CHEMICAL QUANTITIES All Roads Lead to the Mole Chemical Quantities Measuring doughnuts: o1 dozen =...

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CHEMICAL QUANTITIES All Roads Lead to the Mole

Transcript of CHEMICAL QUANTITIES All Roads Lead to the Mole Chemical Quantities Measuring doughnuts: o1 dozen =...

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 Particles

Examples: 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.106.02

mol 123

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

OH molecules1002.6

mol1OH 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 1

CO 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 1

atoms3

CO mol 1

CO 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 = 12.01 + 4(1.01) = 16.05 g/mol

MW Example 2:

Find the MW of Mg(OH)2.

=Mg + 2O + 2H=24.31 + 2(16.00) + 2(1.01)=58.33 g/mol

MW Example 3:

Find the MW of MgSO4•7H2O.

=Mg + S + 4O + 7(H2O)

=24.31 + 32.07 + 4(16.00) + 7(18.02)=246.52 g/mol

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

a) 22.99 gb) 35.45 gc) 58.44 gd) 100.00 g

Atomic mass Na = 22.99 g/mol

Atomic mass Cl = 35.45 g/mol

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(14.01) + 3(16.00) = 76.02 g/mol

1 mole = 76.02 g N2O3

Dinitrogen trioxide = N2O3

3232 ON g547

1.00mol

g02.76ON 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(55.85) + 3(16.00) = 159.70 g/mol

1 mole = 159.70 g Fe2O3

Iron (III) oxide = Fe2O3

3232 OFe mol577.0

159.70g

mol00.1OFe g2.92

What is the mass of 3.7 moles of NaCl?a) 16 gb) 58 gc) 220 gd) 6.0 x 1023 g

gmol

gmol2.216

NaCl 44.58

1

NaCl 7.3

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

a) 1.01 gb) 16.00 gc) 32.32 gd) 18.02 g

Atomic mass H = 1.01 g/mol

Atomic mass O = 16.00 g/mol

Mass of one mole of H2O = 2(1.01) + 16.00 = 18.02 g

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

molg

molg44.8

OH 0.18

1

1

OH 152

2

2

Mole Volume Calculations

1 mole = 22.4 L of gas at STP

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

Volumemol 1

L 22.4

L 22.4

mol 1

Volume Example 1:

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

mol 1

L4.22(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 22.4

mol 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

1mol

L4.22

L

2.054g 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 = 46.01 g/mol (from part a)

Nitrogen = N2 = 2(14.01) = 28.02 g/mol

Fluorine = F2 = 2(19.00) = 38.00 g/mol

Nitrogen dioxide= NO2= 14.01 + 2(16.00) = 46.01 g/mol

Carbon dioxide = CO2 = 12.01 + 2(16.0) = 44.01 g/mol

Ammonia = NH3 = 14.01 + 3(1.01) = 17.04 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 1

atoms1002.6

C 12.01g

mol 1g0.10 23

C atoms 1001.5 23

Mixed Mole Example 2:How many atoms are in 22.0 g of water?

molecule 1

atoms 3

mol 1

molecules1002.6

18.02g

mol 1OH g0.22 232

atoms 1020.2 24