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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.

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the mole

THE MOLE

How many particles are there really?

6.02 x 1023

"Happy Mole Day to You" Chemistry Song

the rule

Mass of a given

amount of substance (g)

Amount of substance (mol)

m

n =

M

Molar Mass

(g mol -1)

The Rule
worked example
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
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
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 example1
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.
solution1

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
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
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.
the rule1

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 example2
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

solution2

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
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 formulas2
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
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 example3
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.
solution3

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
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 formula1
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.

worked example4

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
End of Outcome 1
  • http://www.schooltube.com/video/2370/The-Mole-Worksheet
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