Using the Mole to Calculate % Composition, Empirical Formulas and Molecular Formulas

Illustrating Compound Composition

• So far, we have seen the composition of many chemical compounds in the form of chemical formulas

• Up until now, you believed a chemical formula was just the ratio of number of atoms in a compound

• Chemical formulas are also the ratio of MOLES of atoms in a compound!

• For example:

In 1 MOLE of CO2, there is 1 MOLE of C and 2 MOLES of O

Law of Definite Proportions • In chemistry, a chemical compound always contains

the same elements in certain definite proportions

• Defined by a scientist named Proust in 1799

• Called the Law of Definite Proportions

• Because the atoms of the elements in a compound are combined in constant proportion, they are also combined in a definite proportion by mass

• Examples:

• Sodium chloride is always 39.3% sodium and 60.7% chlorine by mass, no matter what its source

• Water is always 11.2% hydrogen and 88.8% oxygen by mass

More on the Law of Definite Proportions

• In a compound, the ratios by mass of the elements in that compound are fixed independent of the origins or preparation of that compound

• A compound is unique because of the specific arrangement and weights of the elements which make up the compound

• In other words, elements combine in specific ratios to form compounds

• That is, elements combine in whole numbers!

• It is not possible to have a compound with a portion of an atom

A drop of water, a glass of water, and a lake of water all contain hydrogen and oxygen in the same percent by mass.

Example – Atom View

Example – Mass View

How is the Composition of a Compound Measured?

• A chemist could measure composition of a compound in two different ways:

• Count numbers of constituent atoms

• Atoms are too small to see so this method is doesn’t work!

• Percentages (by mass) of its elements

• Molar masses are used because these are measureable!

• So to calculate percent composition of a covalent molecule or an ionic formula unit, you must assume you have one mole of the substance:

% Composition = Molar Mass of Element

Molar Mass of Molecule or Formula Unit× 100

Steps to Calculate % Composition 1. Write correct formula of compound with subscripts (if

necessary)

2. Calculate mass of each element in 1 mole of the compound

• In other words, get the molar mass of each element

3. Calculate the molar mass of compound

4. Find the fraction of the total mass contributed by each element and convert it to a percentage

Molar Mass of Element

Molar Mass of Molecule or Formula Unit× 100

5. Sum the individual mass percent values to make sure they add up to 100%!

Practice! • Carvone is a substance that occurs in two forms having different arrangements of the

atoms but the same molecular formula of C10H14O. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone.

• Step 1: Write correct formula of compound with subscripts C10H14O

• Step 2: Find the masses of each element in 1mole of carvone:

Mass of C in 1 mole = 10 mol ×12.01 g

1 mol= 120.1 g

Mass of H in 1 mole = 14 mol ×1.008 g

1 mol= 14.11 g

Mass of O in 1 mole = 1 mol ×16.00 g

1 mol= 16.00 g

• Step 3: Get molar mass of compound

Mass of 1 mol C10H14O = 120.1 + 14.11 + 16.00 = 150.2 g

Practice!

• Step 4: Find the fraction of the total mass contributed by each element and convert it to a percentage:

Mass percent of C = 120.1 g C

150.2 gC10H14O× 100% = 79.96%

Mass percent of H = 14.11 g H

150.2 gC10H14O× 100% = 9.394%

Mass percent of O = 16.00 g O

150.2 gC10H14O× 100% = 10.65%

• Step 5: Sum the individual mass percent values to make sure they add up to 100%!

Practice!

• What is the percent composition of potassium permanganate?

• Formula of potassium permanganate = KMnO4

• Molar mass of KMnO4:

39.1 + 54.9 + 4 ∙ 15.99 = 157.69 gmol

• % K = 39.1 g K

157.69 g KMnO4× 100 = 24.7%

• % Mn = 54.9 g Mn

157.69 g KMnO4× 100 = 34.8%

• % O = 63.96 g O

157.69 g KMnO4× 100 = 40.5%

Practice!

• What is the percent composition of sodium oxalate?

• What is the mass of bromine in 50.0 grams of potassium bromide? HINT – First, calculate the mass % of bromine then multiply by the mass of sample!

Introducing a New Type of Compound…The Hydrated Salt

• A hydrated salt is an ionic compound that has water molecules trapped within the crystal lattice

• Examples: CuSO4·5 H2O

Copper (II) sulfate pentahydrate

FeCl3·6 H2O Iron (III) chloride hexahydrate

• An anhydrous salt is an ionic compound

without water molecules • Examples:

CaCl2

CuCl2

• The molar mass of hydrated compounds are

larger due to the waters

Using % Composition to Calculate Water in a Hydrated Salt • When a hydrated crystal is heated to very high temperatures, the

crystal can lose water and become dehydrated (anhydrous)

• For example, the chemical reaction for the dehydration of magnesium sulfate is:

MgSO4 ∙ n H2O s → MgSO4 s + n H2O (g)

• The mass of water lost and the mass of the hydrated salt can be used to calculate the percent water in the hydrated salt

% H2O =Mass of H2O lost

Total mass of hydrate× 100

• Can use this information to calculate the empirical formula of the salt

• More on this later!

