#### Transcript How to Find an Empirical Formula

Mark S. Cracolice Edward I. Peters www.cengage.com/chemistry/cracolice

### Chapter 7 Chemical Formula Relationships

Mark S. Cracolice • The University of Montana

### Number of Atoms in a Formula

In writing the formula of a substance, subscript numbers are used to indicate the number of atoms or groups of atoms of each element in the formula unit.

### Number of Atoms in a Formula

How many atoms of each element are in a formula unit of ammonium carbonate?

The formula of the ammonium ion is NH 4 + The formula of carbonate is CO 3 2 – The formula of the ammonium carbonate is (NH 4 ) 2 CO 3 .

Each element in the parentheses is multiplied by two.

Total number of atoms of each element: 2 nitrogen atoms, 8 hydrogen atoms, 1 carbon atom, 3 oxygen atoms

### Atomic Mass

Atomic Mass

The average mass of atoms of an element By definition the mass of a carbon-12 atom is 12 u.

### Molecular & Formula Mass

Molecular Mass

The sum of the atomic masses of each atom in the molecule.

Formula mass

The sum of atomic masses in the formula unit

### Molecular & Formula Mass

What is the formula mass of calcium phosphate?

Calcium ion: Ca 2+ Phosphate ion: PO 4 3 – Calcium phosphate: Ca 3 (PO 4 ) 2 Ca 3 × 40.08 u = 120.24 u P O 2 × 8 × 30.97 u 16.00 u = 61.94 u = 128.00 u Ca 3 (PO 4 ) 2 310.18 u

### Defenition of Mole

The mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 grams of carbon-12

. In 12 g of carbon-12 there are 6.022 x 10 23 atoms 1 mole of any substance = 6.022 x 10 23 units of that substance.

When the mole is used the elementary entities must be specified: atoms, molecules, ions…

A

The number of elementary units in one mole 6.02214179  10 23 units/mol The Avogadro constant is a conversion factor between units and mole.

### Conversion of Mole to Molecules

How many carbon dioxide molecules are in 2.0 moles of carbon dioxide?

2.0 mol CO 2 × 6.02 ´ 10 23 molecules CO 2 mol CO 2 = 1.2 × 10 24 molecules CO 2

### Molar Mass

Molar mass of a substance is the mass in grams of one mole of the substance.

Units: g/mol Molar mass of an element is the mass of the element per mole of its atoms.

### Atomic mass unit and gram

The mass of one atom of carbon-12 is exactly 12 atomic mass units.

The mass of one mole of carbon-12 atoms (6.022 x 10 23 atoms of carbon-12) is exactly 12 grams (6.022 x 10 23 atoms) x (12 u/atom) = 12 grams

6.022 x 10 23 u = 1 gram

### Molar Mass

The mass of one atom of carbon-12 is exactly 12 atomic mass units.

The mass of one mole of carbon-12 atoms is exactly 12 grams.

This leads to the conclusion: The

molar mass

of any substance in grams per mole is

numerically equal

to the atomic, molecular or formula mass of that substance in atomic mass units.

### Molar Mass

Calculate the mass of one NH 3 molecule and the mass of one mole of NH 3 molecules.

14.01 u + 3x1.008 u = 17.03 u The mass of one ammonia molecule is

17.03 u.

To change from molecular mass to molar mass, change the units from u to g/mol:

17.03 g/mol.

One mole of ammonia molecules has a mass of

17.03 g

.

### Mass, # of Moles, # of Units

Molar mass, MM, links mass in grams with the number of moles.

Avogadro’s number, N A , links the number of moles with the number of particles.

### Mass, # of Moles, # of Units

How many molecules are in 454 g of water?

g  moles  molecules 1 mol 18.02

g H 2

O

454g x x 2 6.02

 10 23 molecules H 2

O

mol H 2

O

= 1.52 x 10 25 molecules H 2 O

### Mass↔Moles↔ Units

How many hydrogen atoms are in 1.0 kg of ammonia?

1.0 kg NH 3 × 1000 g NH 3 kg NH 3 × 1 mol NH 3 17.03 g NH 3 × 6.02 ´ 10 23 molecules NH 3 mol NH 3 × 3 atoms H molecule NH 3 = 1.1 × 10 26 atoms H

### Percentage Composition

Percentage

% A = parts of A total parts × 100% The

percentage composition

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

### Percentage Composition

Determine the percentage composition of calcium fluoride Ca F 2 .

