Transcript Chemical Quantities

Chemical
Quantities
The Mole,
% Composition,
Empirical and
Molecular Formulas
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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.
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We count pieces in MOLES.
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.
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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.
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Types of questions

How many oxygen atoms in the
following?
3
– CaCO3
– Al2(SO4)3 12
 How many ions in the following?
– CaCl2
– NaOH
3
2
– Al2(SO4)3
5
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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 = 2.75x1024mc
1
1 mole

How many moles of water is 5.87 x 1022
molecules?
 5.87 x 1022 mc
x 1 mole = 0.0975 mole
1
6.02x1023 mc
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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 = 4.44x1024 atoms
1
1 mole
1 mc

How many moles is 7.78 x 1024 formula
units of MgCl2?
7.78x1024 FU x 1 mole
1
6.02x1024 FU
7
= 12.9 mole
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.
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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
= 28.08 g
1
1mole
 How many moles of magnesium in
24.31 g of Mg?
24.31 g Mg x 1 mole =
1.013 mole
1
24g Mg
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
How many atoms of lithium in 1.00 g
of Li?

1.00 g Li x 1 mole x 6.02x1023 atoms
8.60x1022 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
13.6 g
What about compounds?
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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
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.
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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
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Molar Mass
The generic term for the mass of one
mole.
 The same as gram molecular mass,
gram formula mass, and gram atomic
mass.
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Examples
Calculate the molar mass of the following
and indicate what type it is.
 Na2S
 2 (23) + 32 = 46 + 32 = 78g 1mole Na2S = 78g
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Gram Formula Mass
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N2O4
 2(14) + 4(16) = 28 + 64 = 92g
1 mole N2O4 = 92g
Gram Molecular Mass
C
16
12g
1 mole C = 12 g
Gram Atomic Mass
Molar Mass Cont.
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Ca(NO3)2
40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g
1 mole Ca(NO3)2 = 164g Gram Formula Mass
C6H12O6
 6(12) + 12(1) + 6(16) = 70 + 12 + 96 = 180g
 1 mole C6H12)6 = 180g Gram Molecular Mass
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(NH4)3PO4
3(14) + 12(1) + 31 + 4(16) = 42+12+31+64 = 149g
1 mole (NH4)3PO4 = 149g Gram Formula Mass
Using Molar Mass
Finding moles of compounds
Counting pieces by weighing
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Molar Mass
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
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.
For example
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How many moles is 5.69 g of NaOH?
For example
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How many moles is 5.69 g of NaOH?

5.69 g

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


For example

How many moles is 5.69 g of NaOH?
mole 

5.69 g


g 
 need to change grams to moles
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For example

How many moles is 5.69 g of NaOH?
mole 

5.69 g


g 
 need to change grams to moles
 for NaOH
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For example

How many moles is 5.69 g of NaOH?
mole 

5.69 g


g 
 need to change grams to moles
 for NaOH
 1mole Na = 23g 1 mol O = 16.00 g
1 mole of H = 1 g
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For example
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How many moles is 5.69 g of NaOH?
mole 

5.69 g


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
For example
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How many moles is 5.69 g of NaOH?
1 mole 

5.69 g


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
For example

How many moles is 5.69 g of NaOH?

5.69 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

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1 mole 
 = 0.142 mol NaOH
40.00 g 
Examples
How many moles is 4.56 g of CO2?
 4.56g CO2 x 1 mole
= 0.104 moles
1
44gCO2

How many grams is 9.87 moles of
H2O?
 9.87 moles x 18g H2O
= 178g
1
1 mole
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Examples

How many molecules in 6.8 g of
CH4?

6.8g CH4 x 1 mole x 6.02x1023 mc
1
16g CH4
1mole
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49 molecules of C6H12O6 weighs how
much?
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49 mc x 1 mole x 180g C6H12O6 =
1 6.02x1023 mc
1 mole
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8820 g____= 1.5x10-20 g
6.02x1023
= 2.56x1023 mc
Gases and the Mole
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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
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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.
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Examples
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What is the volume of 4.59 mole of CO2 gas at
STP?
4.59mole x 22.4L
= 102.816 = 103L
1
1mole
How many moles is 5.67 L of O2 at STP?
5.67L x 1 mole
= .523moles
1
22.4L
What is the volume of 8.80g of CH4 gas at STP?
8.80g CH4 x 1mole x 22.4L
= 12.32 = 12.3L
1
16g CH4 1mole
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
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Periodic
Table
Moles
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Mass
Volume
Periodic
Table
Moles
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Mass
Volume 22.4 L
Periodic
Table
Moles
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Mass
Volume 22.4 L
Periodic
Table
Moles
Representative
Particles
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Mass
Volume 22.4 L
Periodic
Table
Moles
6.02 x
23
10
Representative
Particles
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Mass
Volume 22.4 L
Periodic
Table
Moles
6.02 x
23
10
Representative
Particles
Atoms
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Mass
Volume 22.4 L
Periodic
Table
Mass
Moles
6.02 x
23
10
Representative
Particles
Atoms
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Ions
Percent Composition
Like all percents
 Part x 100 %
whole
 Find the mass of each component,
 divide by the total mass.
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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 /33.3 = .8709 x 100 = 87.09%
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S
+ 4.30g /33.3 = .1291 x 100 = 12.91%
33.3g
Getting % from the formula
If we know the formula, assume you
have 1 mole.
 Then you know the pieces and the
whole.
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Examples

Calculate the percent composition of
C2H4?

C 2(12g)=24 /28 = .8571 x 100 = 85.71%
H 4(1g) = +4 /28 = .1429 x 100 = 14.29%
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28g
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Example
Calculate the percent composition of
Aluminum carbonate.
 Al2(CO3)3
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Al 2(27g)= 54 /234
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C 3(12g)= 36 /234 = .15.38 x 100 = 15.38%
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O 9(16)= 144 /234 = .6154 x 100 = 61.54%
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234g
= .2308 x 100 = 23.08%
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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
Calculate the mass of nitrogen in 25g of
(NH ) CO .
42 3
 Calculate the mass of nitrogen in 25g of (NH4)2CO3.
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N 2(14g)
H 8(1g)
C 1(12g)
O 8(16g)
= 28
=
8
= 12
= +128
176g (NH ) CO
42 3
 %N = 28/176 x 100 = 15.91%
 15.91% x 25g = 4.0g N
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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
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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
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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.
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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.
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Example
 Calculate
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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
1gH
 45.11 g N x 1mol N = 3.222 mole N
14 g N
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
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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
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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
1gH
 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

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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
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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
59  Molecular Formula = C8H10N4O2

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
1gH
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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
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