Empirical and Molecular Formulas 6.2 Empirical and Molecular Formulas

Molecular/True Formula Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular (true) formula. Empirical Formula Molecular/True Formula Name CH C2H2 acetylene C6H6 benzene CO2 Carbon dioxide CH2O C5H10O5 ribose

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.

Learning Check 1) CH2O 2) C2H4O2 3) C3H6O3 A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. What is a molecular formula for CH2O? 1) CH2O 2) C2H4O2 3) C3H6O3

Calculating Empirical Formulas Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Learning Check A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula. Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Solution STEP 1 Determine the moles of each element. 7.31 g Ni x 1 mol Ni = 0.125 mol of Ni 58.69 g Ni 20.0 g Br x 1 mol Br = 0.250 mol of Br 79.90 g Br STEP 2 Divide by the smallest number of moles. 0.125 mol Ni = 1 mol of Ni 0.125 0.250 mol Br = 2 mol of Br Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Solution (continued) STEP 3 Use the lowest whole-number ratio of moles as subscripts. NiBr2 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Converting Decimals to Whole Numbers When the number of moles for an element is a decimal greater than 0.1, but less than 0.9 multiply the moles by a small integer to obtain whole numbers Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Percent Composition Using 100 g Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Finding the Empirical Formula “Percent to mass Mass to mole Divide by small Multiply ‘til whole”

Finding the Empirical 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. “Percent to Mass” - state mass percentages as grams in a 100.00 g sample of the compound. Cl 71.65 g C 24.27 g H 4.07 g

2. “Mass to Moles” 71.65 g Cl x 1 mol Cl = 2.02 mol Cl 35.5 g Cl 24.27 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

4. Write the simplest or empirical formula CH2Cl “Divide by Small” Find the smallest whole number ratio by dividing each mole value by the smallest mole values: Cl: 2.02 mol = 1 Cl 2.02 mol C: 2.02 mol = 1 C H: 4.04 mol = 2 H 4. Write the simplest or empirical formula CH2Cl (“Multiply ‘til whole”)

Learning Check Aspirin is 60.0% C, 4.5 % H, and 35.5 % O. Calculate its empirical (simplest) formula. “Percent to mass Mass to mole Divide by small Multiply ‘til whole” Basic Chemistry Copyright © 2011 Pearson Education, Inc.

“Percent to Mass & Mass to Mole” STEP 1 Calculate the moles of each element. 100.0 g aspirin contains 60.0% C or 60.0 g of C, 4.5% H or 4.5 g of H, and 35.5% O or 35.5 g of O. 60.0 g C x 1 mol C = 5.00 mol of C 12.01 g C 4.5 g H x 1 mol H = 4.5 mol of H 1.008 g H 35.5 g O x 1mol O = 2.22 mol of O 16.00 g O Basic Chemistry Copyright © 2011 Pearson Education, Inc.

“Divide by Small” STEP 2 Divide by the smallest number of moles. 5.00 mol C = 2.25 mol of C 2.22 4.5 mol H = 2.0 mol of H 2.22 mol O = 1.00 mol of O Basic Chemistry Copyright © 2011 Pearson Education, Inc.

“Multiply ‘til Whole” STEP 3 Use the lowest whole-number ratio of moles as subscripts. To obtain whole numbers of moles, multiply by a factor, in this case x 4. C: 2.25 mol of C x 4 = 9 mol of C H: 2.0 mol of H x 4 = 8 mol of H O: 1.00 mol of O x 4 = 4 mol of O Using these whole numbers as subscripts, the simplest formula is C9H8O4 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Finding the Molecular Formula Multiplier: molar mass = a whole number empirical mass To get Molecular Formula, first calculate Empirical Formula, then multiply all subscripts by the multiplier Note: If your multiplier = 1, the molecular formula = empirical formula eg. if multiplier = 2, the MF = 2 x EF

Back to the problem…. We calculated Empirical Formula = CH2Cl & were given the molar mass = 99.0 g/mol 5. Calculate EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49.5 g/mol 6. Calculate Multiplier: Molar mass = 99.0 g/mol = multiplier = 2 Empirical mass 49.5 g/mol 7. Molecular formula = Empirical Formula x multiplier (CH2Cl) x 2 = C2H4Cl2

Learning Check 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 S = 0.853 mol & divide by 0.853 = 1 S N = 0.857 mol & divide by 0.853 = 1 N Cl = 1.71 mol & divide by 0.853 = 2 Cl Empirical Formula (EF) = SNCl2 Empirical Mass (EM) = 117.1 g/mol Given Molar Mass = 351 g/mol Multiplier = 351g/mol = 3,  so MF = S3N3Cl6 117.1g/mol