Solution: In one mole of Ca F 2 1x(40.08 g Ca) + 2x(19.00 g F) = 78.08 g CaF 2 (Solution continued on the next slide)

### Percentage Composition

(40.08

g Ca) 78.08

g CaF 2  100 = 51.33 % Ca 2x(19.00g

F) 78.08

g CaF 2 

100 = 48.67% F

Check: 51.33% + 48.67% = 100.00%

### Empirical Formula

Empirical Formula The simplest ratio of atoms of the elements in a compound.

The empirical formula of C 2 H 4 is CH 2 .

Likewise, the empirical formula of C 3 H 6 is CH 2 .

All compounds with the general formula C n H 2n have the same empirical formula and therefore the same percentage composition.

### Empirical Formula

Write the empirical formulas of benzene, C 6 H 6 , and octane, C 8 H 18 .

Look for the simplest whole-number ratio of elements: For C 6 H 6 , the 6/6 ratio can be reduced to 1/1: CH.

For C 8 H 18 , the 8/18 ratio can be divided by 2 on top and bottom to be reduced to 4/9: C 4 H 9 .

### Find Empirical Formula

To find empirical formula, you need to find the ratio of atoms of the elements ratio of atoms = ratio of moles of atoms

How to Find an Empirical Formula

1.

2.

3.

4.

5.

Find the

masses

the compound.

of different elements in a sample of Convert the masses into

moles

of atoms.

Determine the

ratio of moles

of atoms.

Express the moles ratio as the smallest possible

ratio of integers

.

Write the empirical formula, using the number in the integer ratio as the

subscript in the formula

.

### Find Empirical Formula

What is the empirical formula of a compound that has 85.6% carbon, 14.4 % hydrogen?

Solution: It is usually helpful to organize the calculations in a table with the following headings:

Element Grams Moles Mole Ratio Formula Ratio Empirical Formula

### Find Empirical Formula

Element Grams Moles

C 85.6

Mole Ratio Formula Ratio

85.6

12.01

g C g/mol C 7.13

7.13

1

Empirical Formula

H 14.4

7.13

14.4

g H 1.008

g/mol H 14.3

1 14.3

7.13

2.01

2 CH 2

### Find Empirical Formula

What is the empirical formula of a compound that analyzes as 20.0% carbon, 2.2% hydrogen, and 77.80% chlorine?

Mole Formula Empirical Element Grams Moles Ratio Ratio Formula

C 20.0

1.67

1 3 H Cl 2.2

77.8

2.2

1.3

4 2.19 1.31 4 C 3 H 4 Cl 4

### Molecular Formula

The molecular formula of a compound can be found by determination of the number of empirical formula units in the molecule.

### molar mass of compound molar mass of empirical formula

How to Find the Molecular Formula

1.

Determine the empirical formula of the compound.

2.

Calculate the molar mass of the empirical formula unit.

3.

Divide the molar mass of the compound by the molar mass of the empirical formula unit to get n, the number of empirical formula units per molecule.

### Molecular Formula

What is the molecular formula of a compound with the empirical formula C 2 H 5 and a molar mass of 58.12 g/mol?

The molar mass of the empirical formula unit is 2(12.01 g/mol C) + 5(1.008 g/mol H) =29.06 g/mol The number of empirical formula units per molecule is 58.12

g/mol 29.06

g/mol = 2 (C 2 H 5 ) 2 = C 4 H 10

### Find Molecular Formula

An unknown compound is found to be 40.0% of carbon, 6.71% of hydrogen and the remainder is oxygen. The molar mass of the compound is 180.16 g/mol. Find the empirical and molecular formulas of the compound.

### First Find Empirical Formula

Element Grams Moles Mole Ratio Formula Ratio Empirical Formula

C H O 40.0

6.71

53.3

3.33

1 6.66

2 3.33 1 1 2 1

CH 2 O

Calculate Molar Mass/Empirical Formula Mass The molar mass of the empirical formula unit is (12.01 g/mol C) + 1(1.008 g/mol H) + (16.00 g/mol O) = 30.03 g/mol The number of empirical formula units per molecule is 180.16

g/mol = 6 30.03

g/mol (CH 2 O) 6 = C 6 H 12 O 6

### Molecular Formula

Another way to find molecular formula is to consider one mole of the compound (180.16 g of compound). The numbers of moles are also numbers of atoms in the molecule.

The masses of carbon, hydrogen and oxygen are : 40.0% x 180.16 g = 72.06 g of C 6.71% x 180.16 g = 12.09 g of H 53.3% x 180.16 g = 96.03 g of O

### Molecular Formula

Element Grams Moles Molecular Formula

C 72.06

6 H O 12.09

12 96.03 6

C 6 H 12 O 6

### Homework

7, 15, 21, 23, 43, 57, 61, 63, 65, 68, 74.