Transcript Lab #6 - Chem-108

Lab #6
Key Terms
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Atomic Weight
Molecular Weight
Gram Atomic & Molecular Weight
Avogardro’s Number
Moles
Formula Weight
Simplest Empirical Formula
Law of Definite Composition
Stoichiometery
What is Atomic Weight?
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This is the mass of an atom.
Each element’s atom has a different mass.
How is it measured?
In Amus (atomic mass units).
How do we determine the mass of an atom?
We use the periodic table.
Atomic Mass Units
 Example # 1
What is the mass of one Iron atom?
55.8 amus.
What is the of a Copper atom?
63.55 amus
Ionic Vs. Molecular Weight
 Ionic Weight is the mass of all of the
components in an ionic compound.
 Molecular Weight is the mass of all of the
components in a molecular compound.
 Both are determined the same way.
 For now on if both Molecular Wt refers to both
ionic and molecular compounds.
What is a Mole?
 6.02X10 23 parts.
 We use the mole concept to talk about how
many parts not mass.
 If have a mole of computers we have 6.02X10 23
computers.
 This number is used just like how we say we
have 12 parts in a dozen.
 A mole of cars has a different mass then a mole
of potatoes, but each mole contains 6.02X1023
respective parts.
Grams Vs. Amu
 To determine how many grams are in one mole
of any ELEMENT all we have to do is look at the
periodic table.
 55.8 grams = 1 mole of iron.
 So when we are talking about moles the units
are grams, and when we are talking about
atoms the units are in amus.
 Remember the reason for using moles is to
make the numbers easier to use.
Example (1)
 1 mole of Na+1 ions reacts with 1 mole of Cl-1
ions to form one mole of Sodium Chloride.
 This is easier then saying 6.02X 10 23 sodium
ions reacted with 6.02X 10 23 chloride ions to
form 6.02X 10 23 atoms
of sodium chloride
We use the formula:
Grams
Moles 
Molecular Wt.
Determine the number of moles
in .64g grams of NaCl?
.64gNaCl
moles 
 0.017 moles of NaCl
58g/mole
Note :
1
1
mole
 mole
Formula Weight
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If the formula is wrong then the mass will be
wrong. Per means 1 more oxygen then the
base acid.
 Base= HClO3 + 1 oxygen
ANS : HClO4
 H= 1.0 g/mol
 Cl = 36.00 g/mol
 O= 16.00 g/mol
Answer: 100.00 g/mol
What is the percent composition
of all of the elements in HIO?
 Add the masses up for a grand total of
143.90g/mol
 1.0g/mol/ 143.90 X100 = 0.69%
 126.90/143.90 X 100 = 88.0%
 16.0g/mol/ 143.90 X 100 = 11.0%
Types of Formulas
 Simplest Empirical
 This tells us the number of different elements
present in a compound and the lowest
possible ratio of these elements.
 True Molecular Formula
 True molecular formula is always a true
multiple of the simplest formula.
Stoichiometery Rules:
1. We need a balanced equation
2. Start with the number of grams given and convert to
to moles
3. Use your mole ratio and number of grams produced
to determine the number of moles asked for.
4. Convert this number of moles back to grams.
 Given 8.0g of CH4 how many grams of H2O
can be produced?
CH4+ 2O2→CO2 +2H2O
(1) Balance Equation
CH4+ 2O2→CO2 +2H2O
(2) Convert given grams to moles
Given 8.0g of CH4 how many grams of
H2O can be produced?
grams
moles 
mw
8g of CH 4
 .5moles of CH 4
16g of CH 4 /1mole of CH 4
Note :
1
1
mole
 mole
(3) Use mole to mole ratio
CH4+ 2O2→CO2 +2H2O
mole ratio
1mole of CH4 reacts with 2 moles O2
to produce 1 mole of CO2 and 2 moles of H2O
 From our problem: Given 8.0g of CH4 how many grams
of H2O can be produced?
CH4+ 2O2→CO2 +2H2O
We know 1 mole of CH4 will produce 2 moles of H2O
you want
to make our ratio :
started with
2moles of H 2 0
So our mole to mole ratio is
1moles of CH 4
.5 moles of CH 4 2moles of H 2 0
1
1moles of CH 4
Now Do the math:
.5X 2 = 1 mole of H20 can be produced
(4) Convert moles back to grams
grams
moles 
mw
mw  moles  grams
18 grams of H 2 O
1 mole of H 2 O  18 grams of H 2 O produced
1 mole of H 2 O
Lab # 6
 Obtain a sample of salt and determine its
composition.
 This will be accomplished by determining
the % oxygen in your sample.
 Review how to use a crucible.
Table A
 Wash, dry, cool and then determine the mass of
your empty crucible.
 Add the salt and record the mass of salt and
crucible.
 Once the weight is determined add a pinch of the
catalyst to your crucible.
 Heat crucible according to directions in lab.
 Allow crucible to cool and determine the final
mass
Table B
Determine the percent oxygen in the salts listed.
 For KClO2
 Mass of each K = 39g, Cl = 36.g and O =16 and
2 present so 2 X16 = 32g
 Total mass: 39g+ 36g +32g = 107g.
32g massof oxy gen
X100  30%
107g total mass
Table C
M KClOX  M Crucible&salt  M Crucible
M
O 2 lost
of O lost  M
-M
 M ofMCrucible
Salt and Catalyst - M
2
of CrucibleSalt and Catalyst
of Crucible and residue
of Crucible and residue
Mass of O 2 lost
% Oxygen 
X 100
Mass of KClO x Used
Due Next Week
 Complete Lab: pgs 44-45
 Complete homework and exercise for lab # 6:
46-47pgs
 Study for quiz # 2