Percent Composition (Section 11.4) Helps determine identity of unknown compound –Think CSI—they use a mass spectrometer Percent by mass of each element in a compound

To calculate percent composition H2O –Use molar masses to calculate mass of compound 2 x x 16.0 = 18.0 g/mol –Use molar masses to calculate mass of each individual element H = 2 x 1.0 = 2.0 g/mol O = 1 x 16 = 16.0 g/mol –Calculate using the formula H = (2 / 18) x 100 = 11.1 % Hydrogen O = (16/18) x 100 = 88.9 % Oxygen (notice to check your work: = 100%)

Percent Composition of Water H = (2 / 18) x 100 = 11.1 % O = (16/18) x 100 = 88.9 % Even though there are 2 hydrogen atoms in water, most of the mass of water comes from the 1 oxygen atom

Percent Composition Practice KNO 2 Percent composition of N –Molar mass of N = 1 x 14 = 14 g/mol –Molar mass of compound = 1 x x x 16 = 85.1 g/mol –% by mass of N = (14 / 85.1) x 100 = 16.45%

Calculate the percent composition –Phosphorus in Al 2 (PO 4 ) 3 P = 3 x 31.0 = 93 g/mol Al 2 (PO 4 ) 3 = 2 x x x 16 = 339 g/mol 27.43% P = (93 / 339) x 100 = 27.43% –Chlorine in LiCl Cl = 1 x 35.5 = 35.5 g/mol LiCl= 1 x x 35.5 = 42.4 g/mol 83.73% Cl = (35.5 / 42.4) x 100 = 83.73% –Calcium in CaSO 4 Ca = 1 x 40.1 = 40.1 g/mol CaSO 4 = 1 x x x 16 = g/mol 29.44% Ca = (40.1 / 136.2) x 100 = 29.44%

Calculate the percent composition –Sodium in NaCl Na = 1 x 23 g/mol = 23 g/mol NaCl = 1 x x 35.5 = 58.5 g/mol 39.32% Na = (23 / 58.5) x 100 = 39.32% –Sulfur in H 2 SO 4 S = 1 x 32.1 = 32.1 g/mol H 2 SO 4 = 2 x x x 16 = 98.1 g/mol 32.72% S = (32.1 / 98.1) x 100 = 32.72% –Potassium in KNO 3 K = 1 x 39.1 = 39.1 g/mol KNO 3 = 1 x x x 16 = g/mol 38.67% K = (39.1 / 101.1) x 100 = 38.67%

Empirical vs Molecular Formulas Empirical formula = formula with smallest whole number ratio of elements Molecular formula = formula with actual number of atoms of each element EmpiricalMolecular C 4 H 9 C 8 H 18 H 2 O H 6 O 3 We use the percent composition to calculate the empirical and molecular formulas.

How to calculate empirical formula 1.Use mass of elements to calculate moles If given percentages, assume sample is 100 grams. Therefore, percent by mass will equal the actual mass of element. 2.Pick which element has the smallest number of moles 3.Divide each element by the smallest number of moles to get ratio (these numbers become subscripts) 4.Write empirical formula

A blue solid is found to contain 36.43% nitrogen and 63.16% oxygen. What is the empirical formula? If we assume sample is 100 grams, we have g N and g O. 1.Calculate moles of each element N = / 14 = 2.60 mol O = / 16 = 3.95 mol 2.Smallest number of moles = nitrogen with 2.60 mol 3.Divide by smallest number of moles N = 2.60 / 2.60 = 1 x 2 = 2 O = 3.95 / 2.60 = 1.5 x 2 = 3 we must get whole numbers as our answer—that’s why we multiplied by two 4.Empirical formula = N 2 O 3

Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur If we assume sample is 100 grams, we have g Al and g S. 1.Calculate moles of each element Al = / 27 = 1.33 mol S = / 32.1 = 2.00 mol 2.Smallest number of moles = aluminum with 1.33 mol 3.Divide by smallest number of moles Al = 1.33 / 1.33 = 1 x 2 = 2 S = 2.00 / 1.33 = 1.5 x 2 = 3 we must get whole numbers as our answer—that’s why we multiplied by two 4.Empirical formula = Al 2 S 3

Find the empirical formula for a substance that contains 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen If we assume sample is 100 grams, we have g C, 8.16 g H, and g O. 1.Calculate moles of each element C = / 12 = 4.05 mol H = 8.16 / 1 = 8.16 mol O = / 16 = 2.70 mol 2.Smallest number of moles = oxygen with 2.70 moles 3.Divide by smallest number of moles C = 4.05 / 2.70 = 1.5 x 2 = 3 H = 8.16 / 2.70 = 3.0 x 2 = 6 O = 2.70 / 2.70 = 1.0 x 2 = 2 we must get whole numbers as our answer—that’s why we multiplied by two 4.Empirical formula = C 3 H 6 O 2