File - Mr. Webb Science

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Chapters 10
Chemical Quantities
Conversion Factors

Conversion factor – A fraction equal to 1
that is used to change one unit into
another.
(When the numerator = denominator,
a fraction equals 1.)
Dimensional Analysis

Dimensional Analysis – A problem
solving method where conversion factors
are used to cancel unwanted units.
Conversion Examples
a) Convert $25 to nickels.
$25
4 quarters
5 nickels
$1
1 quarter
25 * 4 * 5 / 1 / 1 = 500 nickels
Conversion Example 2

Convert 180 days to seconds.
180 days 24 hours
1 day
60 minutes 60 seconds
1 hour
1 minute
180 * 24 * 60 * 60 / 1 / 1 / 1 = 15,552,000 seconds
or 1.6 x 107 seconds
Common Conversions to
Know
1 base (m, l, g) = 100 centi
.
1 base (m, l, g) = 1000 milli
.
1 kilo
= 1000 base units (m, l, g)
Convert 125 cm to km.
1m
1 km
125 cm

100 cm
1000 m
125 / 100 / 1000 = 0.00125 km or 1.25 x 10-3 km
Conversion Example 4

Convert 15 m/s to km/hr.
15 m
1s
1 km
60 seconds
60 minutes
1000 m
1 minute
1 hour
15 * 60 * 60 / 1000 = 54 km/hr
Measuring Matter

How do we describe how much of
something we have?
By count, by mass, by volume.
 We use words like “dozen” to talk about an
amount.
 In chemistry, we use the MOLE.

Mole

Mole – SI unit for measuring an amount of
a substance.
1 mole = 6.02 x 1023 representative particles

A “particle” will either be:

An atom, a molecule or a formula unit

Avogadro’s Number = 6.02 x 1023

Representative particles = smallest unit
that still has all the characteristics of that
substance.
Representative Particles

What is the representative particle of :
•
atom
Element (ex. Cu): ___________
Exception: The representative particle of the
7 diatomic elements is a molecule. (ex. H2)
•
molecule
Covalent compound (ex. H2O): _________
•
formula unit
Ionic Compound (ex. NaCl): ___________
Conversions

4 moles Ca =
4 moles Ca
atoms Ca.
6.02 x 1023 atoms Ca
1 mole Ca
= 2.41 x 1024 atoms Ca
Conversions

5 x 1018 atoms Cu =
5 x 1018 atoms Cu
moles Cu.
1 mole Cu
6.02 x 1023 atoms Cu
= 8.3 x 10-6 moles Cu
Conversions

9.2 moles F2 =
9.2 moles
molecules F2?
6.02 x 1023 molecules F2
1 mole
= 5.5 x 1024 molecules F2
Conversions

9.2 moles F2 =
9.2 moles F2
6.02 x 1023 mlcls F2
1 mole
= 1.1 x 1025 atoms F
atoms F?
2 atoms F
1 molecule F2
Conversions

3.4 moles C2H4 =
total atoms?
3.4 moles C2H4 6.02 x 1023 mlcls C2H4
1 mole C2H4
= 1.22 x 1025 atoms
6 atoms
1 mlcl C2H4
Molar Mass

Molar Mass – The mass of one mole of an
element or compound.

Molar mass of a compound = the sum of the
masses of the atoms in the formula.

Use the atomic masses in grams/mol on the
periodic table.
Molar Mass

Find the molar mass of each:
 Sr
87.6 g/mol

MgBr2
(24.3) + (2 x 79.9) = 184.1 g/mol

Ba3(PO4)2
(3 x 137.3) + (2 x 31) + (8 x 16) = 601.9 g/mol
Mole–Gram Conversions
1 mole = molar mass (in grams)
5.3 moles LiOH = ___________ grams LiOH
(Molar mass LiOH : 7 + 16 + 1 = 24 g/mol)
5.3 moles LiOH
24 g LiOH
1 mole LiOH
= 127.2 grams LiOH
Gram-Mole Conversions

68 grams F2 =
68 grams F2
moles F2?
1 mole F2
38 grams
68 / 38 = 1.8 moles F2
STP

STP = Standard Temperature & Pressure
Standard Temp  0oC
 Standard Press  1 atm

(See Reference Tables)
Molar Volume of a Gas

Avogadro’s Hypothesis: equal volumes of
gases at the same temperature and
pressure contain equal numbers of
particles.

At STP, 1 mole of any gas occupies a
volume of 22.4 L.
1 mole = 22.4 L (of a gas at STP)
Mole-Volume Conversions

5.4 moles He =
5.4 moles He
L He at STP?
22.4 L He
1 mole He
5.4 x 22.4 = 120.96 L He
Mole-Volume Conversions

5.4 moles CH4 =
5.4 moles CH4
L CH4 gas at STP?
22.4 L CH4
1 mole CH4
5.4 x 22.4 = 120.96 L CH4
Volume-Mole Conversion

560 L SO3 =
560 L SO3
mol SO3 at STP
1 mole SO3
22.4 L SO3
560 / 22.4 = 25 mole SO3
Molar Mass-Density
Conversions
grams
Density =
liters
grams
Molar Mass =
mole
A gaseous compound composed of sulfur
and oxygen has a density of 3.58 g/L at
STP. What is the molar mass of this gas?
3.58 g
L
22.4 L
1 mole
3.58 x 22.4 = 80.2 g/mole
Molar Mass-Density
Conversion

