Welcome to the MOLE

What is a mole?

This is not such a bad

mole, but not what we

need to discuss. . .

This little stinker is just plain mean and ugly. . .

and the Mole People are in the dark and clueless. . . . .

We are discussing the Mole used in Chemistry:

Avogadro’s Number (NA)

or 6.02214179 x 10 23

shortened to

6.022 x 1023

This amount = 1 mole

The mole is a way to describe the number of something without writing a huge number.

• It is similar to common terms like dozen, gross or even π in geometry

• 1 mole of anything – atoms, molecules, cockroaches or even galaxies will number 6.022 x 1023

• We need a number like this since atoms and molecules are extremely small and so many take up such a small space

Key Equations – KNOW THESE!

Divide by Molar Mass x by NA

Grams Number

(mass) of Moles of of

Substance Substance Particles

x by Molar Mass Divide by NA

Molar Mass = Atomic Weight in Grams per Mole

(g/mol)

• Molar Mass is the mass (g) of 1 mole of an element– For example – Na is 22.989 amu which is

22.989 g, and 1 mole of Na = 22.989 g

– CO2 is 1 Carbon at 12.011 g and 2 Oxygens at 15.999 each; therefore the molar mass of CO2 = Σ of 1 C + 2 O ≈ 44 g

• Mass (m) = # mols x # g

• # mols = m / g

• # Mols = m / Molar Mass

• % Composition by Mass = m element x 100

m compound

• # Mols = Concentration (Molar) x Volume (L)

• Volume = # mols / Concentration

Here’s a trick – to find the needed equation, just cover up the wanted result and what is left is the equation!

Mass Mols

Atomic X Mols Concentration X Volume Mass

The Mole Concept• Atomic Mass (or atomic weight) is the

mass of the element in amu (μ)– It is the number under the elemental symbol– Simply make this number into grams (g)– This represents the mass (m) of 1 mole of that

element; and/or the m of 6.022 x 1023 atoms of that substance!

– 1 mole of any gas = 22.4 L

• Mole (mol) – is the # of atoms, ions, molecules that is equal to NA

• Molar Mass (M) – this is the m in g of 1 mol of a substance (g/mol)

– Example – Manganese = 54.94 μ, thus its

M = 54.94 g/mol;

and 54.94 g of Mn will contain 6.022 x 1023 atoms

and this is equal to 1 mol of Mn

Mass to Mole Calculations

• Remember – each element has a differentamu and thus, 1 mol of

each will differ in mass

The Mass of a Mole– Uses the C 12 isotope as its standard– H = μ of 1; or 1/12 of 1 atom of C 12– He has μ of 4 or 4/12 (1/3) of a C 12 atom– Remember – atomic masses use isotopes and their %

abundance in nature to calculate – and the closer to a whole number – the fewer the isotopes

I. Molar Mass of Substance = Grams Substance

1 mol Substance

Therefore:

1. Mols of A = grams of A given x 1 mol A

gram A

2. Mass of B = Mols of B given x gram B

1 mol B

Examples:

• 3 mol Mn = ? Grams

3 mol Mn X 54.9 g Mn = 165 g Mn

1 mol Mn

• 25 g Au = ? Mols

25 g Au x 1 mol Au = 25 = 0.127 mol Au

196.97 g Au 196.97

• 0.127 mols Au = ? Atoms

0.127 mol Au x 6.022 x 1023 = 7.65 x 1022 atoms Au

1 mol Au

II. Moles to Mass

• Mols (given) x # grams = mass 1 molExample:

– 2.5 mol of (C3H5)2S has what mass?M = 1 mol S = 32.07 g

6 mol C = 6 x 12.01 = 72 g 10 mol H = 10 x 1 = 10 g Σ 114.07 g/mol2.5 mol x 114.07 g = ≈ 286 g 1 mol

We just used the bottom left of the diagram!

Divide by Molar Mass x by NA

Grams Number

(mass) of Moles of of

Substance Substance Particles

x by Molar Mass Divide by NA

Molar Mass = Atomic Weight in Grams per Mole

(g/mol)

III. Mass to Moles with Compounds

• Example:m of Ca(OH)2 = 325 g (rounded off)

M = ?

# mols = ?

M = 1 mol Ca = 40.08 g

2 mol O = 2 x 16 = 32 g

2 mol H = 2 x 1 = 2 g

Σ 74.096 g/mol

Given m of 325 g Ca(OH)2 x 1 mol Ca(OH)2 = 4.3 mol

M of 74.096 g

IV. Mass (g) to Particles

• Mols x NA = # Particles

• Example:– m = 35.6 g of AlCl3

What is the number of Al+3 and Cl- ions?

M = Al 26.981 g/mol Cl 35.452 g/mol x 3 = 106.356

Σ 133.337

Mols Al = m given = _____ mols x NA = _____ Al ions 26.981 g/mol

Mols Cl = m given = _____ mols x NA = _____ Cl ions 106.356 g/mol

Continued

• And: 35.6 g AlCl3 = 0.267 mol

133.337 g/mol AlCl3

• 0.267 mol AlCl3 x (6.022 x 1023) =

1.6 x 10 23 molecules

V. Percent Composition of Compounds

• Mass Element (m) x 100 = % by mass

Mass Cmpd (M)

Example:H2O; what percent is H and what percent is O?

