Percent CompositionEmpirical Formulas and Molecular Formulas

Quantification in Chemistry

• Homework: Chapter 11• pg 333 problems 46-50• pg 335 problems 51-53• pg 337 problems 54-57

Warm Up-10/17/13

1. Determine the percent composition of phosphoric acid H3PO4

2. A compound contains 36.84% nitrogen and 63.16% oxygen. What is the empirical formula for this compound?

3. A liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula?

Percent Composition

Analytical chemist work to identify elements in compounds and to determine their percent by mass.

To determine the percentage of an element in a compound you do the following:

mass of element X 100 mass of compound

For Example:A 100-gram sample of a compound contains 55

grams of element X and 45 grams of element Y. What are the percentages of mass for each element?

X: 55g element X X 100 = .55 X 100 = 55% 100 g of compoundand,Y: 45g element Y X 100 = .45 X 100 = 45% 100 g of compound

Percent Composition: Example 2

Determine the Percent Composition in the Compound Sodium Hydrogen Carbonate (NaHCO3).

1 mol Na = 22.99 g Na1 mol H = 1.008 g H1 mol C = 12.01 g C1 mol O = 15.99 x 3 g O (48g) 84.01 g/mol NaHCO3

Percent Composition: Example 2 (continued)Percent Na = 22.99 g Na = .2737 X 100= 27.37% Na 84.01 g/mol NaHCO3

Percent H = 1.008 g H = .012 X 100 = 1.2% H 84.01 g/mol NaHCO3

Percent C = 12.01 g C = .143 X 100 = 14.3% C 84.01 g/mol NaHCO3

Percent O = 15.99 x 3 g O = .5714 X 100 = 57.1% O 84.01 g/mol NaHCO3

Empirical FormulasAn empirical formula is the formula for a

compound with the smallest whole number ratio of its elements.

• Example 1: The percent composition for sulfur oxide is 40.05 % sulfur and 59.95 % oxygen. What is the empirical formula for sulfur oxide?

40.05 g S X 1 mol = 1.29 mol S and 32.07g/mol S 59.95 g O X 1 mol = 3.747 mol O and 15.99g/mol O

Empirical FormulasExample 1 (continued)

1.29 mol S / 1.29 = 1 mol S

3.747 mol O / 1.29 = 3 mol O

The empirical formula for sulfur oxide is:

SO3

Empirical Formulas

Example 2-Determine the empirical formula for methyl acetate (CHO) if the compound has the following percent composition: 48.64% C, 8.16% H and 43.20% O.

48.64g C x 1 mol C = 4.050 mol C 12.01g/mol C8.16g H x 1 mol H = 8.10 mol H 1.008g/mol H43.20g O x 1 mol O = 2.70 mol O 15.99g/mol O

Empirical Formulas Example 2 (continued)

4.050 mol C / 2.70 = 1.5 mol C 8.10 mol H / 2.70 = 3 mol H and 2.70 mol O / 2.70 = 1 mol OMultiply the numbers of moles to turn them into

whole number ratios4.050 mol C / 2.70 = 1.5 mol C x 2 = 3 mol 8.10 mol H / 2.70 = 3 mol H x 2 = 6 mol 2.70 mol O / 2.70 = 1 mol O x 2 = 2 mol• The formula for methyl acetate is: C3H6O2

Warm Up-10/15/13

1. Rubbing alcohol was found to contain 60.0 % carbon, 13.4 % hydrogen, and the remaining mass was due to oxygen. What is the empirical formula of rubbing alcohol?

2. A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine. What is the empirical formula of the indium compound?

Molecular Formulas• Molecular Formulas specify the actual number of

atoms that are in a given compound. • Example 1: What is the molecular formula for methyl

acetate given that its experimental mass was determined to be 144 g/mol?

mass of empirical formula : C3H6O2

C = 12.01 x 3 = 36.03 144 g/mol)/(72 g/mol) = 2 (n)

H = 1.008 x 6 = 6.048 2 x C3H6O2 = C6H12O4

O = 15.99 x 2 = 31.98

72 g/mol