# Laws of Definite & Multiple Proportions ```Laws of
Definite
&amp; Multiple
Proportions
Law of Definite Proportions
A compound is always composed
of the same elements in the
same proportion by mass
Use percent mass to calculate
proportion of each element
% Mass
(aka % Composition)
Mass of element * 100 = % element
mass of compound
• must have as many percents as you have
elements
• %’s should total 100 (appx)
Example #1
A 20.00 g sample of sugar was
analyzed by mass. 8.44g was
carbon, 1.30g was hydrogen, and
the rest was oxygen. Determine
its percent composition.
Example #1 - A 20.00 g sample of sugar was analyzed by mass. 8.44g
was carbon, 1.30g was hydrogen, and the rest was oxygen. Determine its
percent composition.
% C = 8.44g
20.00g
% H = 1.30g
20.00g
*100 = 42.20%
*100
= 6.50%
20.00
Example #1 - A 20.00 g sample of sugar was analyzed by mass. 8.44g
was carbon, 1.30g was hydrogen, and the rest was oxygen. Determine its
8.44
percent composition.
- 1.30
10.26 g of O
% O = 10.26 g
20.00 g
*100 = 51.30%
Since the Law of Definite Proportions
states that a compound is always
composed of the same elements in the
same proportion by mass:
This sugar compound is always
42.20% carbon, 6.50 %
hydrogen, and 51.30% oxygen,
regardless of the sample size.
Law of Multiple Proportions
The masses of one element
which combine with a fixed mass
of the second element are in a
ratio of simple whole numbers.
This is the third postulate of
Dalton's atomic theory.
Use ratios of mass of element 1 to mass
of element 2; then compare the mass
ratios.
Example: Two different
compounds are formed by the
elements carbon and oxygen. The
first compound contains 42.9%
carbon and 57.1% oxygen. The
second compound contains
27.3% carbon and 72.7%
oxygen.
Show that this is consistent with
the Law of Multiple Proportions.
Since you are given %’s and you need
grams, ASSUME that you have 100 g
samples of each cmpd.
(100 is chosen to make calculations
easier with %’s)
Therefore, in 100 g of the first
compound, there are 57.1 g O and
42.9 g C.
The mass of O per gram C is:
57.1 g O
42.9 g C
= 1.33 g O
gC
In 100 g of the second compound,
there are 72.7 g O and 27.3 g C.
The mass of O per gram C is:
72.7 g O
27.3 g C
= 2.66 g O
gC
Compare the ratios of O to C in
cmpd 2 to cmpd 1 (hint: easier to
put larger value in numerator)
The mass of O per gram C is:
2.66 g O / g C
1.33 g O / g C
=
2
This simple whole-number ratio
means that the masses of oxygen
that combine with carbon are in a
2:1 ratio.
The whole-number ratio is
consistent with the Law of Multiple
Proportions.
```