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amount of substance (g)

Amount of substance (mol)

m

n =

M

Molar Mass

(g mol -1)

The RuleWorked 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.

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

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.

2.27

72.7

4.54

= 2.27

= 1

= 4.54

= 2

12.0

2.27

16.0

2.27

Solution27.3

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.

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

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