The Mole IMathematical Relationships

with Chemical Formulas

mole (mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.

This amount is 6.022x1023. The number is called Avogadro’s number and is abbreviated as N.

One mole (1 mol) contains 6.022x1023 entities (to four significant figures)

Water 18.0 gCaCO3

100.1 g

Oxygen 32.0 g

Copper 63.5 g

One mole of common substances.

molar mass (MM) = mass of one mole of a substance

• 1 mol Ca = 40.1 g = 6.02 x 1023 Ca atoms

• 1 mol Cu = 63.5 g = 6.02 x 1023 Cu atoms

• 1 mol Hg = 200.6 g = 6.02 x 1023 Hg atoms

Molar Mass• 1 mol Cl2 = 71.0 g

= 6.02 x 1023 Cl2 molecules = 2(6.02 x 1023) Cl atoms

• 1 mol (NH4)2SO4

= 6.02 x 1023 units of (NH4)2SO4 = 15(6.02 x 1023) atoms

= [2(14.0g) + 8(1.0g) + 1(32.1g) + 4(16.0g)] = 132.1 g

Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol)

Oxygen (O)

Mass/mole of compound

6 atoms

96.00 g

Carbon (C) Hydrogen (H)

Atoms/moleculeof compound

Moles of atoms/mole of compound

Atoms/mole ofcompound

Mass/moleculeof compound

6 atoms 12 atoms

6 moles of atoms

12 moles of atoms

6 moles of atoms

6(6.022 x 1023) atoms

12(6.022 x 1023) atoms

6(6.022 x 1023) atoms

6(12.01 amu) =72.06 amu

12(1.008 amu) =12.10

amu

6(16.00 amu)

=96.00 amu

72.06 g 12.10 g

Calculating molesA from gramsA

• How many moles of Ca are in 22 g of Ca?

1 mol Ca = 40.1 g so

40.1 g1 mol Ca or

40.1 g 1 mol Ca

1 mol Ca22 g Ca x 0.55 mol Ca

40.1 g Ca

Calculating gramsA from molesA

• How many g of (NH4)2SO4 are in 0.0335 mol of (NH4)2SO4 ?

1 mol (NH4)2SO4 = 132.1 g so

24 4

24 4

132.1 g1 mol (NH ) SO or

132.1 g 1 mol (NH ) SO

2 24 4 4 424 4

132.1 g0.0335 mol (NH ) SO x = 4.43 g (NH ) SO

1 mol (NH ) SO

Mass % of element X =

moles of X in formula x molar mass of X

molecular (or formula) mass of compound x 100

Mass percent from the chemical formula

Calculating the Mass Percents and Masses of Elements in a Sample of Compound

Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. What is the mass percent of each element in glucose?

Per mole glucose there are: 6 moles of C 12 moles H 6 moles O

Calculating the Mass Percents and Masses of Elements in a Sample of Compound

6 mol C x 12.01 g C

1 mol C = 72.06 g C 12 mol H

x

1.008 g H 1 mol H

= 12.096 g H

6 mol O x 16.00 g O 1 mol O

= 96.00 g O

M = 180.16 g/mol

mass percent of C =

72.06 g C180.16 g

glucose

x 100 = 39.99 mass %C

mass percent of H =

12.096 g H180.16 g

glucose

x 100 = 6.714 mass %H

mass percent of O =

96.00 g

180.16 g glucose

x 100 = 53.29 mass %O

Empirical and Molecular Formulas

Empirical Formula -

Molecular Formula -

The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms.

The formula of the compound as it exists, it may be a multiple of the empirical formula.

C1H2O1

C6H12O6

Determining the Empirical Formula from Masses of Elements

Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound?

2.82 g Na1 mol Na22.99 g

Na

= 0.123 mol Na

4.35 g Cl

1 mol Cl

35.45 g Cl= 0.123 mol Cl

7.83 g O

1 mol O

16.00 g O= 0.489 mol O

Na1 Cl1 O3.98 NaClO4

NaClO4 is sodium perchlorate.

Divide each mole value by thesmallest value to get the whole number subscripts.

Determining a Molecular Formula from Elemental Analysis and Molar Mass

During physical activity, lactic acid (MM=90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O.

(a) Determine the empirical formula of lactic acid.(b) Determine the molecular formula.

Determining a Molecular Formula from Elemental Analysis and Molar Mass

Assuming there are 100. g of lactic acid, the constituents are:

40.0 g C 6.71 g H

53.3 g O1 mol C

12.01g C

1 mol H

1.008 g H

1 mol O

16.00 g O

3.33 mol C

6.66 mol H 3.33 mol O

C 3.33 H6.66 O3.33

3.33 3.33 3.33CH2O empirical

formula

mass of CH2O

molar mass of lactic acid 90.08 g

30.03 g3 C3H6O3 is the

molecular formula

Divide each mole amount by the smallest value…

Divide the molar mass by the mass of the empirical formula

Calculating moleB <-> moleA

• How many moles of hydrogen are contained in 2.8 mol of (NH4)2SO4?

