Empirical and Molecular Formulas

Empirical vs Molecular Formula The Molecular Formula (MF) gives the actual number of each type of atom present. The Empirical Formula (EF) gives the lowest whole- number ratio of the atoms present. Example: C 2 H 6 and C 3 H 9 – they have the same EF CH 3 yet have very different MF

Timberlake LecturePLUS3 Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true)Name CHC 2 H 2 acetylene CHC 6 H 6 benzene CO 2 CO 2 carbon dioxide CH 2 OC 5 H 10 O 5 ribose

Timberlake LecturePLUS4 An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound.

EF vs. MF When you have an ionic compound: the EF = MF For some molecular compounds: the EF = MF, but not always

Determining the EF Remember: The EF is the lowest whole-number ratio of the moles of each atom present. Example: CH 4 has 1 mol C atoms 4 mol H atoms

Determining the EF If a compound consists of: 62.1% C 13.8% H 24.1% N The percentages are based on MASS not MOLES We can compare moles, not masses

Determining the EF In order to determine the ratio of C:H:N, we need to know the mole ratio Step 1: Convert % of each into grams Make it easy on yourself, assume a sample size of g 62.1g C 13.8g H 24.1g N

Determining the EF Now that you know how many grams of each atom you have: Step 2: Convert grams to moles using the molar mass of each 62.1g C 13.8g H 24.1 g N

Determining the EF Now that you have the mol ratio, you need to make them Whole-Numbers. Step 3: Divide each mol by the smallest mol value from step # mol C 13.7 mol H 1.72 mol N

Determining the EF This results in 3 mol C, 8 mol H and 1 mol N therefore the EF = C 3 H 8 N

Timberlake LecturePLUS12 Learning Check EF-1 A. What is the empirical formula for C 4 H 8 ? 1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. What is a molecular formula for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

Timberlake LecturePLUS13 Solution EF-1 A. What is the empirical formula for C 4 H 8 ? 2) CH 2 B. What is the empirical formula for C 8 H 14 ? 1) C 4 H 7 C. What is a molecular formula for CH 2 O? 1) CH 2 O2) C 2 H 4 O 2 3) C 3 H 6 O 3

Timberlake LecturePLUS14 Learning Check EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN 4 3) S 4 N 4

Timberlake LecturePLUS15 Solution EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S 4 N 4 If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.

Timberlake LecturePLUS16 Empirical and Molecular Formulas molar mass = a whole number = n simplest mass n = 1 molar mass = empirical mass molecular formula = empirical formula n = 2 molar mass = 2 x empirical mass molecular formula = 2 x empirical formula molecular formula = or > empirical formula

EF to MF To go from the EF to the MF, you need two additional pieces of information: 1 – calculate the mass from your EF 2 – You must be given the mass of the MF (X) EF mass = MF mass X = MF mass / EF mass

EF to MF Example: You found the EF to be HO, and the MF mass = 34.02g 1.Calculate the EF mass = 17.01g 2.Calculate X = 34.02g / 17.01g 3.X = 2 4.EF = HO MF = H 2 O 2

Timberlake LecturePLUS19 Learning Check EF-3 A compound has a formula mass of and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9

Timberlake LecturePLUS20 Solution EF-3 A compound has a formula mass of and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 2)C 6 H 8 O 6 C 3 H 4 O 3 = 88.0 g/EF g =

Timberlake LecturePLUS21 Learning Check EF-4 If there are g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4 ? 1) C 7 H 6 O 4 2) C 14 H 12 O 8 3) C 21 H 18 O 12

Timberlake LecturePLUS22 Solution EF-4 If there are g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4 ? 3) C 21 H 18 O g O = 3 x O 4 or 3 x C 7 H 6 O g O in EF

Timberlake LecturePLUS23 Finding the Molecular Formula A compound is Cl 71.65%, C 24.27%, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be 99.0 g/mol. 1. State mass percents as grams in a g sample of the compound. Cl gC g H 4.07 g

Timberlake LecturePLUS24 2. Calculate the number of moles of each element g Cl x 1 mol Cl = 2.02 mol Cl 35.5 g Cl g C x 1 mol C = 2.02 mol C 12.0 g C 4.07 g H x 1 mol H = 4.04 mol H 1.01 g H

Timberlake LecturePLUS25 Why moles? Why do you need the number of moles of each element in the compound?

Timberlake LecturePLUS26 3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values: Cl: 2.02 = 1 Cl 2.02 C: 2.02 = 1 C 2.02 H: 4.04 = 2 H Write the simplest or empirical formula CH 2 Cl

Timberlake LecturePLUS27 5. EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = n = molar mass/empirical mass Molar mass = 99.0 g/mol = n = 2 E M 49.5 g/EM 7.Molecular formula (CH 2 Cl) 2 = C 2 H 4 Cl 2

Timberlake LecturePLUS28 Learning Check EF-5 Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

Timberlake LecturePLUS29 Solution EF g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O

Timberlake LecturePLUS30 Solution EF g C x 1 mol C = 5.00 mol C 12.0 g C 4.5 g H x 1 mol H = 4.5 mol H 1.01 g H 35.5 g O x 1mol O= 2.22 mol O 16.0 g O

Timberlake LecturePLUS31 Divide by the smallest # of moles mol C = ________________ ______ mol O 4.5 mol H = ________________ ______ mol O 2.22 mol O = ________________ ______ mol O Are are the results whole numbers?_____

Timberlake LecturePLUS32 Divide by the smallest # of moles mol C = ___2.25__ 2.22 mol O 4.5 mol H = ___2.00__ 2.22 mol O 2.22 mol O = ___1.00__ 2.22 mol O Are are the results whole numbers?_____

Timberlake LecturePLUS33 Finding Subscripts A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts. (1/2) 0.5 x 2 = 1 (1/3)0.333 x 3 = 1 (1/4)0.25 x 4 = 1 (3/4)0.75 x 4 = 3

Timberlake LecturePLUS34 Multiply everything x 4 C: 2.25 mol C x 4 = 9 mol C H: 2.0 mol Hx 4 = 8 mol H O: 1.00 mol O x 4 = 4 mol O Use the whole numbers of mols as the subscripts in the simplest formula C 9 H 8 O 4

Timberlake LecturePLUS35 Learning Check EF-6 A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?

Timberlake LecturePLUS36 Solution EF mol S /0.853 = 1 S mol N /0.853 = 1 N 1.71 mol Cl /0.853 = 2 Cl Empirical formula = SNCl 2 = g/EF Mol. Mass/ Empirical mass 351/117.1 = 3 Molecular formula = S 3 N 3 Cl 6