Download presentation

Presentation is loading. Please wait.

Published byGeorge Clemence Modified over 2 years ago

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 = MASS element x 100% MASS compd 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 (%) Carbon8.48.4 x 100 = 42% 20.0 Hydrogen1.31.3 x 100 = 6.5% 20.0 Oxygen10.310.3 x 100 = 51.5% 20.0 Total (Sucrose) 20.0100%

5
Percent by Mass (500g sucrose) ElementAnalysis by mass (g) Percent by mass (%) Carbon210.5210.5 x 100 = 42% 500.0 Hydrogen32.432.4 x 100 = 6.5% 500.0 Oxygen257.1257.1 x 100 = 51.5% 500.0 Total (Sucrose) 500.0100%

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 XY 2 ? A: Mass Y XY = 2 ( Mass Y XY 2 ) = 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%

Similar presentations

Presentation is loading. Please wait....

OK

Determining Chemical Formulas

Determining Chemical Formulas

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on acute coronary syndrome guidelines Ppt online open file A ppt on storage devices Download ppt on law and social justice Ppt on current account deficit united Ppt on blood stain pattern analysis terminology Ppt on obesity prevention coalition Ppt on leadership skills training Visuals ppt on tenses Ppt on brain tumor