Welcome to science intervention!
Today we are revising how to calculate
empirical formula
DO NOW:
Work out the relative masses of these elements
and compounds.
The first one has been done for you
1)
2)
3)
4)
5)
6)
7)
8)
H2O ( 1 + 1 + 16= 18)
Mg
CaCO3
H2SO4
CaO
0
SO2
02
Today you will need your
calculators
Today we are learning how to calculate
empirical formula
Answers
1) H2O ( 1 + 1 + 16= 18)
2)Mg = 24
3)CaCO3= 100
4)H2SO4= 98
5)CaO= 56
6)0= 16
7)SO2 = 64
8)02 = 32
Today we are learning how to calculate
empirical formula
The empirical formula is the simplest ratio
of the different atoms in it.
For example, ethane is has the formula
C2H6.
Therefore the empirical formula is CH3.
You can work out the empirical formula of a
compound if you know the mass of each
element in it.
Today we are learning how to calculate empirical formula
Lets look at an example question….
What is the empirical formula of a compound that
contains 27.3% carbon and 72.7% oxygen by mass?
Step 1: Write down the symbols for the elements
in the compound
C
Step 2: Underneath each element, write down the
masses/percentages in the question
27.3
72.7
27.3
12 = 2.3
72.7
16 =4.5
Step 3: Divide each mass by the relative formula
mass of the element
Step 4: Divide all the answers by the smallest
answer to give the ratio…
2.3 = 1
2.3
So the formula is CO2!
O
4.5= 2
2.3
Don’t forget
to write the
formula!
Today we are learning how to calculate
empirical formula
Past exam question
Compound X had the following percentage
composition by mass:
10.8% magnesium, 31.8% chlorine and 57.4%
oxygen.
Relative atomic masses: Mg = 24; Cl = 35.5; O =
16
Calculate the empirical formula of compound X.
Answer: MgCl2O8!
(Put the elements in the order
given in the question)
Hints:
1. Write the
elements!
2. Write the
masses/percentages
under each element
3. Divide the
number’s by the
atomic masses of
those elements
4. Divide by the
smaller of the two
numbers
Write the formula!
Today we are learning how to calculate
empirical formula
Past exam question
A compound is called phosgenite.
Analysis of this compound shows that it
contains:
76.0% lead (Pb) 13.0% chlorine (Cl) 2.2%
carbon (C) 8.8% oxygen (O)
Calculate the empirical formula of this
compound.
Relative atomic masses:
Pb = 207; Cl = 35.5; C = 12 ; O = 16
Answer: Pb2Cl2CO
Hints:
1. Write the
elements!
2. Write the
masses/percentages
under each element
3. Divide the
number’s by the
atomic masses of
those elements
4. Divide by the
smaller of the two
numbers
Write the formula!
Today we are learning how to calculate
empirical formula
Past exam question
A sample of the solvent used in one
perfume contained 0.60 g of carbon,
0.15 g of
hydrogen and 0.40 g of oxygen.
Relative atomic masses: H = l; C = 12; O
= 16.
Calculate the empirical (simplest)
formula of the solvent.
Answer: C2H6O!
Hints:
1. Write the
elements!
2. Write the
masses/percentages
under each element
3. Divide the
number’s by the
atomic masses of
those elements
4. Divide by the
smaller of the two
numbers
Write the formula!
Today we are learning how to calculate
empirical formula
Past exam question- Read the hint!
Calculate the simplest (empirical)
formula of this substance:
Hint: If the
70 % of iron (Fe) and 30 % of oxygen (O) ratio is .5 (eg
2.5), don’t
round up!
Double both
Relative atomic masses: O = 16; Fe = 56.
numbers
Eg 2.5: 3
Answer: Fe2O3!
becomes 5: 6
Today we are learning how to calculate
empirical formula
Lets look at a
trickier one…
Calculate the simplest (empirical)
formula of this substance:
70 % of iron (Fe) and 30 % of oxygen (O)
Relative atomic masses: O = 16; Fe = 56.
Answer: Fe2O3!