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

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

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

Calculating Empirical Formula An oxide of aluminum is formed by the reaction of g of aluminum with g of oxygen. Calculate the empirical formula. 3. Find ratio by dividing each element by smallest amount of moles moles Al = mol Al moles O = mol O 4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds) O = x 2 = 3 Al = x 2 = 2 therefore, Al 2 O 3

Calculating Empirical Formula A g sample of cobalt reacts with g chlorine to form a binary compound. Determine the empirical formula for this compound g Co g mol -1 Co = mol Co g Cl g mol -1 Cl = mol Cl mol Co mol Cl = 2= 1 CoCl 2

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

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains g lead, g of hydrogen, g of arsenic, and g of oxygen. Calculate the empirical formula for lead arsenate g Pb 1 mol Pb g Pb = mol Pb gH 1 mol H g H = mol H g As 1 mol As g As = mol As g Fe 1 mol O g O = mol O

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains g lead, g of hydrogen, g of arsenic, and g of oxygen. Calculate the empirical formula for lead arsenate mol Pb mol H mol As mol O = mol Pb = 1.00 mol H = mol As = mol O PbHAsO 4

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 g of Nylon-6 the masses of elements present are g C, g n, 9.80 g H, and g O. Step 2: g C g mol -1 C = mol C g N g mol -1 N = mol N 9.80 g H 1.01 g mol -1 H = 9.72 mol H g O g mol -1 O = mol O

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: mol C = mol C mol N = mol N 9.72 mol H = 11.0 mol H mol O = mol O 6:1:11:1 C 6 NH 11 O

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 g. What is the compound’s molecular formula? Step 1: Molar Mass P = 2 x g = 61.94g O = 5 x 16.00g = g g Step 2: Divide MM by Empirical Formula Mass g g = 2 Step 3: Multiply (P 2 O 5 ) 2 = P 4 O 10

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 = g H = 1.01 g g 78 g/mol g/mol = 6 (CH) 6 = C 6 H 6

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

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

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

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