% H = 2 x 1 (the molar mass of H) = 2 x 100 = 11.2%

(the molar mass of H2O) 18

Thus, all compounds equal 100%, so 100 – 11.2 = 88.8 % for O

• What is the % of C and O in CO2?

g C x 100 = 12.01 C x 100 = 27.29%

total g CO2 44.01 g CO2

32 g O x 100 = 72.71 %

44.01 g CO2

Example:

• H3PO4 (aq) (Phosphoric Acid)

% H = 3 g H x 100 =3%

M = 98 g

% P = 31 g P x 100 = 32%

98 g

% O = 64 g O x 100 = 65%

98 g

m Compound = H (3 x 1) + P (1 x 31) + O (4 x 16) = 98 g

VI. Mole Ratios• Given Vitamin C (ascorbic acid) with the following

percentages, determine formula:40.92 % C 4.58 % H 54.5 % O

Set up with unknown moles (n):

nC = 40.92 g C / 12.01 g C = 3.4 mol C

nH = 4.58 g H / 1.00 g H = 4.5 mol H

nO = 54.5 g O / 16 g O = 3.4 mol O This is the Mole Ratio

• Set Mole Ratio values as subscripts• Divide each by the lowest value:

C 3.4 / 3.4 H 4.5 / 3.4 O 3.4 / 3.4 =

C 1 H 1.33 O 1

• The 1.33 on H needs to be Δ’d into integer– Do this by multiplying until closest to a whole

number• 1.33 x 2 = 2.66• 1.33 x 3 = 3.99 which can be rounded off to 4

– So the “magic” number is 3 – must multiply all subscripts by 3

• Result is C1 x 3H1.33 x 3O1 x 3 or C3H4O3!

What is the molecular formula that has 92.2% C and 7.8%H. The molar mass is 52.1.

• First – assume a 100 g sample of the substance • The element’s percentages are assumed to be masses

(g)• Determine the moles of elements in compound:

92.2 g C x 1 mol C = 7.68 mol C 12.01 g C7.8 g H x 1 mol H = 7.72 mol H 1.01 g H

• Divide all mols by the lowest value7.68 C = 1 mol C 7.72 H = 1.01 mol H7.68 7.68

Continued

• The Empirical Formula is C1H1 (this is the basic form)

• To get the Molecular Formula:– Molar mass of the Empirical Formula is 12.01 +

1.01 = 13.02 g/mol– Molar mass of unknown is 52.1 g/mol

• So:Molar Mass Compound = Whole # to multiply

Molar Mass Emp. Formula subscripts to get

molecular formula

52.1 g/mol = 4

13.02 g/mol thus, (CH) x 4 = C4H4

Once again, a molecular formula calculation:Given 38.7% C, 9.7% H, and 51.6% O with a

molecular formula mass of 62.0 g. What is the true molecular formula?

First - find the empirical formula (this is CH3O)

Find the formula mass: C 1 x 12 = 12

H 3 x 1.01 = 3.03

O 1 x 16 = 16.0

Σ 31.0

• Divide the molecular mass by the empirical formula mass:

62 g (given) / 31 g (mass of emp. form.) = 2

• Multiply each subscript by (n) or 2 in this example. . .

Thus, the molecular formula:

(CH3O)(n) (CH3O)(2) C2H6O2

VII. Hydrates• Hydrates are

substances that include

H2O in their

formulas, but are

not wet!

– Hydration – adding H2O

– Dehydration – removing it

– Anhydrous – no H2O present

• Methane Hydrate is found on the ocean’s floor– The methane will burn – but the water in it

keeps the skin from burning!

• The methane molecule (CH4) is in a cage of water molecules

• There is 1 mole CH4 per 5.75 mols H2O

• It is found at depths of 300 meters or more

• There is an estimated 1,300 trillion cubic feet of methane hydrate in the oceans

• However – the problem is that methane is one of the major greenhouse gases which contributes to global warming – so more study on retrieval and use is needed

Example:• Barium Chloride Hydrate

– Mass = 5 g. How many H2O per molecule?

– BaCl2•____H2O

• The sample is heated and the result is 4.26 g anhydrous BaCl2

– The difference between the 5 g hydrate and the 4.26 anhydrate is .74 g H2O

So. . . . . . . . .

A common hydrate is. . .

MgSo4•7H2O

Magnesium Sulfate

Heptahydrate

4.26 g BaCl2 = 0.0205 mols BaCl2

M = 208.23 g/mol

0.74 g H2O = 0.041 mol H2

18.02 g/mol

# H2O = x = mols H2O = 0.041 = 2

mols cmpd 0.0205

Thus: BaCl2•2H2O or barium chloride dihydrate

Summary