M OLE / S TOICHIOMETRY By: Tyler Lewis, Michael Stylc and Erich Johnson

D EFINITION OF S TOICHIOMETRY The relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers.

A VOGADRO ’ S N UMBER 6.02 x 10^23

W HAT IS A MOLE ? The mole is a unit of measurement used in chemistry to express amounts of a chemical substance.

M OLAR M ASS Definition: The mass, in grams, of one mole of a substance

M ASS TO M OLE CONVERSION 1 mol = g-formula-mass (periodic table) Problem #1: How many moles in 28 grams of CO 2 ? Gram-formula-mass of CO 2 1 C = 1 x g = g 2 O = 2 x g =32.00 g g/mol 28 g CO 2 x 1 mole CO 2 = 0.64 mole CO g CO 2

Problem #2: How many moles in 480. grams of Fe 2 O 3 ? 2 Fe = 2 x g = g 3 O = 3 x g = 48.0 g g/mol 480. gram Fe 2 O 3 x 1 moleFe 2 O 3 = 3.01 mole Fe 2 O gram Fe 2 O 3

Problem #3: Find the number of moles of argon in 452 g of argon. Ar = 1 x g = g g/mol 452 gram Ar x 1 mole Ar = 11.3 mole Ar gram Ar

P ARTICLE TO M OLE CONVERSION 1 mol = 6.02 x particles Problem #1: How many moles in 39.0 x particles of CO 2 ? 39.0 x particles CO 2 x 1 mole CO 2 = 6.48 x mole CO x particles CO 2

Problem #2: How many moles are 1.20 x particles of phosphorous? 1.20 x particles P x 1 mole P = 19.9 mole P 6.02 x particles P

Problem #3: How many moles in 48.0 x particles of Cs ? 48.0 x particles Cs x 1 mole Cs = 7.97 mole CS 6.02 x particles Cs

V OLUME TO M OLE CONVERSION 1 mol = 22.4 L for a gas at STP Problem #1: How many moles in 22.4 Liters of CO 3 ? 22.4 L CO 3 x 1 mole CO 3 = 1 mole CO L CO 3

Problem #2: How many moles of argon atoms are present in 11.2 L of argon gas at STP? 11.2 L Ar x 1 mole Ar =.500 mole CO L Ar

Problem #3: How many moles of argon atoms are present in L of Xenon gas at STP? L Xe x 1 mole Xe = mole Xe 22.4 L Xe

M OL -M OL CALCULATIONS N H 2 ---> 2 NH 3 Problem #1: if we have 2 mol of N 2 reacting with sufficient H 2, how many moles of NH 3 will be produced? ratio to set up the proportion: That means the ratio from the equation is: NH 3 N mole N 2 x 2 mole NH 3 = 4mole NH 3 1 mole N 2 Initial amountFinal Amount

N H 2 ---> 2 NH 3 Problem #2: Suppose 6 mol of H 2 reacted with sufficient nitrogen. How many moles of ammonia would be produced? 6 mole H 2 x 2 mole NH 3 = 4mole NH 3 3 mole H 2

N H 2 ---> 2 NH 3 Problem #2: We want to produce 2.75 mol of NH 3. How many moles of nitrogen would be required? 2.75 mole NH 3 x 1 mole N 2 = 1.38 mole NH 3 2 mole NH 3

M ASS -M ASS CALCULATION 2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #1: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl 3 ? gram AuCl 3 x 1 mole AuCl 3 =.211 mole AuCl grams AuCl 3 The molar ratio of chemical The amount of moles of initial chemical.211 mole AuCl 3 x 3 mole Cl 2 =.3165 mole Cl 2 2 mole AuCl 3 The mole ratio of chemicals.3165 mole Cl x 70.9 gram Cl 2 = 22.4 gram Cl 2 1 mole Cl 2

3 AgNO 3 + AlCl 3 --> 3 AgCl + Al(NO 3 ) 3 Problem #2: Calculate the mass of AgCl that can be prepared from 200. g of AlCl 3 and sufficient AgNO 3 ? 200. gram AlCl 3 x 1 mole AlCl 3 = 1.50 mole AlCl grams AlCl mole AlCl 3 x 3 mole AgCl = 4.5 mole AgCl 1 mole AlCl mole AgCl x gram AgCl = 645. gram Cl 2 1 mole AgCl

