Experiment #5. Avagadro’s Number 6.02 x 10 23 And The Standard Deviation x S. Introduction. Purpose : Relate a gram formula weight to Avogadro’s number by using the dimensional analysis approach and determine the correct number of significant figures in your applications.
Avagadro’s Number6.02 x 10 23
And The Standard Deviation
Amu – We use if we are talking about the mass of an atom. Ex. the mass of an iron atom is 55.8 amus.
Gram – We use if we are talking about the mass of a mole
or fraction of a mole. Ex. The mass of one mole of iron is
55.8 grams.Amu Vs. Gram
Ion Ex. the mass of an iron atom is 55.8 amus.
Not on the Periodic Table
On Periodic Table
Have a charge; can be
positive or negative.
Do not have a charge
Do not React
Only one atom
Canbe more then one atom
Equal number of
Protons & Electrons
Unequal number of
Protons and ElectronsAtomVs. Ion
Mole Ex. the mass of an iron atom is 55.8 amus.
6.02X10 23 Atoms
Larger mass measured in grams
Very small mass
measured in Amus
Larger in size
Can be seen
Very Small in size
Can’t be seenAtom Vs. Mole
Formula weight – the mass of the collection of atoms represented by a chemical formula. For Ex. water contains two hydrogen atoms and one oxygen atom.
1 x 16.00 (mass of O) = 16.00
2 x 1.01 (mass of H) = + 2.02
Formula Weight 18.02
The formula weight tells us the mass of one mole of our substance.
GFW – Gram Formula Weight Ex. the mass of an iron atom is 55.8 amus.
This refers to the mass of an ion or
Ex. Mass of a Cl-
GAW – Gram Atomic Weight
This refers to the mass of a atom
Ex. Mass of a Cl atom
GMW Gram Molecular Weight
This refers to the mass of a molecule or molecular compound
Ex. Mass of one molecule of H2O
Convert 10 m to mm
Remember 1 m = 1000mm
They both mean the same thing it all depends on what you need.
What you are given Ex. the mass of an iron atom is 55.8 amus.
always over 1
The 1 is a filler.
The unit you want to eliminate is placed in the lower right so it will cancel out with your original unit.
Mult. left to right
divide top to bottomDimensional Analysis
X going and / going Ex. the mass of an iron atom is 55.8 amus.Keep in mind
Find the # of moles in Ex. the mass of an iron atom is 55.8 amus.
the sample of Tin. Given
the sample has a mass
of 29.04gEx#1 pg 51
To determine the g formula wt. Ex. the mass of an iron atom is 55.8 amus.
determine the mass of all atoms present.
To determine #g / moleculethink about the units
Think how can we compare grams and molecules? We need to use moles.
So # grams per mole / number of parts in 1 mole
To determine the number of atoms per mole
add up the number of atoms in the formulaChart on pg 55
X = Avg Ex. the mass of an iron atom is 55.8 amus.Standard Deviation
Mass of Beaker Ex. the mass of an iron atom is 55.8 amus.
Mass of Beaker & water
Density of H20 = 0.9982 g/mL
Use V= M/D to determine volume deliveredStandard Deviation Ex. 1 mL Pipette
#1 Ex. the mass of an iron atom is 55.8 amus.
X avg.= 1.02 mL
(b) 0 = 0
(d) 0 = 0
(e) 0 = 0
Take Square Root of Step #5
Final Answer: +/- 0.007 mL.
N= number of trials