1. Mass Ratios & Law of Multiple Proportions Miss Fogg Fall 2015

2. Mass Ratios Atomic mass = mass of an atom measured in amu (atomic mass units) Mass of oxygen = 15.999  16.00amu Mass of hydrogen = 1.008  1.01amu

3. Mass ratios Mass of element X : Mass of element Y How to calculate mass ratios: ◦ 1) Find the mass of each element ◦ 2) Divide each elemental mass by the smallest element mass ◦ 3) Express as ratio and round

4. Example: H 2 O Find the ratio of oxygen to hydrogen in H 2 O ◦ 1) Find the mass of each element O: 16.00amu x 1 atom = 16.00amu H: 1.01amu x 2 atom =2.02amu

5. Find the ratio O : H ◦ 2) Divide each elemental mass by the smallest element mass O: 16.00amu ÷ 2.02 = 8.0 H: 2.02amu ÷ 2.02 = 1.0 Example: H 2 O

6. Find the ratio O : H ◦ 3) Express as ratio and round O: 8.0  8 H: 1.0  1 (Law of definite proportions) In H 2 O, oxygen and hydrogen always have a ratio of “16 to 2” or “8 to 1” by mass 8 : 1 Example: H 2 O

7. Example: KCl KCl always contains one atom of K for every one atom of Cl Find the ratio K : Cl ◦ 1) Find the mass of each element K: 39.10amu x 1 atom = 39.10amu Cl: 35.45amu x 1 atom =35.45amu

8. Example: KCl Find the ratio K : Cl ◦ 2) Divide each elemental mass by the smallest element mass K: 39.10amu ÷ 35.45 = 1.103 Cl: 35.45amu ÷ 35.45= 1.000

9. Example: KCl Find the ratio K : Cl ◦ 3) Express as ratio and round K: 1.103  1.1 Cl: 1.000  1 (Law of definite proportions) In KCl, potassium and chlorine always have a ratio of “39.09 to 35.45” or “1.1 to 1” by mass. 1.1 : 1

10. Law of Multiple Proportions Dalton’s Law (1804): When 2 elements form different compounds, the mass ratio of the elements in one compound is related to the mass ratio in the other by a small whole number

11. Nitrogen & Oxygen You have more than one molecule made up of the same elements N 2 Onitrous oxide NOnitric oxide NO 2 nitrogen dioxide OO O ON N N N

12. Nitrogen & Oxygen N 2 Onitrous oxide Mass nitrogen: 14.01amux 2 = 28.02amu Mass oxygen: 16.00amux1= 16.00amu ONN

13. Nitrogen & Oxygen N 2 Onitrous oxide Mass nitrogen: 28.02amu Mass oxygen: 16.00amu How many grams of N are used every 1g of O? 28.02amu nitrogen= ? 16.00amu oxygen 1g oxygen ONN 1.751g nitrogen

14. Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide NO 2 nitrogen dioxide OO O ON N N N

15. Nitrogen & Oxygen NOnitric oxide Mass nitrogen: 14.01amux 1 = 14.01amu Mass oxygen: 16.00amux1= 16.00amu ON

16. Nitrogen & Oxygen NOnitric oxide Mass nitrogen: 14.01amu Mass oxygen: 16.00amu How many grams of N are used every 1g of O? 14.01amu nitrogen= ? 16.00amu oxygen 1g oxygen ON 0.8763g nitrogen

17. Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide0.876g NO 2 nitrogen dioxide OO O ON N N N

18. Nitrogen & Oxygen NO 2 nitrogen dioxide Mass nitrogen: 14.01amux 1 = 14.01amu Mass oxygen: 16.00amux 2= 32.00amu ONO

19. Nitrogen & Oxygen NO 2 nitrogen dioxide Mass nitrogen: 14.01amu Mass oxygen: 32.00amu How many grams of N are used every 1g of O? 14.01amu nitrogen= ? 32.00amu oxygen 1g oxygen ON 0.4378g nitrogen O

20. Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide0.876g NO 2 nitrogen dioxide0.438g OO O ON N N N How are all these numbers related?

21. Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide0.876g NO 2 nitrogen dioxide0.438g OO O ON N N N How are all these numbers related? Law of multiple proportions! 1.75 ÷ 0.876 = 1.998 = 2 0.876 ÷ 0.438 = 2 1.75 ÷ 0.438 = 3.995 = 4

22. Ch3 Homework - Due tomorrow (10/7): Assessment Problems Page 76: 20-24; 30 Page 83: Problems 60-69; 74-75