Chemical Masses and
Formulas
L.O: To know how to work out the
empirical formula of compounds
Working out the formula from masses that have reacted
 We
found 5.5g of manganese reacted with 3.2g of
oxygen. What is the formula of the oxide of
manganese formed? (Atomic. Mass Mn=55: O=16)
Substance
Manganese oxide
1. Elements
Mn
2. Mass of each element
(g)
5.5
3. Mass / Atomic Mass
4. Ratio
5. Formula
O
3.2
3.2/16 =0.20
5.5/55 =0.10
1:2
MnO2
 We
found 3.2g of copper reacted with 0.8g of
oxygen. What is the formula of the oxide of copper
that was formed? (At. Mass Cu=64: O=16)
Substance
Copper oxide
1. Elements
Cu
2. Mass of each element
(g)
3.2
3. Mass / Atomic Mass
4. Ratio
5. Formula
O
0.8
0.8/16 =0.05
3.2/64 =0.05
1:1
CuO
Activity
 A chloride
of silicon was found to have the following %
composition by mass: Silicon 16.5%: Chlorine 83.5%
(Atomic. Mass Si=28: Cl=35.5)
Substance
1. Elements
2. Mass of each element
(g per 100g)
3. Mass / Atomic Mass
4. Ratio
Divide biggest by
smallest
5. Formula
Silicon Chloride
Si
Cl
83.5
16.5
16.5/28 =0.59
83.5/35.5 =2.35
Cl÷Si = (2.35 ÷ 0.59) = (3.98)
Ratio of Cl:Si =4:1
SiCl4
Activity
 Calculate
the formula of the compounds formed when the
following masses of elements react completely:
(Atomic. Mass Si=28: Cl=35.5)
Element 1
Element 2
Atomic Masses
Formula
FeCl3
Fe = 5.6g
Cl=106.5g
Fe=56 Cl=35.5
K = 0.78g
Br=1.6g
K=39: Br=80
KBr
P=1.55g
Cl=8.8g
P=31: Cl=35.5
PCl5
C=0.6g
H=0.2g
C=12: H=1
CH4
Mg=4.8g
O=3.2g
Mg=24: O=16
MgO