# THE MOLE - PowerPoint PPT Presentation

1 / 22

THE MOLE. How many particles are there really? 6.02 x 10 23 "Happy Mole Day to You" Chemistry Song. Mass of a given amount of substance (g). Amount of substance (mol). m. n =. M. Molar Mass (g mol -1 ). The Rule. Worked Example.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

THE MOLE

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

## THE MOLE

How many particles are there really?

6.02 x 1023

"Happy Mole Day to You" Chemistry Song

Mass of a given

amount of substance (g)

Amount of substance (mol)

m

n =

M

Molar Mass

(g mol -1)

### Worked Example

• Calculate the mass of 0.35 mol of magnesium nitrate (Mg(NO3)2.

• What is the formula we use?????

• How do we work out M

### Solution

• m = n x M

• (Mg(NO3)2 = 24.3 + 2(14.0 + 3 x 16.0)

= 24.3 + 2 x 62.0

= 148.3 g mol-1

So:

m(Mg(NO3)2= n(Mg(NO3)2 x M(Mg(NO3)2

= 0.35 mol x 148.3 g mol-1

= 52 g

Why only two significant figures?

### Counting by Weighing

• If we have a known mass of a substance can we possibly know the number of particles present?

• We can by combining both the equations we have learnt so far.

• m = n x M and N = n x NA

• How do we do this??

### Worked Example

• a) Calculate the amount (in mol) of CO2 molecules present in 22 g of carbon dioxide.

• b) What is the number of molecules present in this mass of CO2.

m

n =

M

m(CO2)

n(CO2) =

M(CO2)

22 g

n(CO2) =

44.0 g mol-1

### Solution

• Since

M(CO2) = 12.0 + 2 x 16.0

= 44.0 g mol-1

= 0.50 mol

### Solution Part b)

b) Since N = n x NA, where N is the number of specified particles:

N(CO2) = n(CO2) x NA

= 0.50 x 6.02 x 1023

= 3.0 x 1023

So, 22 g of carbon dioxide contains 3.0 x 1023 ??

Atoms, ions or molecules?

Molecules

### Percentage Composition

• The values for molar masses of elements in compounds can be used to calculate the percentage composition of a compound once its formula is known.

• This type of calculation is important in chemistry.

• If for example, a company is producing aluminium from alumina (Al2O3), the management will want to know the mass of aluminium that can be extracted given a quantity of alumina.

Mass of element in 1 mole of compound

% by mass

of the element

x 100

=

Mass of 1 mole of compound

### The Rule

• Percentage Composition

### Worked Example

• Calculate the percentage of aluminium in alumina (Al2O3).

• Solution:

Step 1. Find the molar mass of Al2O3

M(Al2O3) = (2 x 27) + (3 x 16)

= 102 g mol-1

Mass of Al in 1 mole of Al2O3

54

x 100

=

x 100

=

% of Al in Al2O3

102

Mass of 1 mole of Al2O3

### Solution

• Step 2:

Find the percentage of aluminium in the alumina.

Since 1 mol Al2O3 contains 2 mol Al atoms:

Mass of aluminium in 1 mol (102 g) of Al2O3 = 2 x 27 g

= 54 g

= 52.9%

The company management, therefore, knows that aluminium

comprises about 53% by mass of any sample of alumina.

### Empirical formulas

• The empirical formula of a compound is the formula that gives the simplest whole number ratio, by number of moles, of each element in the compound.

• You may have wondered how chemists have determined the formulas of compounds such as water, carbon dioxide, nitrogen dioxide and other compounds.

### Empirical Formulas

• Empirical formulas are determined experimentally, usually by determining the mass of each element present in a given mass of compound.

• To determine the empirical formula of a compound, therefore, an experimentally determined ratio by mass must be converted to a ratio by numbers of atoms.

• This is done by calculating the amount (in mol) of each element

### Empirical Formulas – the process

Step 1: Measure the mass (m) of each

element in the compound

Step 2: Calculate the amount in mole (n)

of each element in the compound

Step 3: Calculate the simplest whole number ratio of moles

of each element in the compound

Step 4: Determine the empirical formula of the compound

### Worked example

• A compound of carbon and oxygen is found to contain 27.3% carbon and 72.7% oxygen by mass. Calculate the empirical formula of the compound.

27.3

2.27

72.7

4.54

= 2.27

= 1

= 4.54

= 2

12.0

2.27

16.0

2.27

### Solution

27.3

72.7

1

2

The empirical formula of this compound is therefore CO2

### Molecular Formula

• While an empirical formula gives the simplest whole number ratio of each element in a compound, a molecular formula gives the actual number of atoms in one molecule of the compound.

• Note that the empirical formula for a compound can be the same or different from its molecular formula

### Molecular Formula

The molecular formula is always a whole number multiple of the

empirical formula. A molecular formula can be obtained from the

empirical formula if the molar mass of a compound is known.

molar mass of compound

78 g mol-1

=

=

molar mass of one unit

13 g mol-1

### Worked example

• A compound has the empirical formula CH. The molar mass of this compound is 78 g mol-1. What is the molecular formula of this compound?

• Solution

The molecule must contain a whole number of (CH) units.

The molar mass of a (CH) unit is 13 g mol-1.

If the compound has a molar mass of 78 g mol-1, then:

The number of CH units in a molecule

= 6

The molecular formula of the compound is, therefore, 6 x CH

Ie C6H6

### End of Outcome 1

• http://www.schooltube.com/video/2370/The-Mole-Worksheet