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
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.
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.
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.= 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