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1
Chemical
Quantities
The Mole, % Composition,
Empirical and Molecular Formulas
2
How you measure how much?
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.
3
Moles Defined as the number of carbon
atoms in exactly
12 grams of carbon-12.
1 mole is 6.02 x 1023 particles. Treat it like a very large dozen 6.02 x 1023 is called Avagadro’s
number.
4
Representative particles
The smallest pieces of a substance. For a molecular compound it is a
molecule. For an ionic compound it is a
formula unit. For an element it is an atom.
5
Types of questions How many oxygen atoms in the
following?–CaCO3
–Al2(SO4)3 How many ions in the following?
–CaCl2–NaOH–Al2(SO4)3
12
3
3
2
5
6
Types of questions using the equality;1 mole = 6.02 x 1023
How many molecules of CO2 are the in
4.56 moles of CO2 ?
4.56 mole x 6.02x1023 mc = 1 1 mole
How many moles of water is 5.87 x 1022 molecules?
5.87 x 1022 mc x 1 mole =
1 6.02x1023 mc
2.75x1024mc
0.0975 mole
7
Types of questions using the equality;1 mole = 6.02 x 1023
How many atoms of carbon are there in
1.23 moles of C6H12O6 ?
1.23 moles x 6.02x1023 mc x 6 atoms =
1 1 mole 1 mc
How many moles is 7.78 x 1024 formula
units of MgCl2?
7.78x1024 FU x 1 mole = 12.9 mole
1 6.02x1024 FU
4.44x1024 atoms
8
Measuring Moles Remember relative atomic mass? The amu was one twelfth the mass
of a carbon 12 atom. Since the mole is the number of
atoms in 12 grams of carbon-12, the decimal number on the periodic
table is also the mass of 1 mole of those atoms in grams.
9
Gram Atomic Mass The mass of 1 mole of an element in
grams. 12.01 grams of carbon has the same
number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron.
We can right this as 12.01 g C = 1 mole
We can count things by weighing them.
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Examples How much would 2.34 moles of
carbon weigh? 2.34 moles C x 12 g C
1 1mole How many moles of magnesium in
24.31 g of Mg?24.31 g Mg x 1 mole =
1 24g Mg
= 28.08 g
1.013 mole
11
How many atoms of lithium in 1.00 g of Li?
1.00 g Li x 1 mole x 6.02x1023 atoms
1 7 g Li 1 mole
How much would 3.45 x 1022 atoms of U weigh?
3.45x1022 atoms U x 1 mole x 238 g U
1 6.02x1023atoms 1 mole
8.60x1022 atoms
13.6 g
12
What about compounds? in 1 mole of H2O molecules there are two
moles of H atoms and 1 mole of O atoms To find the mass of one mole of a
compound determine the moles of the elements they
have Find out how much they would weigh add them up
13
What about compounds? What is the mass of one mole of CH4?
1 mole of C = 12 g 4 mole of H x 1 g = 4g
1 mole CH4 = 12 + 4 = 16g
The Gram Molecular mass of CH4 is 16.05g
The mass of one mole of a molecular compound.
14
Gram Formula Mass The mass of one mole of an ionic
compound. Calculated the same way.
What is the GFM of Fe2O3?
2 moles of Fe x 56 g = 112 g 3 moles of O x 16 g = 48 g The GFM = 112 g + 48 g = 160g
15
Molar Mass The generic term for the mass of one
mole. The same as gram molecular mass,
gram formula mass, and gram atomic mass.
16
Examples Calculate the molar mass of the following
and indicate what type it is.
Na2S
2 (23) + 32 =
N2O4
2(14) + 4(16) =
C
46 + 32 = 78g
Gram Formula Mass
28 + 64 = 92gGram Molecular Mass12g
1mole Na2S = 78g
1 mole N2O4 = 92g
1 mole C = 12 g Gram Atomic Mass
17
Molar Mass Cont. Ca(NO3)2
40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g 1 mole Ca(NO3)2 = 164g
C6H12O6
6(12) + 12(1) + 6(16) = 70 + 12 + 96 = 180g 1 mole C6H12)6 = 180g Gram Molecular Mass
(NH4)3PO4
3(14) + 12(1) + 31 + 4(16) = 42+12+31+64 = 149g 1 mole (NH4)3PO4 = 149g Gram Formula Mass
Gram Formula Mass
18
Using Molar Mass
Finding moles of compounds
Counting pieces by weighing
19
Molar Mass The number of grams of 1 mole of
atoms, ions, or molecules.
