Percent Composition Review The percent composition of elements in any compound can be determined using the comparing ratio equation: n x molar mass of element molar mass of compound x 100 n is the number of moles of the element in 1 mole of the compound %C = 2 x (12.011 g) 46.07 g x 100 = 52.14% C2H6O %H = 6 x (1.008 g) 46.07 g x 100 = 13.13% %O = 1 x (15.99 g) 46.07 g x 100 = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

Percent Composition Review Penicillin, the first of a now large number of antibiotic was discovered accidentally by the Scottish bacteriologist Alexander Aeming in 1928, but he was never able to isolate it as a pure component. This and similar antibiotics have saved millions of lives that might have been lost due to infections. Penicillin-F has the formula C14 H20 N2 SO4. Compute the mass percent of each element. %C = 14 x (12.011 g) 312.39 g x 100 = 53.8 % %S = 1 x (32.066 g) 312.39 g x 100 = 52.14% %H = 20 x (1.008 g) 312.39 g x 100 = 6.45% 10.26% %O = 4 x (15.999 g) 312.39 g x 100 = 13.13% %N = 2 x (14.007 g) 312.39. g x 100 = 8.98% 20.49%

Determining the Formula of a Compound empirical formula – smallest whole number multiple of the components molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH If the molar mass for any compound is known one can get the chemical or molecular formula by the following: a. calculate empirical mass = ∑ atomic mass b. find the multiplication factor = MM/EM c. molecular formula = empirical formula x factor Note: if MM = EM molecular formula = empirical formula

Divide through by the smallest number of mols Determining the Formula of a Compound Caffeine a stimulant found in coffee, tea, and chocolate, contain 49.48 % C, 5.159% H, 28.87 % N, and 16.49 % O by mass has a molar mass 194.2 g/mol. Determine the molecular formula of caffeine. Assume 100g of caffeine mC = 49.48 g nc = 49.48/12.011 = 4.12 mol C mH = 5.159 g nH =5.159/1.008 = 5.118 mol H mN = 28.87 g nN = 28.87/14.007 = 2.06 mol N mO = 16.49 g nO = 16.49/15.999 = 1.03 mol O 4 1.03 5 1.03 2 1.03 1 1.03 Divide through by the smallest number of mols

empirical formula ---C4 H5 N2 O Empirical mass = (4)(12.011) + (5)(1.008) + (2)(14.007) + (1)(15.999) 48.044 + 5.04 + 28.014 + 15.999 = 97.097 MM/EM = 194.2/ 97.097 = 2 molecular formula ---(C4 H5 N2 O)2 molecular formula ---C8 H10 N4 O2

Empirical Formulas from Percent Composition We can use percent composition data to calculate empirical formulas. Assume that you have 100 grams of sample. Benzene (molar mass 78.00 g/mol) is 92.2% carbon and 7.83% hydrogen, what is the empirical formula. If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen.

Empirical Formulas from Percent Composition Calculate the moles of each element: 1 mol C 12.011 g C 92.2 g C × = 7.68 mol C 1 mol H 1.008 g H 7.83 g H × = 7.77 mol H Divide by the smallest number to get the formula. 7.68 = 1.00 1.01 7.77 1 1

Molecular Formulas The empirical formula for benzene is CH. This represents the ratio of C to H atoms of benzene. The actual molecular formula is some multiple of the empirical formula, (CH)n. Benzene has a molar mass of 78.00 g/mol. Find n to find the molecular formula. = 78.00 g/mol 13.019 g/mol n n = 6 so . . . . . . the molecular formula is C6H6.

Review The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule. The molecular formula of benzene is C6H6 The empirical formula of benzene is CH. The molecular formula of octane is C8H18 The empirical formula of octane is C4H9.

Conclusions The percent composition of a substance is the mass percent of each element in that substance. The empirical formula of a substance is the simplest whole number ratio of the elements in the formula. The molecular formula is a multiple of the empirical formula.