Download mole and stoikiometri

MOLE AND STOIKIOMETRI
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 We
measure mass in grams. g
 We measure volume in liters. L
 We
count pieces in MOLES.
mol
MOLE CONCEPT
A Carbon-12 atom
has a mass of 12amu
C C
C C C
C
C CC CC
C
C C
CC C
C C C C
C C
C
C CC C C C
2
C
= 12amu
A mole of Carbon atoms
has a mass of 12 grams
Gram Atomic Mass

Equals the mass of 1 mole of an element

Written on the Periodic Table

12.01 grams of C has the same number of pieces as
1.008 grams of H and 55.85 grams of iron.

We can write this as 12.01 g C = 1 mole C

We can count things by weighing them.
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The Mole Concept
Atomic Mass
Mole of
in grams
atoms
A mole of aluminum = 27.0g
A mole of gold
A mole of silver
A mole of boron
= 197g
= 108g
= 10.8g
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5
If a Helium atom is 4 times
heavier than a Hydrogen atom,
then a dozen Helium atoms is 4 times
heavier than a dozen Hydrogen atoms.
He
H
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And a mole of Helium atoms is 4 times
heavier than a mole of Hydrogen atoms.
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So, the number of atoms in a mole
is always the same
The number of atoms in a mole is called,
Avogadro's Number
Avogadro's Number = 6.022 x
23
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Mole – Particle Conversions
The mole concept allows us to count
atoms, molecules, formula units and
ions.
1 mole of Al = 6.022 x 1023 atoms of Al
So how many atoms in 3 mol of Al?
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Examples
 How
much would 2.34 moles of
carbon weigh?
 How
many moles of magnesium is
24.31 g of Mg?
 How
many atoms of lithium is 1.00 g
of Li?
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What about compounds?
 In
1 mole of H2O molecules there are 2
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
Types of Molar Mass
• Gram Molecular Mass (GMM)- the total
number of atoms in a molecular compound
• Cl2 = 2 x 35.4 g = 70.8 g
• CH4= (1 x 12.0g) + (4 x 1.0g) = 16g
• Gram Formula Mass (GFM)- the total
number of atoms in an ionic compound
• NaCl = (1 x 23.0g) + (1 x 35.4g) = 58.4g
• Ca(OH)2= (1 x 40.0g) + (2 x (16.0g+1.0g))= 74g
Molar Mass
 Molar
mass is the generic term for
the mass of one mole of any
substance (in grams)
 The same as:
1) gram molecular mass
 2) gram formula mass
3) gram atomic mass
Examples
 Calculate
the molar mass of the
following and tell what type it is:
Na2S
= 78.1 g
N2O4
= 92.0 g
Ca(NO3)2 = 164.1 g
= 180.0 g
C6H12O6
(NH4)3PO4 = 149.0 g
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
We need to compare them at the
same temperature and pressure.
Standard Temperature and
Pressure
0ºC
and 1 atm pressure
0oC = 32oF= 273 K
1 atm= 101.3 kPa = 760 mmHg (torr)

At STP 1 mole of gas occupies 22.4 L

Called the molar volume

1 mole = 22.4 L of any gas at STP
Examples
What
is the volume of 4.59
mole of CO2 gas at STP?
4.59 mol CO2 22.4 L CO2
1 mol CO2
=103 L CO2
How
many moles is 5.67 L
of O2 at STP?
5.67 L O2
1 mol O2
22.4 L O2
=0.253 mol O2
Stoichiometry
“Stoichiometry” is Greek for “Measuring
Elements”
It starts with a balanced equation
2H2 + O2 2H2O

2 moles of hydrogen reacts with

1 mole of oxygen

Forming 2 moles of water.
Stoichiometry
The coefficients tell us how many moles of each substance. Not
Grams!
For example:
2H2 + O2 2H2O
2 g of H2 + 1 g of O2
= 2 g of H O
2
3 g of reactants can’t make only 2 g of products
Mass of a Product
The Law of Conservation of Mass applies
Convert the moles to grams and the
equation does work.
2H2 + O2  2H2O
2 moles H2
2.02 g H2
1 mole H2
= 4.04 g H2
1 mole O2 32.00 g O2
= 32.00 g O2
1 mole O2
36.04 g H2+O2
Mass of a Product
2H2 + O2  2H2O
2 moles H2O
18.02 g H2O
= 36.04 g H2O
1 mole H2O
2H2 + O2  2H2O
36.04 g H2 + O2= 36.04 g H2O
2Na + Cl2  2NaCl
How many grams of sodium are
needed to react with .071g of
chlorine gas?
2Na (s) + Cl2(g) 2NaCl (s)
Mole Ratio
Mass Ratio
2
23 x 2=
46
1
2
35.5 x 2 x 1= 58.5 x 2=
71
117
0.071 g Cl2 1 mol Cl2 2 mol Na 46 g Na
71 g Cl2 1 mol Cl2 2 mol Na
= 0.046 g Na
Chemical Yield
/
3. Percent yield – a percentage ratio between the actual yield and
the theoretical yield.
% Yield =
Actual Yield
Theoretical Yield

% yield tells us how “efficient” a reaction is.

% yield can not be bigger than 100 %.
x 100
Example

According to your calculations the theoretical yield for the
production of NaCl is 13.6 grams. In the laboratory your actual yield
is 11.8 grams of NaCl. What is the percent yield?
11.8 g NaCl
x 100
13.6 g NaCl
Answer = 86.7 %
Calculating Percent
Composition of a Compound
 Like
all percent problems:
Part
whole
x 100 %
 Find the mass of each
component,
 then divide by the total mass.
Calculating Percent
Composition of a Compound

Find the mass percent for the elements in Sodium
hydrogen carbonate:

NaHCO3
-molar mass is 84 g/mol
-mass of: Na= 23g
H= 1g
C= 12g
O= 3 x 16= 48g
%C = (12g/84g) x 100= 14.3%
%Na = (23g/84g) x 100= 27.4%
%H = (1g/84g) x 100= 1.2%
%O = (48g/84g) x 100= 57.1%
The Empirical Formula
The
lowest whole number ratio of
elements in a compound.
The molecular formula = the
actual ratio of elements in a
compound.
The two can be the same.
CH2 is an empirical formula
C2H4 is a molecular formula
C3H6 is a molecular formula
H2O is both empirical & molecular
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.
Calculating Empirical
 We
can get a ratio from the
percent composition.
 Assume you have a 100 g.
 The percentages become grams.
 Convert grams to moles.
 Find lowest whole number ratio
by dividing by the smallest mole
value
Example

Calculate the empirical formula of a compound composed of
38.67 % C, 16.22 % H, and 45.11 %N.

Assume 100 g

38.67 g C x 1mol C

16.22 g H x 1mol H
= 16.09 mole H
1.01 g H

45.11 g N x 1mol N
= 3.219 mole N
14.01 g N
= 3.220 mole C
12.01 g C
Example

The ratio is 3.220 mol C
mol N

The ratio is 16.09 mol H = 5 mol H
N

=C H N

Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its
empirical formula?
1 5 1
= 1 mol C
3.219 mol N
3.219 mol N
1
1 mol
= CH5N