
equals 1 mole
 This applies to ALL substance: elements,
molecular compounds, ionic compounds
 Use molar mass to convert between mass and
moles
 Ex: Mass, in grams, of 6 mol of MgCl2 ?
mass of MgCl2 = 6 mol MgCl2
= 571.26 g MgCl2
92.21 g MgCl2
1 mol MgCl2
VOLUME AND THE MOLE
 Volume varies with changes in temperature &
pressure
 Gases are predictable, under the same
physical conditions
 Avogadro’s hypothesis helps explain:
equal volume of gases, at the same temp and
pressure contains equal number of particles
 Ex: helium balloon
 Gases vary at different temperatures, makes
it hard to measure
 Because of variation use STP
 Standard Temperature and Pressure
 Temperature = 0° C
 Pressure = 1 atm (atmosphere) or 101.3 kPal
Molar Volume
 At STP:1 mole, 6.02 x 1023 atoms, of any gas has
a volume of 22.4 L
1 mole gas = 22.4 L gas
 Called Molar Volume
 Used to convert between # of moles and vol of a
gas @ STP
 Ex: what is the vol of 1.25 mol of sulfur gas
1.25 mol S 22.4 L = 28.0 L
1 mol
MOLAR MASS FROM DENSITY
 Different gases have different densities
 Density of a gas measured in g/L @ a specific
temperature
 Can use the following formula to solve :
grams = grams X 22.4 L
mole
L
1 mole
 Ex: Density of gaseous compound containing oxygen
and carbon is 1.964 g/ L, what is the molar mass?
 grams = 1.964 g X
22.4 L then you solve
mole 1 L
1 mole
= 44.o g/mol
Atoms,
molecules
, etc.
Molarity
 Def: the concentration of a solution. How
many moles/liter
 Can be used to calculate # of moles of a
solute
 Ex: Household laundry bleach is a dilute
aqueous solution of sodium hypochlorite
(NaClO). How many moles of solute are
present in 1.5 L of 0.70 M NaClO?
Calculating Percent Composition
of a Compound
 Like all percent problems: a part ÷ the whole
1. Find the mass of each of the components (the
elements)
2. Next, divide by the total mass of the compound
3. Then X 100 % = percent
Formula:
% Composition = Mass of element X 100%
Mass of compound
Method #1: % Comp When Actual Masses are Given
A compound is formed when 9.03 g of Mg combines
completely with 3.48 g of N.
What is the percent composition of the compound?
1. First add the 2 mass of the 2 compounds to reach
the total mass 9.03 g Mg + 3.48 g N =
12.51 g Mg3N2
1. Find the % of each compound
% Mg= 9.03 g Mg X 100% = 72.2 %
12.51 g Mg3N2
% N= 3.48 g N X 100%
= 27.8 %
12.51 g Mg3N2
Method 2: % Comp When Only The
Formula is Known
 Can find the percent composition of a compound
using just the molar mass of the compound and the
element
 % mass=mass of the element 1 mol cmpd X100%
molar mass of the compound
 Example:
Find the percent of C in CO2
12.01 g C X 100% = 27.3% C
44.01 g CO2
Can find O % by subtracting 27.3% from 100%
Using % Composition
 Can use % composition as a conversion factor just
like the mole
 After finding the % comp. of each element in a
cmpd. can assume the total compound = 100g
 Example: C= 27.3%
27.3 g C
O= 72.7 %
72.7 g O
 In 100 g sample of compound there is 27.3 g of C & 72.7 g
of O
How much C would be contained in 73 g of CO2?
73 g CO2 27.3 g C
= 19.93 g C
100 g CO2
EMPIRICAL FORMULAS
 Empirical formulas are the lowest
WHOLE number ratios of
elements contained in a
compound
REMEMBER…
 Molecular formulas tells the actual number of
of each kind of atom present in a molecule of
the compound
 Ex:
H2O2
HO
Molecular
Formula
Empirical
Formula
CO2
CO2
Molecular
Formula
Empirical
Formula
For CO2 they are the same
 Formulas for ionic compounds are
ALWAYS empirical (the lowest whole
number ratio = can not be reduced)
 Examples:
NaCl MgCl2 Al2(SO4)3 K2CO3
Simplest whole number ratio for NaCl
 A formula is not just the ratio of atoms, it is also the
ratio of moles
 In 1 mole of CO2 there is 1 mole of carbon and 2 moles
of oxygen
 In one molecule of CO2 there is 1 atom of C and 2
atoms of O
 Formulas for molecular compounds MIGHT be
empirical (lowest whole number ratio)
 Molecular:
H2O
C6H12O6
H2O
CH2O
C12H22O11
(Correct formula)
 Empirical:
(Lowest whole
number ratio)
C12H22O11
CALCULATING EMPIRICAL
 We can get a ratio from the percent
composition
1. Assume you have a 100 g sample the
percentage become grams (75.1% = 75.1 grams)
2. Convert grams to moles
3. Find lowest whole number ratio by dividing
each number of moles by the smallest value
Example calculations
 Calculate the empirical formula of a compound
composed of 38.67 % C, 16.22 % H, and 45.11 %N
 Assume 100 g sample, so
38.67 g C x 1 mol C = 3.22 mole C
12.0 g C
16.22 g H x 1 mol H = 16.22 mole H
1.0 g H
45.11 g N x 1 mol N = 3.22 mole N
14.0 g N
*Now divide each value by the smallest value
…Example 1
 The ratio is 3.22 mol C = 1 mol C
3.22 mol N 1 mol N
 The ratio is 16.22 mol H = 5 mol H
3.22 mol N 1 mol N
C1H5N1 which is = CH5N
MORE PRACTICE
 A compound is 43.64 % P and 56.36 % O
What is the empirical formula?
PO3
 Caffeine is 49.48% C, 5.15% H, 28.87% N and
16.49% O
What is its empirical formula?
C4H5N2O
EMPIRICAL TO MOLECULAR
 Since the empirical formula is the lowest ratio,
the actual molecule would weigh more
 Divide the actual molar mass by the empirical
formula mass – you get a whole number to
increase each coefficient in the empirical
formula
EXAMPLE
 Caffeine has a molar mass of 194 g, what is its
molecular formula?
1. Find the mass of the empirical formula, C4H5N2O
2. Divide the molar mass by the empirical mass:
194.0 g/mol =2
97.1 g/mol
3. Now multiply the entire empirical formula by 2
2(C4H5N2O) =
C8H10N4O2
final molecular formula