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# Experiment #5 - PowerPoint PPT Presentation

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.

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And The Standard Deviation

x  S

• 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.

• Compute the standard deviation for the volume of water delivered by a 10 mL Graduate Cylinder.

• Amu Vs. Gram –

• Atom Vs. Ion –

• Atom Vs. Mole –

• Formula Weight –

• GFW,GAW,GMW –

• Dimensional Analysis –

• 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.

Atoms

Not on the Periodic Table

On Periodic Table

Have a charge; can be

positive or negative.

Do not have a charge

Do not React

React

Only one atom

Canbe more then one atom

Equal number of

Protons & Electrons

Unequal number of

Protons and Electrons

AtomVs. Ion

Mole Ex. the mass of an iron atom is 55.8 amus.

Atom

6.02X10 23 Atoms

1Atom

Larger mass measured in grams

Very small mass

measured in Amus

Larger in size

Can be seen

Very Small in size

Can’t be seen

Atom Vs. Mole

Formula Weight Ex. the mass of an iron atom is 55.8 amus.

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

ionic compound

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

How do Avagadro’s number, Ex. the mass of an iron atom is 55.8 amus. Mole and GAW Apply to each other?

• One mole of Tin has a mass of 118.69g.

• One atom of Tin has a mass of 118.69 Amu.

• One mole of Tin contains 6.02X1023 atoms.

Dimensional Analysis Ex. the mass of an iron atom is 55.8 amus.

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.

# 1

The unit you want to eliminate is placed in the lower right so it will cancel out with your original unit.

# 2

# 3

Mult. left to right

divide top to bottom

Dimensional 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 29.04g

Ex#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 formula

Chart on pg 55

X = Avg Ex. the mass of an iron atom is 55.8 amus.

Standard Deviation

• This is how much your value is off from the actual answer. Can be used as a correction factor.

• Ex. A 1mL pipette delivers 1mL + or - .007mL

• So the pipette delivers .9993mL or 1.007 mL.

Mass of Beaker Ex. the mass of an iron atom is 55.8 amus.

110.51g

110.51g

110.51g

110.51g

110.51g

Mass of Beaker & water

109.50g

111.53g

109.50g

109.50g

111.53g

M H2O

1.01g

1.02g

1.01g

1.01g

1.02g

Density of H20 = 0.9982 g/mL

Vol. of

pipet

1.01mL

1.02mL

1.01mL

1.02mL

1.02mL

Use V= M/D to determine volume delivered

Standard Deviation Ex. 1 mL Pipette

#1 Ex. the mass of an iron atom is 55.8 amus.

X avg.= 1.02 mL

(Xi-Xavg.)2

(a).01 =1.0X10-4

(b) 0 = 0

(c).01 =1.0X10-4

(d) 0 = 0

(e) 0 = 0

#3

Xi-Xavg.

(a)1.01-1.02=.01

(b)1.02-1.02= 0

(c)1.01-1.02=.01

(d)1.02-1.02= 0

(e)1.02-1.02= 0

#2

#5

∑(Xi-X avg.)2/N-1

#4

∑(Xi-X avg.)2

Take Square Root of Step #5

#6

N= number of trials

Due Next Week Ex. the mass of an iron atom is 55.8 amus.

• Pages 51 & 53 we are determining the number of atoms / ions or moles that are asked for.

• YOU MUST SHOW ALL WORK FOR CREDIT!!

• Page 55 complete the table and show all work.

• Pages 61 determine the St. Dev. for 1 mL pipette ( I do as a example)

• Page 63 determine the St. Dev. for a 10mL. graduated cylinder.