TIME FOR A DEHYDRATION LAB!

What is an Empirical Formula?

• As mentioned, percent composition allows you to calculate the empirical formula of a compound

• An empirical formula is the simplest whole number ratio of elements in a compound

• How is an empirical formula different than what we’ve seen before?

• Up until now, you’ve been working with molecular formulas!

• A molecular formula is the actual ratio of elements in a compound

Differentiating between Empirical and Molecular Formulas

• C2H4 and C3H6 are both molecular formulas • To find the empirical formulas, find the lowest whole number

ratio between atoms

• So, both C2H4 and C3H6 have empirical formulas of CH2

• Note - Empirical formulas and molecular formulas can be the same! • Example is H2O

• Example: • Molecular: C6H12O6

• Empirical: CH2O

• Example: • Molecular: CH4N

• Empirical: CH4N

Using % Composition to Determine Empirical Formula of a Compound

1. Pretend that you have a 100 gram sample of the compound • In other words, change the % to grams

2. Convert the grams to moles for each element • Use Mole Road Map

3. Write the number of each element as a subscript in a chemical formula • Keep each number as a decimal at this point!

4. Divide each subscript by the smallest number

5. Multiply the result by some integer to get rid of any fractions • May not be necessary

Practice! • Calculate the empirical formula of a compound composed of 38.67

% C, 16.22 % H, and 45.11 % N

• Step 1: Pretend that you have a 100 gram sample of the compound

• Step 2: Convert the grams to moles for each element

Example (continued)

• Step 3: Write the number of each element as a subscript in a chemical formula

• Step 4: If we divide all of these by the smallest subscript, it will give us the empirical formula

C3.22H16.09N3.22

CH5N

Practice!

• Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? What is the empirical molar mass?

Using an Empirical Formula to Determine the Molecular Formula

• Since the empirical formula is the lowest ratio, the actual molecule would have a bigger mass

• Molecular formula can always be obtained by multiplying the empirical formula by some whole number

• To do so, follow the steps below:

1. Calculate the empirical molar mass from the empirical formula

2. Divide the actual molecular molar mass (usually given in the problem) by empirical molar mass

• Gives a whole number

3. Multiply empirical formula by the whole number to get the molecular formula

Practice!

• Look back at the previous problem. Caffeine has a molar mass of 194 g/mol. What is its molecular formula?

• A compound is known to be composed of 71.65 % Cl, 24.27% C, and 4.07% H. Its molecular molar mass is known to be 98.96 g/mol. What is its empirical formula? What is its molecular formula?

A REAL-WORLD DETERMINATION OF EMPIRICAL FORMULA

Real-world determination of Empirical Formula • Combustion analysis is

one of the most common methods for determining empirical formulas

• A weighed compound is burned in oxygen and its products are analyzed by a gas chromatogram

• It is particularly useful for analysis of hydrocarbons! • Products are CO2 and H2O 26

More on Combustion Analysis • Combustion Analysis

• The technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion

X + O2 → CO2 + H2O

• Example Data

• 0.250 g of compound X produces 0.686 g CO2 and 0.562 g H2O

27

Determining empirical formulas from Combustion Analysis

X + O2 → CO2 + H2O

• Step 1: Find the mass of C & H that must have been present in X

• Multiply masses of products by percent composition (decimal form) of the

products

C: 0.686 g x (12.01 g/44.01 g) = 0.187 g C

H: 0.562 g x [(2 x 1.008 g)/18.02 g]= 0.063 g H

28

Combustion Analysis

X + O2 → CO2 + H2O

• Step 2: Add masses of products and compare to mass of original compound

0.187 g C + 0.063 g H = 0.250 g total

• Based on original compound mass of 2.90 g, compound X must contain only C and H!!

• Step 3: Find the number of moles of C and H

C: 0.187 g x mole/12.01 g = 0.0156 moles C

H: 0.063 g x mole/1.008 g = 0.063 moles H

29

Combustion Analysis

X + O2 → CO2 + H2O

• Step 4: Write the formula using the mole numbers as subscripts

C.0156H.063

• Step 5: Divide by smallest number of moles

C: 0.0156/0.0156 = 1

H: 0.063/0.0156 = 4

• If these numbers are fractions, multiply each by the same whole

number

30

Combustion Analysis

X + O2 → CO2 + H2O

• Step 6: Rewrite formula with new mole whole numbers

• You now have the empirical formula!

CH4

31