Molar Mass

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Objectives : 

Objectives Define molar mass Find the molar mass of an element using the periodic table. Calculate the molar mass of a compound

Molar Mass : 

Molar Mass 1 mole represents a specific number of particles 6.02 x 1023 particles 1 mole has a certain mass Measure in grams on a balance

Molar mass : 

Molar mass 1 dozen apples have a different mass than 1 dozen paper clips. 1 mole of atoms will have a particular mass, but it will depend on the element 1 mole of carbon atoms has a mass of 12.01 grams

Molar Mass of Elements : 

Molar Mass of Elements Average atomic mass is found on the periodic table for each element. Average atomic mass is measured in amu Mass of an atom Molar mass is measured in grams/mol Mass of a mole of particles

Molar mass of diatomic elements : 

Molar mass of diatomic elements Since there are two atoms in each molecule, the molar mass is twice the average atomic mass Molar mass of O2 2 atoms O per molecule 1 mole O atoms has a molar mass of 16.00 g/mol 1 mole O2 molecules has a molar mass of 32.00 g/mol

Molar mass of polyatomic elements : 

Molar mass of polyatomic elements Example: O3 Average atomic mass of O = 16.00 amu Molar mass of O = 16.00 g/mol Molar mass of O3 3 x 16.00 g/mol = 48.00 g/mol

Molar mass of compounds : 

Molar mass of compounds Sum of the molar masses of the elements in the formula Same calculation for ionic compounds or molecular compounds

Calculating Molar Mass : 

Calculating Molar Mass Write the formula. List the types of atoms. Count each type of atom, using subscripts, and write down the count. Multiply the count by the average atomic mass of the element. Total all the masses.

Sample Problem : 

Sample Problem Barium acetate 1. Ba(C2H3O2)2

Sample Problem : 

Sample Problem Barium acetate Ba(C2H3O2)2 2. Ba C H O

Sample Problem : 

Sample Problem Barium acetate Ba(C2H3O2)2 3. Ba 1 C 4 H 6 O 4

Sample Problem : 

Sample Problem Barium acetate Ba(C2H3O2)2 4. Ba 1 x 137.33 = 137.33 C 4 x 12.01 = 48.04 H 6 x 1.01 = 6.06 O 4 x 16.00 = 64.00

Sample Problem : 

Sample Problem Barium acetate Ba(C2H3O2)2 5. Ba 1 x 137.33 = 137.33 C 4 x 12.01 = 48.04 H 6 x 1.01 = 6.06 O 4 x 16.00 = 64.00 255.43 g/mol

Sample Problem : 

Sample Problem Ammonium sulfate 1. (NH4)2SO4

Sample Problem : 

Sample Problem Ammonium sulfate (NH4)2SO4 2. N H S O

Sample Problem : 

Sample Problem Ammonium sulfate (NH4)2SO4 3. N 2 H 8 S 1 O 4

Sample Problem : 

Sample Problem Ammonium sulfate (NH4)2SO4 4. N 2 x 14.01 = 28.02 H 8 x 1.01 = 8.08 S 1 x 32.07 = 32.07 O 4 x 16.00 = 64.00

Sample Problem : 

Sample Problem Ammonium sulfate (NH4)2SO4 5. N 2 x 14.01 = 28.02 H 8 x 1.01 = 8.08 S 1 x 32.07 = 32.07 O 4 x 16.00 = 64.00 132.17 g/mol