• 1 mol (NH4)2SO4 = 2 mol N, 8 mol H, 1 mol S, 4 mol O

24 424 4

8 mol H2.8 mol (NH ) SO x = 22 mol H

1 mol (NH ) SO

Calculating particles from moles

• Particles can be molecules, formula units (ionic compounds), atoms or ions.

• 1 mol particles = 6.02 x 1023 unitsNe atom

O2 molecule

NH3 molecule

NaCl formula unit

Calculating particles <-> moles

• How many atoms of oxygen are in 0.580 mol of iron (II) nitrate?

• How many moles of sulfur trioxide can be made from 4.0 x 1023 oxygen atoms?

2324

3 23 2

6 mol O 6.02 x 10 O atoms0.580 mol Fe(NO ) x x = 2.09 x 10 O atoms

1 mol Fe(NO ) 1 mol O

23 3323

1 mol SO1 mol O4.0 x 10 O atoms x X = 0.22 mol SO

6.02 x 10 O atoms 3 mol O

Calculating particles <-> g

• How many atoms of potassium are in 23.4 g of potassium carbonate?

• How many g of sulfur hexachloride can be made from 3.33 x 1024 chlorine atoms?

23232 3

2 32 3 2 3

1 mol K CO 2 mol K 6.02 x 10 atoms K23.4 g K CO x x x = 2.04 x 10 atoms K

138.2 g K CO 1 mol K CO 1 mol K

24 66623

6

245.1 g SCl 1 mol SCl1 mol Cl atoms3.33 x 10 Cl atoms x x x = 226 g SCl

6.02 x 10 Cl atoms 6 mol Cl 1 mol SCl

GramA <-> GramB with Formulas

• How many g of oxygen are contained in 458 g of C12H22O11?

12 22 1112 22 11

12 22 11 12 22 11

16.0 g O1 mol C H O 11 mol O458 g C H O x x x = 236 g O

342 g C H O 1 mol C H O 1 mol O

MASS(g)of substance A

MASS(g)of substance A

AMOUNT(mol)of substance A

AMOUNT(mol)of substance A

MOLECULES or ATOMS(or formula units)

of substance A

MOLECULES or ATOMS(or formula units)

of substance A

Avogadro’s number

(particles/mol)

MM (g/mol) of

compound Amolar ratio from

formula subscripts

Summary of the mass-mole-number relationships in a chemical formula.

MASS(g)of substance B

MASS(g)of substance B

AMOUNT(mol)of substance B

AMOUNT(mol)of substance B

MOLECULES or ATOMS(or formula units)

of substance B

MOLECULES or ATOMS(or formula units)

of substance B

Avogadro’s number

(particles/mol)

MM (g/mol) of

compound B

Modified from Silberberg, Principles of Chemistry

Questions to Ask When Solving a Problem

• Are grams mentioned?If yes you will need to use a molar mass

conversion.• Are you comparing two different

substances?If yes you will need a conversion using the

subscripts in a formula.• Are particles (atoms or molecules) involved?If yes you will need a conversion with

Avogadro’s number.You may need any number of these

conversions depending on the problem.

How many atoms of oxygen are required to form 6.8 grams of

aluminum nitrate?• Grams? Yes, need molar mass• Two substances? Yes, need subscripts• Particles? Yes, need Avogadro’s #• So at least 3 conversions are needed…

23233 3

3 33 3 3 3

1 mol Al(NO ) 9 mol O 6.02 x 10 O atoms6.8 g Al(NO ) x x x = 1.7 x 10 O atoms

213.0 g Al(NO ) 1 mol Al(NO ) 1 mol O

How many moles of nitrogen are contained in 45.0 grams of aluminum

nitrate?• Grams? Yes, need molar mass• Two substances? Yes, need subscripts• Particles? No• So, at least 2 conversions are needed…

3 33 3

3 3 3 3

1 mol Al(NO ) 3 mol N45.0 g Al(NO ) x x = 0.634 mol N

213.0 g Al(NO ) 1 mol Al(NO )

Molarity

• Because many reactions happen between substances in solution it is helpful to have a concentration based on moles.

moles of solute (dissolved substance)Molarity (M) =

1 liter of solution

http://fuelcell.com

Calculating the Molarity of a Solution

Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL?

Molarity is the number of moles of solute per liter of solution.

0.715 mol glycine

495 mL soln

1000mL

1 L= 1.44 M glycine

Summary of mass-mole-number-

volume relationships in solution.

MASS (g)of

compoundin solution

AMOUNT (mol)

of compoundin solution

MOLECULES(or formula

units)of compoundin solution

VOLUME (L)of solution

MM (g/mol)

M (mol/L)Avogadro’s number(molecules/mol)