Empirical Formulas & Molecular Formulas Empirical Formulas Molecular Formulas

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

Empirical Formulas Write your own one-sentence definition for each of the following: Empirical formula Molecular formula

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.

Empirical Formulas 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

Solution 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

Molecular Formula 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

Solution 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.

Empirical and Molecular Formulas To obtain the molecular formula, you must first have an empirical formula. You must also have the molar mass or be given enough information to determine the molar mass. With these two values, you can determine the molecular formula.

Empirical and Molecular Formulas molar mass = a whole number = n EF 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

Empirical Formula Empirical Mass Molecular Formula Molecular Mass

Molecular Formula 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

Solution 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 =

Molecular Formula If there are g of O in the molecular formula, what is the true formula if the Empirical Formula 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

Solution 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

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

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

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

Why moles? Why do you need the number of moles of each element in the compound? Recall Avogadro’s number! A mole contains equal numbers of particles. That means we are comparing atoms on an equal basis.

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

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

Empirical Formula 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.

Solution 60.0 g C x ___________= ______ mol C 4.5 g H x ___________ = _______mol H 35.5 g O x ___________ = _______mol O

Solution 60.0 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

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?_____

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 the results whole numbers?_____

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

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 (2/3)0.667x 3=2 (3/4)0.75 x 4 = 3

Molecular Formula 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?

Solution 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