Chemical Formula Worksheet With Answers Page 5
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1
2
3
4
5
6
7
8
95
60.00
4.48
35.53
= 4.996 mol C
= 4.44 mol H
= 2.220 mol O
12.01
1.008
16.00
C
H
O
Pseudoformula:
4.996
4.44
2.220
C
H
O
4.996
4.44
2.220
2.220
2.220
2.220
= C
H
O
x 4 (make whole numbers)
C
H
O
2.25
2
1
9
8
4
As C
H
O
= n = 1 = 180.17, then there is no need to multiply so this is the molecular
9
8
4
formula.
PRACTICE EXAMPLE TWO
Find the empirical formulae of the following compounds which contained:
(a) 5.85 g K; 2.10 g N; 4.80 g O
(b) 3.22 g Na; 4.48 g S; 3.36 g O
(c) 22.0% C; 4.6% H; 73.4% Br (by mass)
(a)_KNO
________________________________________________
2
(b)_ Na
S
O
_____________________________________________________
2
2
3
(c) __C
H
Br_________________________________________________
2
5
a) 5.85/39.1 = 0.15 mole K; 2.1 / 14.01 = 0.15 mole N; 4.8/16 = 0.3 K
N
O
0.15
0.15
0.3
= KNO
2
b) 3.22 / 22.99 = 0.14 mole Na; 4.48 / 32.07 = 0.16 mole S; 3.16 / 16 = 0.21 =
Na
S
O
= Na
S
O
0.14
0.16
0.21
2
2
3
Assume 100g so 22/12.01 = 1.83 mole C; 4.6 / 1.01 = 4.55 mole H; 73.4 / 79.9 =
1.09 mole Br
C
H
Br
= C
H
Br
1.83
4.55
1.09
2
5
______________________________________________________
Benzopyrene has a molar mass of 252 g and an empirical formula of C
H
. What
5
3
is its molecular formula? (C = 12.01, H=1.01) .
5 x 12.01 3 x 1.01 = 63.08__252/63.08 = = 4 so molecular formula = 4 x (C
H
) =
5
3
C
H
___________________________________________________________
20
12
______________________________________________
[ans = C
H
]
20
12
A compound containing 40% carbon, 6.73% hydrogen and 52.28% oxygen was shown to
-1
have a molar mass of 60.06 gmol
by mass spectrometry. Use this information to work
out its molecular formula.
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