1. Formulas and Composition

2. Percent Composition Percent composition lists a percent each element is of the total mass of the compound Percent composition lists a percent each element is of the total mass of the compound H 2 O H 2 O  Total mass is 18.02  Mass of O is 16, so O is 16/18.02 or 88.79%  Mass of H is 2.02, so H is 2.02/18.02 or 11.21%

3. You Practice - Iron (III) Sulfate Fe 2 (SO 4 ) 3 Fe 2 (SO 4 ) 3 2*Fe = 111.70 g/mol 2*Fe = 111.70 g/mol 3*S = 96.21 g/mol 3*S = 96.21 g/mol 12*O = 192.00 g/mol 12*O = 192.00 g/mol Total mass = 399.91 g/mol Total mass = 399.91 g/mol Fe = 111.70/399.91 => 27.93% Fe = 111.70/399.91 => 27.93% S = 96.21/399.91 => 24.06% S = 96.21/399.91 => 24.06% O = 192.00/399.91 => 48.01% O = 192.00/399.91 => 48.01%

4. Hydrates Some compounds come with water attached Some compounds come with water attached These are known as hydrates These are known as hydrates The water is not part of the molecule, just attached to the molecule The water is not part of the molecule, just attached to the molecule A hydrate has a specific amount of water attached to each molecule A hydrate has a specific amount of water attached to each molecule  Iron (II) Sulfate is FeSO 4 * Iron (II) Sulfate is FeSO 4 *7H 2 O  There are 7 water molecules attached to one Iron (II) Sulfate

5. Molar Mass of Hydrates Finding the molar mass will include the mass of the attached water molecules Finding the molar mass will include the mass of the attached water molecules Iron (II) Sulfate is FeSO 4 *7H 2 O Iron (II) Sulfate is FeSO 4 *7H 2 O  Fe + S + 4*O + 14*H + 7*O

6. % Composition of Hydrates Iron (II) Sulfate is FeSO 4 *7H 2 O Iron (II) Sulfate is FeSO 4 *7H 2 O 2*Fe = 111.70 g/mol 2*Fe = 111.70 g/mol 3*S = 96.21 g/mol 3*S = 96.21 g/mol 12*O = 192.00 g/mol 12*O = 192.00 g/mol 7*H 2 O = 126.14 g/mol 7*H 2 O = 126.14 g/mol Total mass = 526.05 g/mol Total mass = 526.05 g/mol Fe = 111.70/526.05 => 21.23% Fe = 111.70/526.05 => 21.23% S = 96.21/526.05 => 18.29% S = 96.21/526.05 => 18.29% O = 192.00/526.05 => 36.50% O = 192.00/526.05 => 36.50% H 2 O =126.14/526.05 => 23.98% H 2 O =126.14/526.05 => 23.98%

7. Molecular Formula So far, we have been dealing with molecular formulas So far, we have been dealing with molecular formulas These are the complete formula for a molecule of the ionic compound These are the complete formula for a molecule of the ionic compound Ionic compounds are always written in the reduced form – we use NaCl, not Na 2 Cl 2 Ionic compounds are always written in the reduced form – we use NaCl, not Na 2 Cl 2 This is not true for covalent molecules This is not true for covalent molecules

8. Glucose is C 6 H 12 O 6 Glucose is C 6 H 12 O 6 This is the molecular formula as is shows the complete number of each element in the molecule This is the molecular formula as is shows the complete number of each element in the molecule This formula can be reduced to CH 2 O This formula can be reduced to CH 2 O The reduced formula, if it is not the molecular formula, is called the empirical formula The reduced formula, if it is not the molecular formula, is called the empirical formula

9. You practice – write the molecular and empirical formulas Copper (II) Phosphite Copper (II) Phosphite Sulfurous Acid Sulfurous Acid Copper (II) Chloride Dihydrate Copper (II) Chloride Dihydrate Sucrose (C 12 H 22 O 12 ) Sucrose (C 12 H 22 O 12 )

10. Using % Composition to find the empirical formula Hydrogen Peroxide has a molar mass of 34.02 g/mol Hydrogen Peroxide has a molar mass of 34.02 g/mol H is 5.94%, O is 94.06% H is 5.94%, O is 94.06% Using this, we can assign masses to each element based on the % Using this, we can assign masses to each element based on the % 94.06 g O and 5.94 g H 94.06 g O and 5.94 g H Now we convert these to moles by dividing by molar mass Now we convert these to moles by dividing by molar mass

11. 94.06 g O (1 mol O/16 g O) = 5.88 mol O 94.06 g O (1 mol O/16 g O) = 5.88 mol O 5.94 g H (1 mol H/1.01 g H) = 5.88 mol H 5.94 g H (1 mol H/1.01 g H) = 5.88 mol H Therefore the empirical formula is HO Therefore the empirical formula is HO Using this, and the molar mass, we can find the molecular formula Using this, and the molar mass, we can find the molecular formula x * (HO) = 34.02 x * (HO) = 34.02 x* (17.01) = 34.02 x* (17.01) = 34.02 x=2 x=2 Formula = H 2 O 2 Formula = H 2 O 2

12. You Try It This compound is 12.63% Li, 29.15% S, and 58.22% O. Its molar mass is 109.92 g/mol. This compound is 12.63% Li, 29.15% S, and 58.22% O. Its molar mass is 109.92 g/mol. Find the empirical and molecular formulas Find the empirical and molecular formulas