UNIT 5: THE MOLE

KEY TERMS

• Avogadro’s number - The number of atoms or molecules in one mole of a substance, equal to 6.023 × 1023

• Dimensional analysis - The practice of checking relations among physical quantities by identifying their dimensions

• Empirical formula - A formula giving the proportions of the elements present in a compound

• Formula unit - It is the lowest whole number ratio of ions represented in an ionic compound

• Molar mass - The number of grams of a substance in 1 mole

• Mole - The basic unit of substance. It is the amount of substance that contains as many carbon atoms as there are atoms in 0.012 kg of carbon-12

• Molecular formula - A chemical formula that shows the number of atoms of each element in a molecule of a compound

• Molecular mass - The combined mass of all the elements in a compound

• Percent composition - Expresses the mass ratio between different elements in a compound

• Stoichiometry - The relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers

MOLAR MASS•  The mass in grams of one mole of a pure substance

• Examples:

• The molar mass of 1 mole of carbon is 12.011 grams

• There are 12.011 grams of carbon in 1 mole

• The molar mass of 1 mole of sodium chloride (NaCl) is the mass of one

mole of sodium + mass of one mole of chlorine

• 22.990 + 35.453 = 58.443 grams

• There are 58.443 grams of NaCl in 1 mole of NaCl

EXAMPLE

• What is the molar mass of (NH4)2SO4?

• There are 2 nitrogen atoms → (2)(14.007) = 28.014 g

• There are 8 hydrogen atoms → (8)(1.008) = 8.064 g

• There is 1 sulfur atom → (1)(32.066) = 32.066 g

• There are 4 oxygen atoms → (4)(15.999) = 63.996 g

132.140

g

• Molar Mass of (NH4)2SO4 = 132.140 g/mol

FORMULA MASS

• The sum of the average atomic mass units (amu) of all the atoms

represented in the formula of any element or compound.

• Same as the molar mass except the units are in atomic mass units

(amu)

PERCENT COMPOSITION

• Percent composition is the percentage by mass of each element in a compound.

• Percent composition = Mass of element x 100% Total mass of compound

• What is the percent composition by mass of carbon in carbon dioxide (CO2)?

C = 12.011 x 1 atom = 12.011 g

O = 15.999 x 2 atoms = 31.998 g

Total Mass = 44.009 g

• Percent composition of carbon = 12.011 g x 100% 44.009 g

= 27.29 % carbon

• This means that there will be 72.71 % of oxygen by mass

EXAMPLE

• What is the percent composition of water?

Percent composition

H = 1.008 x 2 atoms = 2.016 g 11.19 %

O = 15.999 x 1 atoms = 15.999 g 88.81 %

Total Mass = 18.015 g

• The percentage for the entire compound will always add up to 100%

EXAMPLE• What is the percent composition of water in a hydrated compound such as MgSO4 • 7 H2O?

(7 H2O means there are 7 water molecules trapped inside the crystal lattice for every formula unit of the ionic compound)

Percent composition•Mg = 24.305 x 1 atom = 24.305 g 9.9 %•S = 32.066 x 1 atom = 32.066 g 13.0 %•O = 15.999 x 4 atoms = 63.996 g 26.0 %•Total Mass MgSO4 = 120.367 g 48.9 %

•7 x 18.015 g = 126.105 g 51.1 % 246.472 g

• The percentage for the entire compound will always add up to 100%

EXAMPLES

• 1. How many moles of sodium chloride is equivalent to 32.5 grams of sodium chloride?

32.5 grams NaCl x 1 mol NaCl/58.443 g NaCl = 0.556 mol NaCl

• 2. How much mass is represented in 0.425 moles of ammonium sulfate?

0.425 mol (NH4)2SO4 x 132.14 g (NH4)2SO4/1 mol (NH4)2SO4 = 56.160 g (NH4)2SO4

CALCULATING EMPIRICAL FORMULAS• The empirical formula is calculated from the percent composition of a compound. It is the reverse of the process

to find the percent composition

• Example:

• What is the empirical formula of a compound that is 25.9% nitrogen and 74.1 % oxygen?

• Empirical formula = N?O?

• Use following steps:

1. Select sample size of 100g - In 100.0 g of the compound, there are 25.9 g N and 74.1 g O.

2. Convert your mass to moles

25.9 g N x 1 mole N = 1.85 mol N

14.007 g N

74.1 g O x 1 mole O = 4.63 mol O

15.999 g O

EXAMPLE

3. The mole ratio of nitrogen to oxygen is N1.85O4.63

While this is the correct mole ratio, it is not the correct empirical formula because it is not the lowest whole-number ratio. The correct values can be obtained by dividing both molar quantities by the smaller number of moles. This will give a 1 for the element with the smaller number of moles.

Smaller number of moles = 1.85 mol N Therefore,

1.85 moles N 4.63 mol O

----------------- = 1 -------------- = 2.50

1.85 moles N 1.85 mol O

The mole ratio of nitrogen to oxygen is now N1O2.50

But, now this only has one element in a whole number. All of the subscripts have to be whole numbers.

EXAMPLE

•  Now, the mole ratio of nitrogen to oxygen is N1O2.50

How do you fix it?

If you multiply all subscripts by 2, you will get rid of the decimal.

• Your final answer should be N2O5

• Check your answer; the percent composition should be equal to the one given in the

problem

MOLECULAR AND EMPIRICAL FORMULAS

• Molecular Formula – The actual formula of a molecular structure

• Empirical Formula – The lowest whole number ratio of the elements in a molecular structure

• For Ionic Compounds, the Empirical Formula is the Molecular Formula as ionic compounds are always written in the lowest whole number ratio.

Example : Empirical Formula for Magnesium Chloride MgCl2. This is also the Molecular Formula for MgCl2.

• For Covalent Compounds, the Empirical Formula may or may not be the Molecular Formula based on the mass of the molecular (Molecular Mass) structure. The Molecular Formula will be a whole number multiple of the Empirical Formula.

CALCULATING MOLECULAR FORMULAS

• The Molecular Formula is the same as, or a multiple of the Empirical Formula, and is based

on the actual number of atoms of each type in a compound. For example, if the empirical

formula of a compound is C3H8, its molecular formula may be C3H8 , C6H16 , etc.

• You can determine the molecular formula of a compound if you know its empirical formula

and it’s Molecular Mass.

Molar mass of formula to be determined = multiple

Mass of Empirical Formula

• Then, multiply the empirical formula’s subscripts by the multiple and you will obtain the

molecular formula.

EXAMPLE

• Empirical Formula for Hydrogen Peroxide is HO. The molar mass of HO is 17.007 g. The

Molecular Mass for Hydrogen Peroxide is 34.014 g. Therefore the Molecular Formula is H2O2

(the molar mass of the Molecular Formula is twice the molar mass of the Empirical Formula of

Hydrogen Peroxide).

34.014 g

---------- = 2 H1x2O1x2 becomes H2O2

17.007 g

EXAMPLE

• Calculate the molecular formula of the compound whose molar mass is 60.10 g

and empirical formula is CH4N.

Molar Mass of empirical formula = 30.05 g of CH4N

Molar mass of molecular formula to be determined = 60.10 g

60.10 g

-------- = 2

30.05 g

• The molecular formula is C(1x2)H(4x2)N(1x2) which is C2H8N2

THE MOLE

• The amount of matter containing 6.022 x1023 representative particles

• Avogadro’s number

• The number of particles in exactly one mole of a pure substance (6.022 x 1023)

• There are 12 of anything in a dozen

• The mole is a counting system just like the dozen is

• There are 602,200,000,000,000,000,000,000 of anything in one mole!

• Examples:

• 1 mole of sodium atoms = 6.022 x 1023 atoms

• 1 mole of cars = 6.022 x 1023 cars

• 1 mole of molecules = 6.022 x 1023 molecules

MOLE

• The amount of substance that contains 6.022 x 1023 representative particles of that

substance

Form of Matter Representative Particle

Element Atom

Molecular Compound Molecules

Ionic Compound Formula Units

MOLE TO PARTICLE CONVERSION• Dimensional analysis process based on Avogadro’s number

• Since Avogadro’s Number represents the number of particles in one mole of a substance, the number of moles can be determined through dividing by Avogadro’s Number

# of Particles X 1 mole = # of moles Avogadro’s Number

• Since Avogadro’s Number represents the number of particles in one mole of a substance, the number of particles can be determined to moles through multiplying by Avogadro’s Number

# of Moles X Avogadro’s Number = # of particles 1 mole

EXAMPLES• 1. How many moles of copper is equivalent to 1.83 x1023 atoms?

1.83 x1023 atoms Cu x 1 mol Cu/6.022 x1023 atoms Cu = 0.304 mol Cu

• 2. How many molecules are in 1.28 moles of water?

1.28 mol H2O x 6.022 x1023 molecules H2O /1 mol H2O = 7.71 x 1023 molecules H2O