Empirical and Molecular Formulas

Evaluation/Assessment: Objective: Today I will be able to: Calculate an empirical formula from experimental data Derive a molecular formula from experimental data Evaluation/Assessment: Informal Assessment – Monitoring student interactions and questions as they complete the practice problems Formal Assessment – Analyzing student responses to the lab and the empirical/molecular formula practice Common Core Connection Make sense of problem and persevere in solving them Look for and express regularity in repeated reasoning

Lesson Sequence Evaluate: Warm – Up Explain: Empirical and Molecular Formula Notes Elaborate: Empirical and Molecular Formula Practice Evaluate: Exit Ticket

Warm - Up What is the difference between the following pairs of formulas: CO2; C2O4 CH2O; C6H12O6

Objective Today I will be able to: Calculate an empirical formula from experimental data Derive a molecular formula from experimental data

Homework Work on Mole Project Due March 11 Empirical/ Molecular Formula Practice Wear Closed Toe Shoes for Empirical and Molecular Formula Lab Monday

Agenda Warm-Up Empirical and Molecular Formula Notes Empircal and Molecular Formula Practice Exit Ticket

Empirical and Molecular Formula Notes

Empirical Formula

Empirical Formula A formula that gives the simplest whole number ratio of atoms in a compound Example: The empirical formula for hydrogen peroxide is HO (the actual formula is H2O2). This is not the actual formula; it tells you the ratio of hydrogen to oxygen is 1:1

Empirical Formula A compound was analyzed and found to contain 13.5 g of Ca, 10.8 g of O, and .675 g of H. What is the empirical formula?

Empirical Formula Step 1: Find the mole amounts of each element 13.5 g Ca 1 1 mol 40 g Ca = .337 mol Ca x 10.8 g O 1 1 mol 16 g O x = .675 mol O 1 mol 1 g H .675 g H 1 .675 mol H x =

Empirical Formula Step 2: Divide each mole value by the smallest number of moles 13.5 g Ca 1 1 mol 40 g Ca = .337 mol Ca .337 mol x = 1 10.8 g O 1 1 mol 16 g O x = .675 mol O .337 mol = 2 1 mol 1 g H .675 g H 1 .675 mol H .337 mol = 2 x =

Empirical Formula Step 3: Determine the Empirical Formula Ca1O2H2 or…

Empirical Formula CaCl2 Determine the empirical formula for a compound containing 1.203 g of Ca and 2.128 g of Cl 1.203 g Ca 1 1 mol 40 g Ca .03008 mol Ca = ________ x = 1 .03008 mol CaCl2 2.128 g Cl 1 1 mol 35.4 g Cl = .06011 mol Cl x _________ = 2 .03008 mol

Empirical Formula Practice Complete the practice at your desk. Ask Mr. Klotz for help if you have questions!

Molecular Formula

Molecular Formula Is always a whole number multiple of the empirical formula Ex: A 100.0 g sample of TNT is composed of 31.7 g carbon, 2.60 g hydrogen, 18.5 g nitrogen, and 42.3 g oxygen. (Its molar mass is 227 g/mol)

Molecular Formula C2H2NO2 Step 1: Determine the empirical formula 12 g C 2.64 mol C 31.7 g C 1 _________ 2 = = x 1.32 mol 2.60 g H 1 1 mol 1 g H 2.60 mol H x _________ = = 2 1.32 mol C2H2NO2 18.5 g N 1 1 mol 14 g N 1.32 mol N x = _________ = 1 1.32 mol 42.3 g O 1 1 mol 16 g O 2.64 mol O = ________ x = 2 1.32 mol

Molecular Formula Step 2: Determine the empirical formula molar mass C2H2NO2 = 72 g/mol

Molecular Formula 227 g/mol 72 g/mol = 3.15 or 3 Step 3: Divide the molar mass by the empirical formula molar mass 227 g/mol 72 g/mol = 3.15 or 3

Molecular Formula Step 4: Multiply the empirical formula by the ratio of the molar mass to the empirical formula 3 (C2H2NO2) = C6H6N3O6

Molecular Formula Practice Complete the Practice at your desk ask Mr. Klotz for help if you have questions.

Exit Ticket On a scale of 1 to 5, how comfortable are you with empirical and molecular formula calculations?