1. Molar Mass Practice Calculate the molar mass of the following: Hydrogen Fe Iron II sulfate

2. Percent Composition

3. Percent Composition – the percentage by mass of each element in a compound Percent = _______ Part Whole x 100% Percent composition of a compound or = molecule Mass of element in 1 mol ____________________ Mass of 1 mol x 100%

4. Percent Composition Determine the percentage composition of sodium carbonate (Na 2 CO 3 )? Molar MassPercent Composition % Na = 46.0 g 106 g x 100% =43.4 % % C = 12.0 g 106 g x 100% =11.3 % % O = 48.0 g 106 g x 100% =45.3 % Na = 2(23.00) = 46.0 C = 1(12.01) = 12.0 O = 3(16.00) = 48.0 MM= 106 g

5. Percent Composition Determine the percentage composition of ethanol (C 2 H 5 OH)? % C = 52.13%, % H = 13.15%, % O = 34.72% _______________________________________________ Determine the percentage composition of sodium oxalate (Na 2 C 2 O 4 )? % Na = 34.31%, % C = 17.93%, % O = 47.76%

6. Percent Composition Calculate the mass of bromine in 50.0 g of Potassium bromide. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 79.90 g ___________ 119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.

7. Empirical and Molecular Formulas

8. Formulas Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms. Molecular Formula – actual formula of a compound showing the number of atoms present Percent composition allow you to calculate the simplest ratio among the atoms found in compound. Examples: C 4 H 10 - molecular C 2 H 5 - empirical C 6 H 12 O 6 - molecular CH 2 O - empirical

9. Formulas Is H 2 O 2 an empirical or molecular formula? Molecular, it can be reduced to HO HO = empirical formula

10. Calculating Empirical Formula An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula. 1. Determine the number of grams of each element in the compound. 4.151 g Al and 3.692 g O 2. Convert masses to moles. 4.151 g Al 1 mol Al 26.98 g Al =0.1539 mol Al 3.692 g O 1 mol O 16.00 g O = 0.2308 mol O

11. Calculating Empirical Formula When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula. 2.000 g Fe 1 mol Fe 55.85 g Fe = 0.03581 mol Fe 0.573 g O 1 mol O 16.00 g = 0.03581 mol Fe Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g 1 : 1 FeO

12. Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate. 1.3813 g Pb 1 mol Pb 207.2 g Pb = 0.006667 mol Pb 0.00672 gH 1 mol H 1.008 g H = 0.00667 mol H 0.4995 g As 1 mol As 74.92 g As = 0.006667 mol As 0.4267g Fe 1 mol O 16.00 g O = 0.02667 mol O

13. Calculating Empirical Formula The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6. Step 1: In 100.00g of Nylon-6 the masses of elements present are 63.38 g C, 12.38 g n, 9.80 g H, and 14.14 g O. Step 2: 63.38 g C1 mol C 12.01 g C = 5.302 mol C 12.38 g N1 mol N 14.01 g N = 0.8837 mol N 9.80 g H1 mol H 1.01 g H = 9.72 mol H 14.14 g O1 mol O 16.00 g O = 0.8832 mol O

14. Calculating Empirical Formula The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6. Step 3: 5.302 mol C 0.8837 = 6.000 mol C 0.8837 mol N 0.8837 = 1.000 mol N 9.72 mol H 0.8837 = 11.0 mol H 0.8837 mol O 0.8837 = 1.000 mol O 6:1:11:1 C 6 NH 11 O

15. Calculating Molecular Formula A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88g. What is the compound’s molecular formula? Step 1: Molar Mass P = 2 x 30.97 g = 61.94g O = 5 x 16.00g = 80.00 g 141.94 g Step 2: Divide MM by Empirical Formula Mass 238.88 g 141.94g = 2 Step 3: Multiply (P 2 O 5 ) 2 = P 4 O 10

16. Calculating Molecular Formula A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula? C = 12.01 g H = 1.01 g 13.01 g 78 g/mol 13.01 g/mol = 6 (CH) 6 = C 6 H 6