Level Chemistry Transition Unit Worksheet With Answer Key - Healthcote School Page 12
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1711
Empirical and molecular formulae
Task 10:
Empirical formula is the simplest whole number ratio of elements. Divide the percentage or mass by the
Mr of each element in the compound, divide by the smallest number and simplify to give a whole number
ratio.
Element
A
Element
A
Element
A
r
r
r
aluminium
Al
27
hydrogen
H
1
phosphorus
P
31
bromine
Br
80
iodine
I
127
potassium
K
39
calcium
Ca
40
iron
Fe
56
silver
Ag
108
carbon
C
12
lead
Pb
207
sodium
Na
23
chlorine
Cl
35.5
magnesium
Mg
24
sulfur
S
32
copper
Cu
63.5
nitrogen
N
14
zinc
Zn
65
fluorine
F
19
oxygen
O
16
1 Copy and complete the table.
Empirical formula
M
Molecular formula
r
CH
42
C
H
2
3
6
CH
70
C
H
2
5
10
CH
56
C
H
2
4
8
C
H
44
C
H
3
8
3
8
HO
34
H
O
2
2
CH
78
C
H
6
6
2 Find the empirical formula of each of the following substances using the data about composition by
mass.
a H
5%
F
95%
HF
b Na
3.71 g
O
1.29 g
Na
O
2
c Pb
90.7%
O
9.3%
Pb
O
3
4
d C
60.0%
H
13.3%
O
26.7%
C
H
O
3
8
3 3.53 g of iron reacts with chlorine to form 10.24 g of iron chloride. Find the empirical formula for the
iron chloride.
FeCl
3
4 Analysis of a compound consisting of carbon, hydrogen and oxygen showed it to contain 0.273 g C,
0.046 g H, and 0.182 g O. It has a relative formula mass (M
) of 88.
r
a Calculate the empirical formula of the compound.
C
H
O
2
4
b Calculate the molecular formula of the compound.
C
H
O
4
8
2
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