Law of Definite Proportions
 Regardless of the amount, a
compound is always composed of the
same elements in the same proportion
by mass.
 The proportions are found by
calculating the percent by mass.
Percent by Mass
 Based on the law of conservation of
mass
MASS compd = sum of MASSES elements
 % by mass = MASSelement x 100%
MASScompd
Percent by Mass
 Example on page 75
 Sucrose = Carbon, hydrogen, oxygen
 To find percent by mass of each element:
C= (mass C / mass of sucrose) x 100%
H= (mass H / mass of sucrose) x 100%
O= (mass O / mass of sucrose) x 100%
Percent by Mass (20g of sucrose)
Element
Carbon
Hydrogen
Oxygen
Total
(Sucrose)
Analysis by Percent by mass
mass (g)
(%)
8.4
8.4 x 100 = 42%
20.0
1.3
1.3 x 100 = 6.5%
20.0
10.3
10.3 x 100 = 51.5%
20.0
20.0
100%
Percent by Mass (500g sucrose)
Element
Carbon
Hydrogen
Oxygen
Total
(Sucrose)
Analysis by Percent by mass
mass (g)
(%)
210.5
210.5 x 100 = 42%
500.0
32.4
32.4 x 100 = 6.5%
500.0
257.1
257.1 x 100 = 51.5%
500.0
500.0
100%
Law of Definite Proportions
 Therefore, mass percentages of
elements in a compound do NOT
depend on amount.
 Compounds with the same mass
proportions must be the same
compound
Practice Problems
Q: A 78.0 g sample of an unknown
compound contains 12.4g of
hydrogen. What is the percent by
mass of hydrogen in the compound?
A: % Mass H = mass H x 100%
mass comp
= 12.4g x 100%
78.0g
= 15.9%
Practice Problems
Q: If 3.5 g of X reacts with 10.5g of Y to
form the compound XY, what is the
percent by mass of X in the compound?
A: % Mass X = mass X
x 100%
mass XY
= 3.5g
(3.5 + 10.5)g
= 25%
x
100%
Practice Problems
Q: If 3.5 g of X reacts with 10.5g of Y to
form the compound XY, how many
grams of Y would react to form XY2?
A: Mass YXY = 2 ( Mass YXY )
2
= 2 ( 10.5g)
= 21.0 g
Practice Problems
Q: 2 unknown compounds are tested.
Compound 1 contains 15.0g of
hydrogen and 120.0g oxygen.
Compound 2 contains 2.0g of
hydrogen and 32.0g oxygen. Are the
compounds the same?
HINT!! If % Masses = , then they are the same
Practice Problems
A: Compd 1%H =
=
%O =
=
[15.0 / (15.0+120.0)] x 100%
11.1%
[120.0 / (15.0+120.0)] x 100%
88.9%
Compd 2%H = [2.0 / (2.0+32.0)] x 100%
= 5.9%
%O = [32.0 / (2.0+32.0)] x 100%
= 94.1%
NOT THE SAME COMPOUNDS
More Practice
Complete #s 10 – 13 on page 28-29 of
Solving Problems: A Chemistry
Handbook.
More Practice (ANSWERS)
10. a) % Bromine = 72.9%
b) Total = 100%
11.
% Hydrogen = 4.48%
% Carbon = 60.00%
% Oxygen = 35.52%
12. % Sulfur = 33.6%
% Copper = 66.4%
13. Mass of oxygen = 14.6g
Handout Key
1.
2.
3.
4.
5.
7.
Since the mass percentages in the unknown
compound are the same as in sucrose, the
compound must be sucrose.
Since amount does NOT affect mass percentages,
the mass percentage of carbon is 42.40% in 5, 50,
and 500 g of sucrose.
51.30% = [Mass O / 50.00g sucrose] x 100%
Mass O = [51.30 / 100] x 50.00 = 25.65g
42.40% = [Mass C / 100.0g sucrose] x 100%
Mass C = [42.40 / 100] x 100.0 = 42.40g
6.50% = [Mass H / 6.0 g sucrose] x 100%
Mass H = [6.50 / 100] x 6.0 = 0.39g
% Cl= [12.13g Cl / 20.00g salt] x 100% = 60.65%
% Na= [7.87g Na / 20.00g salt] x 100% = 39.4%