What is the density of Krypton gas at STP?
83.8 grams Kr
mole
1 mole
22.4 Liters
83.8 / 22.4 = 3.74 g/L Kr
22.4 Liters
at STP
(gases only)
1 mole
1 mole
Molar Mass
Grams
(use Per.Tble)
6.02 x 1023 particles
23
6.02 x 10
particles
Multi-step Problem: Example 1



If you had 5.0 L of CO2 how many grams would
that be?
Step 1: L  moles
Step 2: moles  grams
5.0 L CO2 1 mole CO2
22.4 L CO2
44.0 g CO2 = 9.8 g CO2
1 mole CO2
Multi-step Problem: Example 2



How many molecules are in 60.0 grams of water?
Step 1: grams  moles
Step 2: moles  molecules
60.0 g H2O 1 mole H2O 6.02 x 1023 mlcls =
18.0 g H2O 1 mole H2O
= 2.0 x 1024 molecules of H2O
Percent Composition


Percent Composition - % by mass of
each element in a compound
Part x 100
Percent =
Whole
Percent Composition


Percent Comp = Mass of 1 element x 100
Mass of compound
Example: Find the mass percent
composition of Al2(SO4)3
Al:
2 x 27 = 54
S:
3 x 32 = 96
O: 12 x 16 = 192
Total Comp. = 342
54 x 100 = 15.8%
% Al:
342
% S:
96 x 100 = 28.1%
342
%O: 192 x 100 = 56.1%
342
Percent Example

Find the percent composition of NiSO3.
Ni:
58.7 g %Ni: 58.7
x 100 = 42.3 %
S:
32 g
138.7
O: (3 x 16) = 48 g
%S: 32 x 100 = 23.1 %
138.7
Total Comp. 138.7
%O: 48 x 100 = 34.6%
138.7
More Percents

Which of the following shows a compound
that is 92.3%C and 7.7%H?
a) C2H4
c) CH4
b) C3H6
d) C6H6
Empirical Formulas

Empirical Formula – The simplest formula.




Shows the smallest whole number ratio of elements in
a compound.
Covalent formulas will not always be empirical.
Example: CH
Molecular Formula – The actual formula.



For ionic compounds – it will be the simplest ratio.
For molecular compounds – it will NOT always be the
simplest ratio.
Example: C6H6
To Calculate Empirical Formula

Calculate the empirical formula of a 2.5
gram compound containing 0.90g Ca and
1.60g Cl.
 Step
1: Convert GRAMS to MOLES.
 Ca:
 Cl:
0.90g
1.60g
1 mole
40.1 g
= 0.0224 mole Ca
1 mole
35.5 g
= 0.0451 mole Cl
Calculating Empirical
Formula

Step 2: DIVIDE the # of moles of each
substance by the smallest number to get
the simplest mole ratio.
Ca: 0.0224 = 1
0.0224
CaCl2
Cl: 0.0451 = 2.01 ~ 2
0.0224
Calculating Empirical
Formulas


Step 3: If the numbers are whole
numbers, use these as the subscripts for
the formula. If the numbers are not whole
numbers, multiply each by a factor that will
make them whole numbers.
Look for these fractions:
0.5  x 2
 0.33  x 3
 0.25  x 4

Empirical Formula Example

What is the empirical formula of a
compound that is 66% Ca and 34% P?
(Assume you have 100 grams of a
compound and replace % with grams.)
Empirical Formula Example

Step 1: grams  moles
Ca: 66g
P

1 mole = 1.646 mole Ca
40.1 g
34g 1 mole = 1.097 mole P
31.0 g
Step 2: Divide by the smallest.
Ca: 1.646 = 1.5
1.097
P: 1.097 = 1
1.097
Empirical Formula Example

Step 3: Multiply until you get whole
numbers.
(If you multiply one factor by a number,
you have to multiply ALL the factors by
that number!)
Ca: 1.5 x 2 = 3
P: 1 x 2 = 2
Ca3P2
Determining Molecular
Formulas

A compound has an empirical formula of
CH2O. Its molecular mass is 180g/mol.
What is its molecular formula?

Step 1: Find the mass of the empirical
formula.
C: 1 x 12 = 12
H: 2 x 1 = 2
O: 1 x 16 = 16
Total: 30
Determining Molecular Formula

Step 2: Divide the molecular mass by the
mass of the empirical formula to get the
“multiplying factor”.
180
=6
30

Step 3: Multiply each of the subscripts in
the empirical formula by this factor to get
the molecular formula.
6 (CH2O)  C6H12O6
Determining Molecular
Formula
Find the molecular formula of ethylene
glycol (CH3O) if its molar mass is 62
g/mol.
Step 1: 12 + (3 x 1) + 16 = 31 g/mol
Step 2: 62 / 31 = 2
Step 3: 2 (CH3O)  C2H6O2
Empirical/Molecular Example

The percent composition of
methyl butanoate is 58.8% C,
9.8% H, and 31.4 % O and its
molar mass is 204 g/mol.
 What is its empirical formula?
 What is its molecular formula?
Empirical/Molecular Example
58.8 g C
1 mol C
12.0 g C
= 4.9 mol C
4.9 = 2.5 x 2 = 5
1.96
9.8 g H
1 mol H
1.0 g H
= 9.8 mol H
9.8 = 5
1.96
x 2 = 10
1.96 = 1
1.96
x2=2
31.4 g O
1 mol O
16.0 g O
= 1.96 mol O
Empirical Formula = C5H10O2
Empirical/Molecular Example
Empirical Formula  C5H10O2
Mass = 5(12) + 10(1) + 2(16) = 102 g/mole
Molecular mass  204 g/mol = 2
Empirical mass  102 g/mol
So molecular formula is 2 x emp. form:
2(C5H10O2) =
C10H20O4
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