The Mole Concept Molar Mass, Conversion Problems, Percentage Composition, Empirical Formulas, Molecular Formulas
The Mole Chemist’s counting number amount of a substance 1 mole is equal to 6.022 x 1023 atoms or particles or molecules (Avogadro’s number) There are always 6.022 x 1023 particles in one mole of ANY substance Ex: 1 mole of Carbon atoms = 6.022 x 1023 atoms
Molar Mass Mass of one mole of any substance Equals the atomic mass on periodic table Units are g/mol Examples: C Mg(OH)2
Chemistry equalities 1 mole = grams = 6.022 x 1023 atoms (cloud these) (Happy mole day )
Percentage Composition Part___ x 100 = Percentage Whole Percentage by mass of each element in a compound Mass of element Total mass of compound X 100 = % of the element
Example: Which has a higher percentage of Hydrogen? H2O C6H12O6
Conversion Factors 1 mole = 6.022 x 1023 1 mole = ~ g ~ g = 6.022 x 1023 Each of these can be written as a fraction
Mole Calculations Uses “standard equalities” to set up a ratio Special notes **Answer will have same # of significant figures as given value. *** Scientific notation uses the EE button on calculator
Common examples: 1. How many eggs in 58 dozen?
Example #2 How many moles are in 8.9 grams of Carbon?
Example #3: How many grams are in 1.38 x 1023 atoms of gold?
Chemistry conversions: Example #2: How many grams of Lithium are in 3.5 moles?
Cancellation Method (Dimensional Analysis) Conversion factors are a fraction where the top and bottom have equal values 1 foot 16 ounces 12 inches 1 pound Given units are on the bottom and unknown units are on the top
Molecular formula Actual formula of a compound Ex: H2O, C6H12O6
Empirical Formula Formula with the lowest whole number ratio of elements Ex: Write empirical formula of C6H12O6 N3O9 H2O C2H6
Calculating List the mass of each element. Assume there is 100 grams if the percentage is given. Step 1: Convert the grams into moles using the molar mass. Step 2: Divide by smallest answer Step 3: To get a whole number ratio Whole number answers are written as subscripts
Example 1. Determine the empirical formula for a compound containing 52% Carbon, 13% Hydrogen, and 35 % Oxygen
Determining molecular formula Molecular Formula Atomic Mass Empirical Formula Mass Multiply subscripts of empirical formula by whole number answer.
Molecular Formula 1. Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu.