The Mole Concept Avagadro’s Number = 6.022 x 10 23.

The Mole Concept

Avagadro’s Number = 6.022 x 1023

Counting Atoms

• Chemistry is a quantitative science - we need a "counting unit."

• The MOLEMOLE

• 1 mole is the amount of substance that contains as many particles (atoms or molecules) as there are in 12.0 g of C-12.

The Mole is DevelopedCarbon Atoms Hydrogen Atoms Mass Ratio

Number Mass (amu) Number Mass (amu) Mass carbon / Mass hydrogen

12

1

12 amu = 12

1 amu 1

24

[2 x 12]

2

[2 x 1]

24 amu = 12

2 amu 1

120

[10 x 12]

10

[10 x 1]

120 amu = 12

10 amu 1

600

[50 x 12]

50

[50 x 1]

600 amu = 12

50 amu 1

(6.02 x 1023) x (12)Avogadro’s number (6.02 x 1023) x (1)

(6.02 x 1023) x (12) = 12

(6.02 x 1023) x (1) 1

Avogadro’s number

Amedeo Avogadro

(1776 – 1856)

1 mole = 602213673600000000000000

or 6.022 x 1023

thousandsmillionsbillionstrillions

quadrillions?

There is Avogadro's number of particles in a mole of any substance.

Particles in a MoleParticles in a Mole

Careers in Chemistry - Philosopher

Q: How much is a mole?A: A mole is a quantity used by chemists to count atoms and molecules. A mole of something is equal to 6.02 x 1023 “somethings.”

1 mole = 602 200 000 000 000 000 000 000

Q: Can you give me an example to put that number in perspective?A: A computer that can count 10,000,000 atoms per second would take 2,000,000,000 years to count 1 mole of a substance.

Counting to 1 Mole

Is that right? A computer counting 10 million atoms every secondwould need to count for 2 billion years to count just a single mole.

Lets look at the mathematics.

365 24 60min 60secsec 1 31,536,000sec

1 1 1 1min

days hoursx year

year day hour

Therefore 1 year has 31,536,000 seconds or 3.1536 x 107 sec. A computer counting 10,000,000 atoms every second could count3.153 x 1014 atoms every year.

Finally, 6.02 x 1023 atoms divided by 3.1536 x 1014 atoms every yearequals 1,908,929,477 years or approximately 2 billion years!

How Big is a Mole?

One mole of marbles would cover the entire Earth (oceans included) for a depth of three miles.

One mole of $100 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 milliontimes.

It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

Avogadro’s Number A MOLE of any substance contains as many elementary units

(atoms and molecules) as the number of atoms in 12 g of the isotope of carbon-12.

This number is called AVOGADRO’s number NA = 6.02 x 1023 particles/mol

The mass of one mole of a substance is called MOLAR MASS symbolized by MM

Units of MM are g/mol Examples

H2 hydrogen 2.02 g/molHe helium 4.0 g/mol

N2 nitrogen 28.0 g/mol

O2 oxygen 32.0 g/mol

CO2 carbon dioxide 44.0 g/mol

1 Mole of Particles

Molar Mass

Molecular Weight and Molar Mass

Molecular weight Molecular weight is the sum of atomic weights of all atoms in the molecule.

Molar mass Molar mass = molecular weight in grams.

example: NaCl has a molecular weight of 58.5 a.m.u.

example: NaCl has a molar mass of 58.5 grams

this is composed of a single molecule of NaCl

this is composed of a 6.02 x1023 molecules of NaCl

The Molar Mass and Number of Particles in One-Mole Quantities

Substance Molar Mass Number of Particles in One Mole

Carbon (C) 12.0 g 6.02 x 1023 C atoms

Sodium (Na) 23.0 g 6.02 x 1023 Na atoms

Iron (Fe) 55.9 g 6.02 x 1023 Fe atoms

NaF (preventative 42.0 g 6.02 x 1023 NaF formula unitsfor dental cavities)

CaCO3 (antacid) 100.1 g 6.02 x 1023 CaCO3 formula units

C6H12O6 (glucose) 180.0 g 6.02 x 1023 glucose molecules

C8H10N4O2 (caffeine) 194.0 g 6.02 x 1023 caffeine molecules

Chemical Equations

Molar Mass and Mole Calculations

m

n M

Molar Mass

• Molar mass is the mass in grams of one mole of particles (atoms, ions, molecules, formula units).

• Equal to the numerical value of the average atomic mass (get from periodic table)

1 mole of C atoms = 12.0 g

1 mole of Mg atoms = 24.3 g

1 mole of Cu atoms = 63.5 g• So, Molar Mass is the atomic mass expressed in

units grams/mole (g/mol)

Molar Mass

Chlorine has an atomic mass of 35.453 amu. So what is its molar mass?

35.453 grams per mole

The number is the same, just change the unit.

