Transcript 1.1.5 Types of Formula

1.1.5 Types of Formula
You will be able to…
• Explain the terms empirical formula and molecular
formula
• Calculate empirical and molecular formula
Empirical Formula
• The empirical formula of a compound is the
simplest formula which represents its
composition
• It shows the elements present and the ratio
of the amounts of the elements present
• To find the empirical formula you need to
find the ratio of the amounts of the elements
present
Worked example 1
Analysis showed that 0.6075 g of magnesium combines with
3.995 g of bromine to form a compound.
[Ar: Mg, 24.3; Br, 79.9.]
Find the molar ratio of atoms:
Mg : Br
0.6075 : 3.995
24.3 : 79.9
0.025 : 0.050
Divide by smallest number (0.025):
Empirical formula is:
1:2
MgBr2
Worked Example 2
Analysis of a compound showed the following percentage composition by mass:
Na: 74.19%; O: 25.81%. [Ar: Na, 23.0; O,16.0.]
100.0 g of the compound contains 74.19 g of Na and 25.81 g of O.
Find the molar ratio of atoms:
Na : O
74.19 : 25.81
23.0 : 16.0
3.226 : 1.613
Divide by smallest number (1.613):
2:1
Empirical formula is:
Na2O
Questions
1.
An 18.000g sample of magnesium oxide was
found by reduction to contain 10.800g of
magnesium. Calculate the empirical formula
2.
A 9.435g sample of a chloride was found to
contain 3.400g of calcium. Calculate the empirical
formula
3.
A sample of a compound contains 91.87g Fe,
79.11g S and 157.90g O. Determine the empirical
formula
Molecular Formula
• The molecular formula is a simple multiple of the empirical
formula e.g
– If the empirical formula is CH2O then the molecular formula
could be
CH2O, C2H4O2, C3H6O3 …….. CnH2nOn
n can be calculated by knowing the empirical formula mass
and the molar mass
n=
Molar mass
Empirical Mass
Worked Example
A compound has an empirical formula of CH2 and a relative
molecular mass, Mr, of 56.0. What is its molecular formula?
Answer
• empirical formula mass of CH2:
14.0
= 12.0 + (1.0 × 2) =
• number of CH2 units in a molecule:
=
56.0
14.0
• molecular formula: (4 × CH2)
=
C 4H8
=4
Questions
1. Benzene has the empirical formula
CH and a molar mass of 78. What is
the molecular formula?
2. Which of the following is an empirical
formula and which a molecular
formula?
C 3 H7
C 3 H6
C4H10
Calculation of percentage
composistion
From the formula of a compound and the
relative atomic masses, the percentage of
each element in the compound can be
determined
This is formally termed the percentage
composition by mass
Example - JFe
• Calculate the percentage composition of nitroglycerine
C3H5O9N3
C=12.01
H=1.01
O= 16.00
N=14.01
1.1.6 Moles and Gas Volumes
You will be able to…
• Calculate the amount of substance, in moles, using
gas volumes
Avogadro’s Law
Equal volumes of gases at the same
temperature and pressure contain equal
numbers of molecules
i.e. 100cm3 of hydrogen at some
temperature and pressure contains exactly
the same number of molecules as 100cm3 of
carbon dioxide
• One mole of different gases
– Note the same volume of each gas has a
different mass
Example
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
•The equation says you need twice as many molecules of
oxygen as you do of methane. According to Avogadro’s
Law, this means you will need twice the volume of oxygen
as of methane.
•So, if you have to burn 1 litre of methane, you will need 2
litres of oxygen.
•You should also produce 1 litre of carbon dioxide
•Because water is a liquid, we don’t know how much of that
we get
Units of volume
• Note:
1 litre = 1dm3 = 1000cm3
n = V/24(dm3)
Problems
•
Take the molar volume to be 24.0dm3 at
rtp.
a) Calculate the mass of 200cm3 of chlorine
gas (Cl2) at rtp
b) Calculate the volume occupied by 0.16g
of oxygen (O2) at rtp.