Transcript Empirical Formula

Calculating Empirical Formula
Using percentage or mass to find
the Empirical Formula
Empirical Formula
• Empirical Formula is the lowest whole
number ratio of elements in a compound.
• Molecular Formula the actual ratio of
elements in a compound
• The two can be the same: H2O.
• Or Different:
– CH2 empirical formula
– C2H4 molecular formula
– C3H6 molecular formula
Calculating Empirical Formula
• Just find the lowest whole number ratio
– C6H12O6 becomes CH2O
– CH4N is the lowest ratio
• It is not just the ratio of atoms, it is also
the ratio of moles of atoms
– In 1 molecule of CO2 there is 1 atom of C and
2 atoms of O
– In 1 mole of CO2 there is 1 mole of carbon and
2 moles of oxygen
Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this
– C is determined from the mass of CO2 produced
– H is determined from the mass of H2O produced
– O is determined by difference after the C and H have been
determined
Calculating Empirical Formula
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If you have mass, skip the next 2 steps
If you have %, pretend that you have a 100
gram sample of the compound.
That is, change the % to grams.
Convert the grams to moles for each element.
Write the number of moles as subscripts in a
chemical formula.
Divide each number by the least number.
Multiply the results to get rid of any fractions.
Example
• Calculate the empirical formula of a
compound composed of 38.7 % C, 16.2 %
H, and 45.1 %N.
• Assume 100 g so:
• 38.7 g C x 1mol C
= 3.22 mole C
12.0 gC
• 16.2 g H x 1mol H
= 16.2 mole H
1.0 gH
• 45.1 g N x 1mol N = 3.22 mole N
14.0 gN
Determining Mole Ratio
• 3.22 mole C
• 16.2 mole H
• 3.22 mole N
•C3.22H16.2N3.22
If we divide all of these by the smallest
one It will give us the empirical formula
Divide by smallest to get
Empirical Formula
• The ratio is 3.22 mol C = 1 mol C
3.22 molN
1 mol N
• The ratio is 16.2 mol H = 5 mol H
3.22 molN = 1 mol N
• C1H5N1 is the Empirical Formula
A compound is 43.64 % P and
56.36 % O. What is the
empirical formula?
• 43.6 g P x 1mol P
31.0 gP
• 56.4 g O x 1mol O
16 gO
P1.4O3.5
= 1.4 mole P
= 3.5 mole O
Divide both by the lowest one
P1.4O3.5
• The ratio is 3.5 mol O = 2.5 mol O
1.4 mol P
1 mol P
P1O2.5
• Multiply the result to get rid of any fractions.
2X
P1O2.5
= P2O5
• Caffeine is 49.5% C, 5.15% H, 28.9% N
and 16.5% O. What is its empirical
formula?
• 49.5 g C
1mol
12 g
• 5.15 g H 1mol
1g
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= 4.1 mol
= 5.15 mol
1mol
28.9 g N
14 g
= 2.06 mol
1mol
16 g
= 1.03 mol
• 16.5 g O
We divide by
lowest (1mol O)
and ratio
doesn’t change
Since they are
close to whole
numbers we will
use this formula
C4.12H5.15N2.06O1
OR C4H5N2O1
empirical mass = 97g
Empirical to molecular
• Since the empirical formula is the lowest
ratio, the actual molecule could weigh
more (by whole number multiples
• Divide the actual molar mass by the mass
of one mole of the empirical formula.
• Caffeine has a molar mass of 194 g. what
is its molecular formula?
•
molar mass
Find x if x 
empirical formula mass
194 g
97 g
2X
C4H5N2O1
C8H10N4O2.
=2
Example
• A compound is known to be composed of
71.65 % Cl, 24.27% C and 4.07% H. Its
molar mass is known (from gas density) is
known to be 98.96 g. What is its molecular
formula?
Example- Data to moles
1mol
• 71.65 Cl
35.5 g
24.27C 1mol
= 2.0mol
= 2.0mol
12 g
4.07 H. 1mol
1g
= 4.0mol
Moles to Empirical Formula
• C2H4Cl2
We divide by
lowest (2mol )
•C1H2Cl1 Lowest ratio = Empirical Formula
would give an empirical wt of 48.5g/mol
Its molar mass is known (from gas density)
is known to be 98.96 g. What is its molecular
formula?
•Empirical to Molecular Formula
would give an empirical wt of 48.5g/mol
Its molar mass is known (from gas density)
is known to be 98.96 g. What is its molecular
formula?
molar mass
x
empirical formula mass =
98.96 g
=2
48.5 g
Molecular Formula
2 X C1H2Cl1
= C2H4Cl2