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Chapter Three
Mass Relationships in Chemical
Reactions
Chapter Three / Mass relationships in Chemical Reactions
Percent Composition of a Compounds
• It possible to determine the empirical formula from the percentage of
elements in the compound.
1- change % to g
2- change g to mole (remember the triangle).
3- divided by the smallest number of moles.
4- if there was fraction after division change to integer subscripts ( multiply by
1 or 2 or 3 etc until you reach integer.
Chapter Three / Mass relationships in Chemical Reactions
Percent Composition of a Compounds
Example 1:
Determine the empirical formula of Vitamin C. it is compose of 40.92% of C,
4.58% of H, and 54.50% of O by mass?
1- we change from % to g
Mass (g)
40.92 g of C, 4.58 g of H , 54.50 g of O
Mole
Molar mass
2- change from g to mole using
(mol)
40.92
nc = 12.011 = 3.407 mol of C
4.58
nH =
1.008 = 4.54 mol of H
54.50
nO =
= 3.406 mol of O
16
Divided by the smallest number of mole which is 3.406
Chapter Three / Mass relationships in Chemical Reactions
Percent Composition of a Compounds
C:
3.407
3.406
≈1
H:
4.54
3.406
= 1.33
O:
3.406
3.406
=1
4- Because number of hydrogen is 1.33 then we start to multiply until we
reach integer this is trail and error procedure:
1 x 1.33 = 1.33
2 x 1.33 =2.66
3 x 1.33 = 3.99 ≈ 4
Then we multiply all element with the same number which is 3 in this
example.
C : 1x3 =3 , O: 1 x 3 = 3 , H =4
Thus the empirical formula is C3H4O3
Chapter Three / Mass relationships in Chemical Reactions
Percent Composition of a Compounds
• Example 2 :
Allicin is the compound responsible of characteristic smell of garlic. an
analysis of the compound gives the following percent composition by
mass: C: 44.4 %, H: 6.21%, S: 39.5%, O: 9.86%. Calculate its empirical
formula?
1- we change from % to g
Mass (g)
44.4 g of C, 6.21 g of H , 39.5 g of S, 9.86 g of O.
Mole
Molar mass
(mol)
2- change from g to mole using
nC =
44.4
12.01
= 3.70 mol of C
nH =
6.21
1.008
= 6.16 mol of H
39.5
9.86
nO =
=
1.23
mol
of
S
= 0.62 mol of O
32.07
16
Divided by the smallest number of mole which is 0.62
nS =
Chapter Three / Mass relationships in Chemical Reactions
Percent Composition of a Compounds
3.70
6.16
H:
≈6
≈ 10
0.62
0.62
1.32
0.62
O:
S:
≈
2
=1
0.62
0.62
Because all the numbers are integer then we do not need to do anything else
and we just write the formula as following :
The empirical formula is C6H10O S2
C: