Mass Ratios & Law of Multiple Proportions Miss Fogg Fall 2015
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Mass Ratios & Law of Multiple ProportionsMiss FoggFall 2015
Mass RatiosAtomic 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
Mass ratiosMass 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
Example: H2O
Find the ratio of oxygen to hydrogen in H2O◦1) Find the mass of each element
O: 16.00amu x 1 atom = 16.00amuH: 1.01amu x 2 atom = 2.02amu
Find the ratio O : H◦2) Divide each elemental mass by the
smallest element mass
O: 16.00amu ÷ 2.02 = 8.0H: 2.02amu ÷ 2.02 = 1.0
Example: H2O
Find the ratio O : H◦3) Express as ratio and round
O: 8.0 8H: 1.0 1
(Law of definite proportions) In H2O, oxygen and hydrogen always have a ratio of “16 to 2” or “8 to 1” by mass
8:1
Example: H2O
Example: KClKCl always contains one atom of K
for every one atom of ClFind the ratio K : Cl
◦1) Find the mass of each element
K: 39.10amu x 1 atom = 39.10amuCl: 35.45amu x 1 atom = 35.45amu
Example: KClFind the ratio K : Cl
◦2) Divide each elemental mass by the smallest element mass
K: 39.10amu ÷ 35.45 = 1.103Cl: 35.45amu ÷ 35.45 = 1.000
Example: KClFind the ratio K : Cl
◦3) Express as ratio and round
K: 1.103 1.1Cl: 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
Law of Multiple ProportionsDalton’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
Nitrogen & OxygenYou have more than one
molecule made up of the same elements
N2O nitrous oxide
NO nitric oxide
NO2 nitrogen dioxideOO
O
ON
N
N
N
Nitrogen & OxygenN2O nitrous oxide
Mass nitrogen: 14.01amu x 2 = 28.02amu
Mass oxygen:16.00amu x1 = 16.00amu
ONN
Nitrogen & OxygenN2O nitrous oxide
Mass nitrogen: 28.02amuMass 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
Nitrogen & Oxygen
Mass of N for 1g ON2O nitrous oxide 1.75g
NO nitric oxide
NO2 nitrogen dioxide
OO
O
ON
N
N
N
Nitrogen & OxygenNO nitric oxide
Mass nitrogen: 14.01amu x 1 = 14.01amu
Mass oxygen:16.00amu x1 = 16.00amu
ON
Nitrogen & OxygenNO nitric oxideMass nitrogen: 14.01amuMass 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
Nitrogen & Oxygen
Mass of N for 1g ON2O nitrous oxide 1.75g
NO nitric oxide 0.876g
NO2 nitrogen dioxide
OO
O
ON
N
N
N
Nitrogen & OxygenNO2 nitrogen dioxide
Mass nitrogen: 14.01amu x 1 = 14.01amu
Mass oxygen:16.00amu x 2 = 32.00amu
ONO
Nitrogen & OxygenNO2 nitrogen dioxide
Mass nitrogen: 14.01amuMass 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
Nitrogen & Oxygen
Mass of N for 1g ON2O nitrous oxide 1.75g
NO nitric oxide 0.876g
NO2 nitrogen dioxide 0.438g
OO
O
ON
N
N
N
How are all these numbers
related?
Nitrogen & Oxygen
Mass of N for 1g ON2O nitrous oxide 1.75g
NO nitric oxide 0.876g
NO2 nitrogen dioxide 0.438g
OO
O
ON
N
N
N
How are all these numbers
related?
Law of multiple proportions! 1.75 ÷ 0.876 = 1.998 = 20.876 ÷ 0.438 = 21.75 ÷ 0.438 = 3.995 = 4
Ch3 Homework - Due tomorrow (10/7):
Assessment ProblemsPage 76: 20-24; 30Page 83: Problems 60-69; 74-75
Related Documents