Molar Mass

• How do we find the molar mass of a compound? – Find the molar mass of each element

in the compound– Determine how many of each element

are in the compound.– Multiply the number of each element

by its molar mass. – Add these mumbers.

Molar Mass

In one mole of Aluminum Bromide, there is 1 mole of

aluminum atoms and 3 moles of bromine atoms.

Aluminum Bromide

AlBr3

Molar Mass

• To find the molar mass of AlBr3, we do the following:

1 mol Al x 26.9815 g/mol Al = 26.9815 g Al

3 mol Br x 79.904 g/mol Br = 239.712 g Br

26.9815 g Al + 239.712 g Br = 266.6935 g AlBr3 per mole or 266.6935 g/mol AlBr3

Mole Calculations

Where, n is the # of moles (mol) m is the mass (g)M is the molar mass (g/mol)

m

n M

n = m / M

M = m/n

m = n x M

Relationship Between Moles, Mass and Particles

Mole Calculations

• What is the mass of 3.0 moles of chromium (III) iodide? CrI3

Mole and # of Particles

6.02x1023 molecules in a mole

A mole contains 6.02 x 1023 particles.

How many particles are in 2.0 moles of particles?

2 mol x 6.02 x 1023 particles/mol= 12.04 x 1023 particles= 1.2 x 1024 particles

When changing moles to number of particles simply Multiply the number of moles of particles by

6.02 x 1023 particles/mol

A mole contains 6.02 x 1023 particles.How many particles are in 4.0 moles of particles?

4 mol x 6.02 x 1023 particles/mol

= 24.08 x 1023 particles


How many particles are in 0.35 moles of particles?

0.35 mol x 6.02 x 1023 particles/mol


A mole contains 6.02 x 1023 particles.How many moles of particles are in 1.02 x 1025 particles

= 1.02 x 1025 particles / 6.02 x 1023

particles/mole

= 16.9 moles

To change # of particles to moles simply

Divide the # of particles by 6.02 x 1023 particles/mole

A mole contains 6.02 x 1023 particles.How many moles of particles are in 4.74 x 1024 particles

= 4.74 x 1024 particles / 6.02 x 1023 particles/mole

= 7.873 moles = 7.87mol

How many moles of particles are in 3.24 x 1022 particles

= 3.24 x 1022 particles / 6.02 x 1023 particles/mole

= 0.05382 moles = 5.38x10-2 moles

A mole contains 6.02 x 1023 particles.How many particles are in 45.6 moles of particles?

44.6 moles x 6.02 x 1023 particles/mol= 24.08 x 1023 particles= 2.68 x 1025 particles

How many moles are in 4.5 x 1021 particles?

4.5 x 1021 particles / 6.02 x 1023 particles/mol= 0.0075 mol

How many molecules are there in 4.78 g of chlorine dioxide? ClO2

a) Find the number of moles of ClO2 first

n = m / M = 4.78 g / 67 g/mol

=0.0713 mol

b) Now convert # of mol into particles

= 0.0713 mol x 6.02 x 1023 particles/mol

=4.3 x 1022 molecules of ClO2

How many molecules are there in 2.3 g of aluminum bromide? AlBr3

a) Find the number of moles of AlBr3

n = m / M = 2.3 g / 267 g/mol

= 0.0086 mol of AlBr3

b) Now convert # of mol into particles

= 0.0086 mol x 6.02 x 1023 particles/mol

= 5.2 x 1021 molecules of AlBr3

What is the mass of 8.2 x 1021 molecules of C2H5?

a) First find the # of moles.

8.2 x 1021 molecules / 6.02 x 1023 molecules/mol

= 0.01362 mol

b) Convert from moles to mass by multiplying the number of moles by the molar mass of C2H5.

MM = (12 x 2 + 5 x 1 ) = 29 g/mol

= (0.01362mol)x 29 g/mol = 0.395 g


a) First find the # of moles.



MM = (12 x 2 + 5 x 1 ) = 29 g/mol

= (8.2 x 1021 molecules / 6.02 x 1023) x 29 g/mol

= 0.395 g


a)First find the # of moles, then find the mass.


(= 6.15mol)


MM = (12 x 4 + 10 x 1 ) = 58 g/mol

1 step:(3.7 x 1024 molecules / 6.02 x 1023) x 58 g/mol = 356 g

= 3.6x102g

Converting Particles, Moles and Mass 1. Find the mass of each of the following:

a) 7.8 x 1026 molecules of N2O4

b) 5.6 moles of C2H5OH

c) 1.56 x 1021 molecules of C3H6

d) 7.4 moles of ClS3

e) 3.6 x 1025 molecules of CO2

2. Find the number of particles in each of the following:

a) 75 g of NF5

b) 0.0034 moles of C3H7OH

c) 2.8 g of C5H10

d) 2.5 moles of FBr3

e) 435 g of CO2

The Mole Concept Avagadro’s Number = 6.022 x 10 23.

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