Chapter 6 Empirical and Molecular Formulas

Chapter 6 Empirical and Molecular Formulas

EMPIRICAL FORMULA A chemical formula that indicates the relative proportions of the elements in a molecule rather than the actual number of atoms of the elements. (An empirical formula may be obtained from percentage composition of elements in a compound.) MOLECULAR FORMULA A chemical formula that indicates the actual number of atoms of the elements in a molecule. (Information in addition to percentage composition of elements is needed to determine a molecular formula.)

Molecular Formulas May Differ from Empirical Formulas Benzene Empirical Formula, CH Molecular Formula, C6H6 Acetylene Empirical Formula, CH Molecular Formula, C2H2

Percent Composition

Mass Percent Composition of Compounds Percentage of each element in a compound by mass Can be determined from 1. the formula of the compound, or 2. the experimental mass analysis of the compound

Mass Percent Composition of Compounds The mass percent composition, or mass percent, of an element is the element s percentage of the total mass of the compound Mass percent of element X = Mass of X in a sample Mass of the sample x 100%

1. Find the mass percent of Na in NaCl 1 x (molar mass of Na) Mass % Na = molar mass of NaCl x 100% 1 x (22.99 g/mol) Mass % Na = x 100 % = 39.33% 58.44 g/mol 22.99 + 35.45 = 58.44

2. Find the mass percent of Cl in C2Cl4F2 4 x (molar mass of Cl) Mass % Cl = x 100% molar mass of C2Cl4F2 4 x molar mass Cl = 4(35.45) = 141.8 g/mol molar mass C2Cl4F2 = 2(12.01)+4(35.45)+2(19.00) = 203.8 g/mol Mass % Cl = x 100% =69.58% 141.8 g/mol 203.8 g/mol

Mass Percent as a Conversion Factor 3. If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na. g Na g NaCl 100 g NaCl 39 g Na 39 g Na 100 g NaCl 2.4 g Na x 100 g NaCl = 6.1538 6.2 g g NaCl 39 g Na

4. Find the mass of sodium in 6.2 g of NaCl g NaCl mol NaCl mol Na g Na 1.00 mol NaCl 58.44 g NaCl 1.00 mol Na 1.00 mol NaCl 22.99 g Na 1.00 mol Na 1.00 mol NaCl 1.00 mol Na 22.99 g Na 6.2 g NaCl x x x = 2.4390 g Na 58.44 g NaCl 1.00 mol NaCl 1.00 mol Na 2.4 g Na

Empirical Formula Simplest, whole-number ratio of the atoms of elements in a compound Can be determined from elemental analysis

Finding an Empirical Formula from Mass Composition 1. Convert grams of elements to moles of elements. 2. Write a pseudoformula using moles as subscripts 3. Divide all by smallest number of moles 4. Multiply all mole ratios by number to make all whole numbers

5. A 33.34 g sample of aspirin contains 20.01 g of carbon, 1.49 g of hydrogen, and 11.84 g of oxygen. Find the empirical formula of aspirin. 20.01 + 1.49 + 11.84 = 33.34

g C g H g O mol C mol H mol O! pseudoformula CxHyOz Manipulate subscripts to obtain whole-number ratio empirical formula CxHyOz

Calculate the moles of each element 20.01 g C x 1.000 mol C = 1.666 mol C 12.01 g C 1.48 g H x 1.000 mol H = 1.48 mol H 1.008 g H 1.000 mol O 11.84 g O x = 0.7400 mol O 16.00 g O Write a pseudoformula C1.666H1.48O0.7400

C1.666H1.48O0.7400 Find the integer mole ratio 0.7400 C2.25H2.00O1.00 Multiply subscripts by factor to give whole number C9H8O4 (x 4)

Finding an Empirical Formula from % Composition 1. Assume 100.0 g of the compound and convert the percentages to grams 2. Convert grams to moles 3. Write a pseudoformula using moles as subscripts 4. Divide all by smallest number of moles 5. Multiply all mole ratios by number to make all whole numbers

6. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70 g/mol) and the rest fluorine (19.00 g/mol) Given: 75.7% Sn, (100 75.3) = 24.3% F Therefore, in 100 g of stannous fluoride there are 75.5 g Sn and 24.3 g of F!! g Sn g F mol Sn mol F pseudo formula empirical formula!

6. Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70 g/mol) and the rest fluorine (19.00 g/mol) Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Sn 75.7g X 1mol Sn 118.7g 0.6377 1 F 24.3g X 1mol F 19.00g 1.279 2.005 0.6377 SnF2

7. Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Given: 72.4% Fe, (100 72.4) = 27.6% O Therefore, in 100 g of magnetite there are 72.4 g Fe and 27.6 g of O!! g Fe g O mol Fe mol O pseudo formula empirical formula!

Practice Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Ratio in Moles Fe 72.4g X 1mol Fe 55.85g 1.296 1 3 O 27.6g X 1mol O 16.00g 1.725 1.33 4 1.296 x 3 Fe3O4

Molecular Formulas The molecular formula is a multiple of the empirical formula. To determine the molecular formula you need to know the empirical formula and the molar mass of the compound.

8. Find the molecular formula of butanedione if its empirical formula is C2H3O and its molar mass (MM) is 86.03 g/mol. Molar Mass (emp. form.) Factor of 2 = 2 x (12.01 gc/molc) + 3 x (1.008 gh/molh) + 1 x (16.00 go/molo) = 43.04 g/mol Molecular formula = C2H3O x 2 = C4H6O2

9. Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (12.01 g C/mol C, 1.01 g H/mol H) C5 = 5(12.01 g) = 60.05 g H3 = 3(1.01 g) = 3.03 g = 63.08 g C5H3? 252 Molecular formula = {C5H3} x 4 = C20H12

Calculating Molecular Formulas for Compounds: Fructose 10. Find the molecular formula for fructose (a sugar found in fruit) from its empirical formula, CH2O, and its molar mass, 180.2 g/mol. The molecular formula is a whole-number multiple of CH2O.

Based on CH2O, empirical formula molar mass = 1(12.01) + 2(1.01) + 1(16.00) = 30.03 g/mol 30.03 g/mol---------???-------->180.2 g/mol

Molecular Formulas May Differ from Empirical Formulas Formaldehyde Empirical Formula, CH 2 O Molecular Formula, CH 2 O Glucose Empirical Formula, CH2O Molecular Formula, C6H12O6 Fructose Empirical Formula, CH2O Molecular Formula, C6H12O6