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CHEMICAL QUANTITIES All Roads Lead to the Mole

CHEMICAL QUANTITIES

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CHEMICAL QUANTITIES. All Roads Lead to the Mole. Measuring doughnuts: 1 dozen = 12 doughnuts (count) 1 dozen = 500 g doughnuts (mass) 1 dozen = 1 box doughnuts (volume). Measuring steam (H 2 O gas): 1 mole = 6.02 x 10 23 H 2 O molecules (count) 1 mole = 18.0 g H 2 O (mass) - PowerPoint PPT Presentation

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CHEMICAL QUANTITIES

All Roads Lead to the Mole

Chemical Quantities

• Measuring doughnuts:o 1 dozen = 12 doughnuts (count) o 1 dozen = 500 g doughnuts

(mass)o 1 dozen = 1 box doughnuts

(volume)• Measuring steam (H2O gas):o 1 mole = 6.02 x 1023 H2O molecules

(count)o 1 mole = 18.0 g H2O (mass)o 1 mole = 22.4 L H2O (volume) at

STP

Dozen: Baker as Mole: Chemist

Count Amedeo Avogadro• 1776-1856• Lawyer who became interested in

math and physics• Discovered that equal volumes of

different gases contained an equal number of particles.

• 9 years after his death, Joseph Loschmidt determined a constant and named it after Avogadro.

AVOGADRO’S CONSTANT= 6.02 x 1023

Not to be confused with the delicious

fruit

AVOGADRO’S CONSTANT= 6.02 x 1023

1 mole = 6.02 x 1023 particles1 mole = molar mass (grams)1 mole (of a gas at STP) = 22.4 L

The MOLE is to chemists as the DOZEN is to bakers.

5 Pound Bag of Sugar contains 6.6 moles Of C12H22O11 1 Liter bottle of Water contains 55.5 moles H206.02 x 1023 Grains of Sand: Would be more than all of the sand on Miami Beach.6.02 x 1023 Blood Cells: Would be more than the total number of blood cells found in every human on earth. 6.02 x 1023 Watermelon Seeds: Would be found inside a melon slightly larger than the moon.6.02 x 1023 Pennies: Would make at least 7 stacks that would reach the moon.6.02 x 1023 Donut Holes: Would cover the earth and be 5 miles (8 km) deep.

MassVolume

R.P.Molecules

Atoms

Formula Units

Ions

Types of Representative Particles

• Molecules (breaks down into atoms)• Atoms• Formula Units (breaks down into ions)• Ions

Naming Representative Particles

Pure Substance

Element

Monotomic

Charged

ION

Neutral

ATOM

Polyatomic

MOLECULE

Compound

Molecular

MOLECULE

Ionic

FORMULA UNIT

Naming Representative ParticlesExamples: Name the representative particle for each substance given. For each molecule, state how many atoms are present. For each formula unit, state how many ions make up the ionic compound.

H+:

Cl2:

C2H6:

Cu(NO3)2:

Al:

NaCl:

ION

MOLECULE – 2 ATOMS per molecule

MOLECULE – 8 ATOMS

FORMULA UNIT – 3 IONS per Form.U.

ATOM

FORMULA UNIT – 2 IONS per Form.U.

Mole R.P. Calculations

1 mole = Avogadro’s Number = 6.02 x 1023 R.P.’s

R.P.mol 1

R.P.106.02 23

R.P.106.02mol 1

23

R.P. Example 1:How many moles are in 1.4 x 1022 molecules of H2O?

OH molecules1002.6mol1OH molecules104.1

223

222

OH mol 023.0 2

R.P. Example 2:How many representative particles are in 2.6 mol CO2?

2

223

2

CO mol 1CO molecules 106.02CO mol 6.2

224 CO molecules 106.1

R.P. Example 3:How many atoms are in 5.2 mol CO2?

22

223

2

CO molecule 1atoms3

CO mol 1CO molecules106.02CO mol 2.5

atoms104.9 24

Molar Mass … a.k.a. Molecular Weight (MW)

molar mass = mass of 1 mole of substance

Molar mass can be determined by adding up the atomic masses from the periodic table.

MW Example 1:

Find the MW of CH4.

= 1C + 4H = 12.0 + 4(1.0) = 16.0 g/mol

MW Example 2:

Find the MW of Mg(OH)2.