We can make conversion factors from these. To change grams of a compound to moles of a compound.
20
For example
How many moles is 5.69 g of NaOH?
21
For example
How many moles is 5.69 g of NaOH?
5 69. g
22
For example
How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles
23
For example
How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles for NaOH
24
For example
How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16.00 g
1 mole of H = 1 g
25
For example
How many moles is 5.69 g of NaOH?
5 69. g mole
g
need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g
1 mole of H = 1 g 1 mole NaOH = 40 g
26
For example
How many moles is 5.69 g of NaOH?
5 69. g 1 mole
40.00 g
need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g 1
mole of H = 1 g 1 mole NaOH = 40 g
27
For example
How many moles is 5.69 g of NaOH?
5 69. g 1 mole
40.00 = 0.142 mol NaOH
g
need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g
1 mole of H = 1 g 1 mole NaOH = 40 g
28
Examples How many moles is 4.56 g of CO2?
4.56g CO2 x 1 mole
1 44gCO2
How many grams is 9.87 moles of
H2O?
9.87 moles x 18g H2O
1 1 mole
= 0.104 moles
= 178g
29
Examples How many molecules in 6.8 g of
CH4? 6.8g CH4 x 1 mole x 6.02x1023 mc
1 16g CH4 1mole
49 molecules of C6H12O6 weighs how
much? 49 mc x 1 mole x 180g C6H12O6 =
1 6.02x1023 mc 1 mole
8820 g____= 1.5x10-20 g
6.02x1023
= 2.56x1023 mc
30
Gases and the Mole
31
Gases Many of the chemicals we deal with
are gases. They are difficult to weigh. Need to know how many moles of
gas we have. Two things effect the volume of a gas Temperature and pressure Scientists compare gases at
Standard Temperature and Pressure
32
Standard Temperature and Pressure
0ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Avogadro’s Hypothesis - at the same
temperature and pressure equal volumes of gas have the same number of particles.
33
Examples What is the volume of 4.59 mole of CO2 gas at
STP? 4.59mole x 22.4L
1 1mole
How many moles is 5.67 L of O2 at STP? 5.67L x 1 mole
1 22.4L
What is the volume of 8.80g of CH4 gas at STP?
8.80g CH4 x 1mole x 22.4L
1 16g CH4 1mole
= 102.816 = 103L
= .523moles
= 12.32 = 12.3L
34
We have learned how to change moles to grams moles to atoms moles to formula units moles to molecules moles to liters molecules to atoms formula units to atoms formula units to ions
35
Moles
MassPeriodic Table
36
Moles
MassVolume Periodic
Table
37
Moles
MassVolume 22.4 L Periodic
Table
38
Moles
MassVolume
Representative Particles
22.4 L Periodic Table
39
6.02 x 1023
Moles
MassVolume
Representative Particles
22.4 L Periodic Table
40
Moles
MassVolume
Representative Particles
6.02 x 1023
Atoms
22.4 L Periodic Table
41
Moles
MassVolume
Representative Particles
6.02 x 1023
Atoms Ions
22.4 L Periodic Table
42
Percent Composition Like all percents Part x 100 %
whole Find the mass of each component, divide by the total mass.
43
Example Calculate the percent composition of
each element in a compound that is 29.0 g of Ag with 4.30 g of S.
Ag 29.0g
S + 4.30g 33.3g
/33.3 = .8709 x 100 = 87.09%
/33.3 = .1291 x 100 = 12.91%
44
Getting % from the formula If we know the formula, assume you
have 1 mole. Then you know the pieces and the
whole.
45
Examples Calculate the percent composition of
C2H4?
C 2(12g)=24
H 4(1g) = +4 28g
/28
/28
= .8571 x 100 = 85.71%
= .1429 x 100 = 14.29%
46
Example Calculate the percent composition of
Aluminum carbonate. Al2(CO3)3
Al 2(27g)= 54
C 3(12g)= 36
O 9(16)= 144
234g
/234
/234
/234
= .2308 x 100 = 23.08%
= .15.38 x 100 = 15.38%
= .6154 x 100 = 61.54%
47
You can also calculate the mass of an element in a given amount of a compound using % composition.
