Empirical + Molecular formula
formula 25/01/2015
Calculate empirical formula from
experimental data.
Quick recap. Molecular and
empirical formula.
Molecular formula is the actual ratio
of atoms (only applies to covalent
molecules)
Empirical formula is the simplest
ratio of atoms (always given for ionic
compounds)
Recap
But what does it mean ?
Copper oxide could be
CuO
or Cu2O
1. What is the Ar of Cu ?
2. What is the Ar of O ?
3. So if 64g of copper reacted with 16g of oxygen what would the empirical
formula of the compound be?
4. If 128g of copper reacted with 16g of oxygen what would the empirical
formula be ?
What if I get horrible numbers?
13.6g of copper combined with 1.7g of oxygen what is the
formula of the copper oxide ?
Copper oxide could be
CuO
atoms
Moles =Mass ÷ Ar
Ratio (divide through by lowest
denominator)
or Cu2O
Cu
O
What if I get percentages?
A sample of a hydrocarbon was found
Copper oxide could be
CuO
atoms
Moles =Mass ÷ Ar
Ratio (divide through by lowest
denominator)
or Cu2O
Cu
O
What is I get percentage data ?
e.g. A compound was found to consist of 85.7 % C and
14.3% hydrogen by mass.
a. What is its empirical formula ?
-just turn % directly into g. just pretend you were given 100g,
(e.g. Whatever % you are given use it as grams).
atoms
Moles =Mass ÷ Ar
Ratio (divide through by lowest
denominator)
C
H
What is I get percentage data ?
Sometimes you might get data in percentage composition.
-just pretend you were given 100g, (e.g. Whatever % you
are given use it as grams).
Example. a molecule contained 40% carbon, 6.7%
hydrogen and 53% oxygen, what is its empirical formula ?
atoms
Moles =Mass ÷ Ar
Ratio (divide through
by lowest denominator)
C
H
O
What if I get experimental data
for the question
1.
Don’t freak out
1. Copper oxide was strongly heated in a boiling tube. The
results are given below.
a. Calculate the empirical formula of the copper oxide.
Mass of empty tube
52.5g
Mass of tube and copper
103.7g
Mass of tube and copper oxide
after heating
110.1g
Find out the masses of each element.
103.7g- 52.5g = 51.2g
51.2g ÷ 64= 0.8
Cu=
O=
110.1g-103.7g- = 6.4g
6.4g ÷ 16= 0.4
atoms
Moles =Mass ÷ Ar
Ratio (divide through by lowest
denominator)
Cu
O
atoms
Moles =Mass
÷ RFM
Ratio (divide
through by
lowest
denominator)
Mass of empty tube
40.5g
Mass of tube and zinc oxide
81g
Mass of tube and copper after
73g
The other thing you might be asked is to go from empirical formula to
molecular formula.
e.g. A compound was found to have 85.7 % C and 14.3%
hydrogen by mass.
a. What is its empirical formula ? CH2
a. It has a relative formula mass of 56, what is its
molecular formula ?
How many times does the
empirical formula “fit into” the
molecular formula.
The other thing you might be asked is to go from empirical formula to
molecular formula.
How many times does the empirical formula “fit into” the
molecular formula.
Empirical formula= CH2O = (12+2+16)=30
Molecular formula= C6H12O6 = 180
CH2O
180 = 6.
30
The other thing you might be asked is to go from empirical formula to
molecular formula.
e.g. A compound was found to have 85.7 % C and 14.3%
hydrogen by mass.
a. What is its empirical formula ? CH2
a. It has a relative formula mass of 56, what is its
molecular formula ?
e.g. A compound was found to have 85.7 % C and 14.3% hydrogen by mass.
a. What is its empirical formula ? CH2
a.
It has a relative formula mass of 56, what is its molecular formula ?
CH2
C4H8
1. Work out empirical formula mass
=(12)+ (2x1) = 14
2. Divide molecular formula mass
By empirical formula mass
=56g ÷ 14g=4
2. Multiply each atom ratio by this number
4
Now try worksheet
Now try question on page 87,
89 and worksheet