2. 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

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

4. 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 26.98 g mol -1 Al =0.1539 mol Al 3.692 g O 16.00 g mol -1 O = 0.2308 mol O

5. 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. 3. Find ratio by dividing each element by smallest amount of moles. 0.1539 moles Al 0.1539 = 1.000 mol Al 0.2308 moles O 0.1539 = 1.500 mol O 4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds) O = 1.500 x 2 = 3 Al = 1.000 x 2 = 2 therefore, Al 2 O 3

6. Calculating Empirical Formula A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical formula for this compound. 4.550 g Co 58.93 g mol -1 Co = 0.07721 mol Co 5.475 g Cl 35.45 g mol -1 Cl = 0.1544 mol Cl 0.07721 mol Co0.1544 mol Cl 0.07721 = 2= 1 CoCl 2

7. 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

8. 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

9. 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. 0.006667 mol Pb 0.00667 mol H 0.006667 mol As 0.02667 mol O 0.006667 = 1.000 mol Pb = 1.00 mol H = 1.000 mol As = 4.000 mol O PbHAsO 4

10. 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 C 12.01 g mol -1 C = 5.302 mol C 12.38 g N 14.01 g mol -1 N = 0.8837 mol N 9.80 g H 1.01 g mol -1 H = 9.72 mol H 14.14 g O 16.00 g mol -1 O = 0.8832 mol O

11. 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

12. Calculating Molecular Formula A white powder is analyzed and found to have an empirical formula of P 2 O 5. 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

13. 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

14. Example: finding the empirical formula for a compound that was found to contain 80% carbon and 20% hydrogen CH 80g 20g Step two : divide each element by its Mr value simply convert each % to grams Step one : Step three : What is the smaller number? Now divide each of these by this smaller number Now this is the empirical formula for the compound just write out the formula Step four : 1 x C 3 x H CH 3

15. Example : Now find the molecular formula for the compound that was found to contain 80% carbon and 20% hydrogen. Given that the compound was found to have a molar mass of 30 gmol -1 Step five: use the empirical formula you just found ie CH 3 and follow the method below Step seven: now multiply your empirical formula by 2 to get the molecular formula - 2 x CH 3 = C 2 H 6 Step six : divide your given molar mass by the molar mass of the empirical formula CH 3

16. A compound was found to have a % composition of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen find the empirical formula

17. OH 65.3g2g Step two : divide each element by its Mr value simply convert each % to grams Step one : Step three : What is the smaller number? Now divide each of these by this smaller number Now round and just write out the empirical formula Step four : A compound was found to have a % composition of 2.0% Hydrogen, 32.7% Sulfur, 65.3% Oxygen find the empirical formula S 32.7g 412 S H2H2 O4O4