Ch 3.4 law of definite proportions 11 12

1. 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.

2. Percent by Mass
 Based on the law of conservation of
mass
 % by mass = MASSelement x 100%
MASScompd
MASS compd = sum of MASSES elements

3. 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%

4. Percent by Mass (20g of sucrose)
ElementAnalysis by
mass (g)
Percent by mass
(%)
Carbon 8.4 8.4 x 100 = 42%
20.0
Hydrogen 1.3 1.3 x 100 = 6.5%
20.0
Oxygen 10.3 10.3 x 100 = 51.5%
20.0
Total
(Sucrose)
20.0 100%

5. Percent by Mass (500g sucrose)
ElementAnalysis by
mass (g)
Percent by mass
(%)
Carbon 210.5 210.5 x 100 = 42%
500.0
Hydrogen 32.4 32.4 x 100 = 6.5%
500.0
Oxygen 257.1 257.1 x 100 = 51.5%
500.0
Total
(Sucrose)
500.0 100%

6. 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

7. 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%

8. 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 x 100%
(3.5 + 10.5)g
= 25%

9. 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 YXY2 )
= 2 ( 10.5g)
= 21.0 g

10. 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

11. Practice Problems
A: Compd 1-
%H = [15.0 / (15.0+120.0)] x 100%
= 11.1%
%O = [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

12. More Practice
Complete #s 10 – 13 on page 28-29 of
Solving Problems: A Chemistry
Handbook.

13. 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

14. Handout Key
1.Since the mass percentages in the unknown
compound are the same as in sucrose, the
compound must be sucrose.
2.Since amount does NOT affect mass percentages,
the mass percentage of carbon is 42.40% in 5, 50,
and 500 g of sucrose.
3.51.30% = [Mass O / 50.00g sucrose] x 100%
Mass O = [51.30 / 100] x 50.00 = 25.65g
4.42.40% = [Mass C / 100.0g sucrose] x 100%
Mass C = [42.40 / 100] x 100.0 = 42.40g
5.6.50% = [Mass H / 6.0 g sucrose] x 100%
Mass H = [6.50 / 100] x 6.0 = 0.39g
7.% Cl= [12.13g Cl / 20.00g salt] x 100% = 60.65%
% Na= [7.87g Na / 20.00g salt] x 100% = 39.4%