• Step 1: calculate the % comp. only of the element you want to find the mass of.
• Step 2: Multiply the elements %, by the mass of the compound given.
Example: Calculate the mass of sulfur in 3.54g of H2S.
MM of H2S = H 2 (1) = 2 S 1(32) = +32 34g H2S % S = 32/34 x 100 = 94.1% S
94.1% x 3.54g = 3.33g S
48
Calculate the mass of nitrogen in 25g of (NH4
)2
CO3
.
Calculate the mass of nitrogen in 25g of (NH4
)2
CO3
.
N 2(14g) = 28
H 8(1g) = 8
C 1(12g) = 12
O 8(16g) = +128
176g (NH4
)2
CO3
%N = 28/176 x 100 = 15.91%
15.91% x 25g = 4.0g N
49
Calculate the mass of magnesium in 97.4g of Mg(OH)2.
Mg 1 (24g) = 24 O 2(16g) = 32 H 2 ( 1g) = +2 58g Mg(OH)2
%Mg = 24/58 x 100 = 41.38% 41.38% x 97.4g = 40.3g Mg
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Empirical Formula
From percentage to formula
51
The Empirical Formula The lowest whole number ratio of
elements in a compound. The molecular formula the actual
ration of elements in a compound. The two can be the same. CH2 empirical formula
C2H4 molecular formula
C3H6 molecular formula H2O both
52
Calculating Empirical Just find the lowest whole number ratio C6H12O6
CH4N It is not just the ratio of atoms, it is also
the ratio of moles of atoms. 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.
53
Calculating Empirical Means we can get ratio from percent
composition. Assume you have a 100 g. The percentages become grams. Can turn grams to moles. Find lowest whole number ratio by
dividing by the smallest moles.
54
Example Calculate the empirical formula of a
compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
Assume 100 g so 38.67 g C x 1mol C = 3.223 mole C
12 g C 16.22 g H x 1mol H = 16.22 mole H
1 g H 45.11 g N x 1mol N = 3.222 mole N
14 g N
55
Example The ratio is 3.223 mol C = 1 mol C
3.222 mol N 1 mol N
The ratio is 16.00 mol H = 5 mol H 3.222 mol N 1
mol N
C1H5N1
56
A compound is 43.64 % P and 56.36 % O. What is the empirical formula?
43.64 g P x 1mol P = 1.408 mole P 31 g P
56.36 g O x 1mol O = 3.523 mole O 16 g O
The ratio is 3.523 mol O = 2.5 mol O 1.408 mol P 1 mol P
Can not have 2.5 atoms! Double P2O5
57
Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?
49.48 g C x 1mol C = 4.123 mole C 12 g C
5.15 g H x 1mol H = 5.15 mole H 1 g H
28.87 g N x 1mol N = 2.062 mole N 14 g N
16.49 g O x 1mol O = 1.031 mole O 16 g O
58
The ratio is 4.123 mol C = 4 mol C 1.031 mol O 1
mol O The ratio is 5.15 mol H = 5 mol H
1.031 mol O 1 mol O
The ratio is 2.062 mol N = 2 mol N 1.031 mol O 1
mol O C4H5N2O1
59
Empirical to molecular Since the empirical formula is the lowest
ratio the actual molecule would weigh more.
By a whole number multiple. Divide the actual molar mass by the the
mass of one mole of the empirical formula.
Caffeine has a molar mass of 194 g. what is its molecular formula?
C4H5N2O1 = 97g 194/97 = 2 Molecular Formula = C8H10N4O2
60
Example A compound is known to be composed of 71.65 % Cl,
24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula?
71.65 g Cl x 1mol Cl = 2.047 mole Cl 35 g Cl
24.27 g C x 1mol C = 2.023 mole C 12 g C
4.07 g H x 1mol H = 4.07 mole H 1 g H
61
The ratio is 2.047 mol Cl = 1 mol Cl 2.023 mol C 1
mol C The ratio is 4.07 mol H = 4 mol H
1.031 mol C 1 mol C
Empirical Formula = CClH4
12+35+4 = 51g 98.96/51 = 2 Molecular Formula = C2Cl2H8