2 Au + 3 Cl 2 ---> 2 AuCl 3 Problem #4: How many grams of AuCl 3 can be made from grams of chlorine? 100. gram Cl 2 x 1 mole Cl 2 = 1.41mole Cl grams Cl mole Cl 2 x 2 mole AuCl 3 =.94mole AuCl 3 3 mole Cl 2.94 mole AuCl 3 x grams AuCl 3 = 285 gram AuCl 3 1 mole AuCl 3

P ARTICLE -P ARTICLE CALCULATION 2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #1: How many particles of chlorine can be liberated from the decomposition of 6.02x10 23 particles of AuCl 3 ? 6.02x10 23 particles AuCl 3 x 1 mole AuCl 3 = 1 mole AuCl x10 23 particles AuCl 3 The partial ratio of chemical The amount of moles of initial chemical 1 mole AuCl 3 x 3 mole Cl 2 = 1.5mole Cl 2 2 mole AuCl 3 The mole ratio of chemicals 1.5 mole Cl 2 x 6.02x10 23 particles Cl 2 = 9.03x10 23 particles Cl 2 1 mole Cl 2

2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #2: How many particles of chlorine can be liberated from the decomposition of 150. particles of AuCl 3 ? 150. particles AuCl 3 x 1 mole AuCl 3 = 2.492x mole AuCl x10 23 particles AuCl x mole AuCl 3 x 3 mole Cl 2 = 3.738x mole Cl 2 2 mole AuCl x mole Cl 2 x 6.02x10 23 particles Cl 2 = 225 particles Cl 2 1 mole Cl 2

3 AgNO 3 + AlCl 3 --> 3 AgCl + Al(NO 3 ) 3 Problem #3: Calculate the particles of AgCl that can be prepared from 132particals of AlCl 3 and sufficient AgNO 3 ? 132. article AlCl 3 x 1 mole AlCl 3 = 2.19x10 24 mole AlCl x10 23 particle AlCl x10 24 mole AlCl 3 x 3 mole AgCl = 6.57x10 24 mole AgCl 1 mole AlCl x10 24 mole AgCl x 6.02x10 23 particle AgCl =396. particle Cl 2 1 mole AgCl

V OLUME -V OLUME CALCULATION 2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #1: How many particles of chlorine can be liberated from the decomposition of 22.4 liters of AuCl 3 ? 22.4 liters AuCl 3 x 1 mole AuCl 3 = 1 mole AuCl liters AuCl 3 The volume ratio The amount of moles of initial chemical 1 mole AuCl 3 x 3 mole Cl 2 = 1.5mole Cl 2 2 mole AuCl 3 The mole ratio of chemicals 1.5 mole Cl 2 x 22.4 liters Cl 2 = 33.6 liters Cl 2 1 mole Cl 2

N H 2 ---> 2 NH 3 Problem #2: Suppose 5.6 liters of H 2 reacted with sufficient nitrogen. How many liters of ammonia would be produced? 5.6 liters H 2 x 1 mole H 2 =.25 moles H liters H 2.25 mole H 2 x 2 mole NH 3 =.167mole NH 3 3 mole H mole NH 3 x 22.4 liters NH 3 = 3.7 liters NH 3 1 mole NH 3

2 Na + Cl 2 ---> 2 NaCl Problem #3: How many liters of Na are required to react completely with 75.0 liters of chlorine? 75. liters Cl 2 x 1 mole Cl 2 = 3.35moles Cl liters Cl mole Cl 2 x 2 mole Na = 6.7 mole Na 1 mole Cl mole Na x 22.4 liters Na = 150. liters Na 1 mole Na

P ERCENT C OMPOSITION F ORMULAS Mass of element in sample of compound X 100= % Element in Compound Mass of sample of compound OR Mass of element in 1 mol of compound X 100= % Element in Compound Molar mass of compound

P ERCENT C OMPOSITION C ALCULATION Given: Formula Cu 2 S Problem #1 : Percent Composition of Sulfur 2 Mol Cu X Cu = g Cu mol Cu 1 mol S X 32.07g S = g S mol S Molar mass of Cu 2 S = g g S x 100 = 20.15% S g Cu 2 S

Problem #2 : Calculate the percent composition of carbon in the following: C 6 H 12 O 6 6 Mol C x g/mole = g C 12 Mol H x g/mole= g H 6 Mol O x g/mole =96.00 g O g C 6 H 12 O g C x 100 = 39.99% C g C 6 H 12 O 6

Problem #3 : Calculate the percent composition of carbon in the following: CO 2 1 Mol C x g/mole = g C 2 Mol O x g/mole =32.00 g O g CO g C x 100 = 27.29% C g CO 26