1. Empirical and Molecular Formulas For compounds: How to calculate Empirical Formula How to calculate Molecular Formula

2. Warmup Calculate the percent composition of these compounds: 1.C 2 H 2 (Acetylene) 2. C 8 H 8 (Styrene) Acetylene Cutting Torch Styrene Foam

3. Learning Objectives I can describe the difference between molecular and empirical formulas I can determine the empirical formula for a compound I can determine the molecular formula for a compound given its empirical formula and the compound’s molar mass.

4. Calculating Empirical Formulas The molecular formula of a compound shows the actual number and kinds of atoms present in a molecule. –example : H 2 O 2 is hydrogen peroxide The empirical formula just gives the lowest whole-number ratio of atoms of the elements in a compound. –If you were to decompose hydrogen peroxide into its elements, you would find one hydrogen for every oxygen, a 1:1 ratio. –therefore the empirical formula for hydrogen peroxide is HO

5. Calculating Empirical Formulas An empirical may or may not be the same as a molecular formula. If the formulas are different, the molecular formula is a simple multiple of the empirical formula. molecular formula empirical formula molar mass of H 2 O 2 hydrogen peroxide H2O2H2O2 HO34.0 g/mol Note: A compound’s molar mass depends on its molecular formula: H 2 O 2 = 2x1.0 + 2x16.0 = 34.0g/mol

6. Calculating Empirical Formulas Example: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? The % composition tells the ratio of the masses of nitrogen atoms to oxygen atoms. fact: percent = grams per 100 grams Step 1: Convert the % composition into mass ratio. –We could say: “In 100g of the compound, there are ______g N and _______g O”

7. Calculating Empirical Formulas Example: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? The % composition tells the ratio of the masses of nitrogen atoms to oxygen atoms. fact: percent = grams per 100 grams Step 1: Convert the % composition into mass ratio. –We could say: “In 100g of the compound, there are 25.9g N and 74.1g O”

8. Calculating Empirical Formulas Example: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? Step 2: Use percent composition values to convert to moles. In 100g of the compound, there are 25.9g N and 74.1g O.

9. Calculating Empirical Formulas Example: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? Step 3: Divide both molar quantities by the smaller number of moles. This will give 1 for the element with smaller moles. Is the final answer N 1 O 2.5 ????? Obviously not!!

10. Calculating Empirical Formulas Example: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? Step 4: Multiply each part of the ratio by a number to convert the fraction to a whole number: The empirical formula is N 2 O 5 --- dinitrogen pentoxide ---

11. You try it now… Calculate the empirical formula of a compound that is 94.1% O and 5.9% H Step 1: Use percent composition values to convert to moles. In 100g of the compound, there are ______ g O and _______g H. Step 2: Use percent composition values to convert to moles. mol O = ______ mol O mol H = ______ mol H Step 3: Divide both molar quantities by the smaller number of moles. This will give 1 for the element with smaller moles. ______ mol O ______ mol H Step 4: Multiply each part of the ratio by a number to convert the fraction to a whole number. empirical formula : ___________

12. You try it now… Calculate the empirical formula of a compound that is 94.1% O and 5.9% H Step 1: Use percent composition values to convert to moles. In 100g of the compound, there are 94.1 g O and 5.9 g H. Step 2: Use percent composition values to convert to moles. mol O = 5.88 mol O mol H = 5.9 mol H Step 3: Divide both molar quantities by the smaller number of moles. This will give 1 for the element with smaller moles..9966 = 1 mol O 1 mol H Step 4: Multiply each part of the ratio by a number to convert the fraction to a whole number. since the numbers in step 3 are whole already, the empirical formula is HO

13. CALCULATING MOLECULAR FORMULAS If we know a compound’s empirical formula and its molar mass, we can determine the molecular formula

14. Calculating Molecular Formulas p194 Different compounds can have the same empirical formula. The empirical formula gives the lowest whole- number ratio of atoms of the elements in a compound. example: molar massempirical formula C 2 H 2 (Acetylene)26.0 g/molCH C 8 H 8 (Styrene)104.0 g/molCH

15. Calculating Molecular Formulas p194 Different compounds can have the same empirical formula. The empirical formula gives the lowest whole- number ratio of atoms of the elements in a compound. example: molar massempirical formula C 2 H 2 (Acetylene)26.0 g/molCH C 8 H 8 (Styrene)104.0 g/molCH

16. Calculating Molecular Formulas You can determine the molecular formula of a compound if you know its empirical formula and its molar mass. step 1: Calculate the empirical formula molar mass. (efm). step 2: Divide the efm into the molar mass to get a whole number. step 3: multiply the empirical formula subscripts to get the molecular formula.

17. Calculating Molecular Formulas Example: Calculate the molecular formula for the compound whose molar mass is 60.0g and empirical formula CH 4 N step 1: empirical formula mass (efm) 12.0 + 4x1.0 + 14.0 = 30.0 g/mol step 2: divide efm into the molar mass to get a whole number: 60.0 / 30.0 = 2 step 3: multiply the empirical formula subscripts to get the molecular formula. CH 4 N subscripts x 2 is molecular formula C 2 H 8 N 2

18. Calculating Molecular Formulas Example: Calculate the molecular formula for the compound whose molar mass is 60.0g and empirical formula CH 4 N step 1: empirical formula mass (efm) 12.0 + 4x1.0 + 14.0 = 30.0 g/mol step 2: divide efm into the molar mass to get a whole number: 60.0 / 30.0 = 2 step 3: multiply the empirical formula subscripts to get the molecular formula. CH 4 N subscripts x 2 is molecular formula C 2 H 8 N 2

19. You try it: Calculating Molecular Formulas The empirical formula for ethylene glycol used in car antifreeze is CH 3 O. Its molar mass is 62.0 g/mol. Find the molecular formula: step 1: Determine the empirical formula molar mass (efm) step 2: divide efm into the molar mass to get a whole number: step 3: multiply the empirical formula subscripts to get the molecular formula.