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

% by mass = MASSelement x 100% MASScompd

MASS compd = sum of MASSES elements

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 Analysis 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%

Percent by Mass (500g sucrose)

Element Analysis 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%

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 ProblemsQ: 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 x 100%

(3.5 + 10.5)g

= 25%

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

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 ProblemsA: 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

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 Key1. 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%