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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