=Mg + 2O + 2H=24.3 + 2(16.0) + 2(1.0)=58.3 g/mol

MW Example 3:

Find the MW of MgSO4•7H2O.

=Mg + S + 4O + 7(H2O)=24.3 + 32.1 + 4(16.0) + 7(18.0)=246.4 g/mol

What is the mass of one mole of table salt (NaCl)?

a) 23.0 gb) 35.5 gc) 58.5 gd) 100.0 g

Atomic mass Na = 23.0 g/molAtomic mass Cl = 35.5 g/molMass of one mole of NaCl = 23.0 + 35.5 = 58.5 g

Mole Mass Calculations

1 mole = molar mass (MW) in grams

Massmol 1

(g) massmolar

(g) massmolar mol 1

Mass Example 1:Mass Example 1: How many grams are in 7.20 moles of dinitrogen trioxide?

MW of N2O3 = 2N + 3O = 2(14.0) + 3(16.0) = 76.0 g/mol

1 mole = 76.0 g N2O3

Dinitrogen trioxide = N2O3

3232 ON g547

1.00molg0.76ON mol20.7

Mass Example 2:Mass Example 2: Find the number of moles in 92.2 g of iron(III) oxide, Fe2O3.

MW of Fe2O3 = 2Fe + 3O = 2(55.8) + 3(16.0) = 159.6 g/mol

1 mole = 159.6 g Fe2O3

Iron (III) oxide = Fe2O3

3232 OFe mol578.0

159.6gmol00.1OFe g2.92

What is the mass of 3.7 moles of NaCl?a) 15.8 gb) 58.5 gc) 216.5 gd) 6.02 x 1023 g

gmolgmol 5.216NaCl 5.58

1NaCl 7.3

What is the mass of one mole of water (H2O)?

a) 3.0 gb) 16.0 gc) 17.0 gd) 18.0 g

Atomic mass H = 1.0 g/molAtomic mass O = 16.0 g/molMass of one mole of H2O = 2(1.0) + 16.0 = 18.0 g

How many moles are in 152 g of water?a) 0.118 molb) 8.44 molc) 2736 mold) 6.02 x 1023 mol

molgmolg 44.8

OH 0.181

1OH 152

2

2

Mole Volume Calculations1 mole = 22.4 L of gas at STP

STP = standard temperature and pressure (0 °C & 1 atm)

Volumemol 1

L 22.4

L 22.4mol 1

Volume Example 1:Volume Example 1: Determine the volume, in liters, of 0.600 mol of SO2 gas at STP.

mol 1L4.22(g)SO mol600.0 2

2SO L4.13

Volume Example 2:

Volume Example 2: Determine the number of moles in 33.6 L of He gas at STP.

L 22.4mol 1He(g) 3.6L3

He mol 50.1

Density

Density = Mass / Volume

When given the density of an unknown gas, one can multiply by the molar volume to find the MW. The MW can allow for identification of the gas from a list of possibilities.

Density Example (part A):

The density of an unknown gas at STP is 2.054 g/L. (a) What is the molar mass?

massmolar memolar voludensity

1molL4.22

L2.054g

g/mol 01.46

Density Example (part B):

The density of an unknown gas is 2.054 g/L. (b) Identify the gas as either nitrogen, fluorine, nitrogen dioxide, carbon dioxide, or ammonia.

MW = 46.0 g/mol (from part a)

Nitrogen = N2 = 2(14.0) = 28.0 g/mol

Fluorine = F2 = 2(19.0) = 38.0 g/mol

Nitrogen dioxide = NO2 = 14.0 + 2(16.0) = 46.0 g/mol

Carbon dioxide = CO2 = 12.0 + 2(16.0) = 44.0 g/mol

Ammonia = NH3 = 14.0 + 3(1.0) = 17.0 g/mol

Mixed Mole Conversions

1 mole = 6.02 x 1023 RP’s = MW = 22.4 L of gas @STP

All Roads Lead to the Mole.

Always convert to units of moles first when converting between grams, liters, and representative particles.

Mixed Mole Example 1:

How many carbon atoms are in a 50.0-carat diamond that is pure carbon? Fifty carats is the same as 10.0 g.

C mol 1atoms1002.6

C 12.0gmol 1g0.10 23

C atoms 1002.5 23

Mixed Mole Example 2:

How many atoms are in 22.0 g of water?

molecule 1atoms 3

mol 1molecules1002.6

18.0gmol 1OH g0.22 23

2

atoms 1021.2 24