MOLE/ STOICHIOMETRY

By: Tyler Lewis, Michael Stylc and Erich Johnson

DEFINITION OF STOICHIOMETRY The relationship between the relative

quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers.

AVOGADRO’S NUMBER

6.02 x 10^23

WHAT IS A MOLE? The mole is a unit of measurement used in

chemistry to express amounts of a chemical substance.

MOLAR MASS Definition:

The mass, in grams, of one mole of a substance

MASS TO MOLE CONVERSION1 mol = g-formula-mass (periodic table) Problem #1: How many moles in 28 grams of CO2 ?Gram-formula-mass of CO2 1 C = 1 x 12.01 g = 12.01 g

2 O = 2 x 16.00 g =32.00 g 44.01

g/mol 28 g CO2 x 1 mole CO2 = 0.64 mole CO2

44.01 g CO2

Problem #2: How many moles in 480. grams of Fe2O3 ?2 Fe = 2 x 55.85 g = 111.7 g3 O = 3 x 16.00 g = 48.0 g 159.7 g/mol 480. gram Fe2O3 x 1 moleFe2O3 = 3.01 mole Fe2O3

159.7 gram Fe2O3

Problem #3: Find the number of moles of argon in 452 g of argon.Ar = 1 x 39.95 g = 39.95 g 39.95 g/mol

452 gram Ar x 1 mole Ar = 11.3 mole Ar 39.95 gram Ar

PARTICLE TO MOLE CONVERSION1 mol = 6.02 x 1023 particles Problem #1: How many moles in 39.0 x 1028 particles of CO2 ? 39.0 x 1028 particles CO2 x 1 mole CO2 = 6.48 x 1028 mole CO2

6.02 x 1023 particles CO2

Problem #2:How many moles are 1.20 x 1025

particles of phosphorous?

1.20 x 1025 particles P x 1 mole P = 19.9 mole P 6.02 x 1023 particles P

Problem #3: How many moles in 48.0 x 1023 particles of Cs ?

48.0 x 1023 particles Cs x 1 mole Cs = 7.97 mole CS 6.02 x 1023 particles Cs

VOLUME TO MOLE CONVERSION1 mol = 22.4 L for a gas at STPProblem #1: How many moles in 22.4 Liters of CO3 ? 22.4 L CO3 x 1 mole CO3 = 1 mole CO3

22.4 L CO3

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 CO3

22.4 L Ar

Problem #3: How many moles of argon atoms are present in 403.2 L of Xenon gas at STP?

403.2 L Xe x 1 mole Xe = 18.00 mole Xe 22.4 L Xe

MOL-MOL CALCULATIONSN2 + 3 H2 ---> 2 NH3

Problem #1: if we have 2 mol of N2 reacting with sufficient H2, how many moles of NH3 will be produced? ratio to set up the proportion: That means the ratio from the equation is:

NH3N2 2

1

2 mole N2 x 2 mole NH3 = 4mole NH31 mole N2

Initial amount Final Amount

N2 + 3 H2 ---> 2 NH3

Problem #2: Suppose 6 mol of H2 reacted with sufficient nitrogen. How many moles of ammonia would be produced?

6 mole H2 x 2 mole NH3 = 4mole NH33 mole H2

N2 + 3 H2 ---> 2 NH3

Problem #2: We want to produce 2.75 mol of NH3. How many moles of nitrogen would be required?

2.75 mole NH3 x 1 mole N2 = 1.38 mole NH3 2 mole NH3

MASS-MASS CALCULATION 2 AuCl3 ---> 2 Au + 3 Cl2

Problem #1: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl3?

64.0 gram AuCl3 x 1 mole AuCl3 = .211 mole AuCl3 303.35 grams AuCl3

The molar ratio of chemical

The amount of moles of initial chemical

.211 mole AuCl3 x 3 mole Cl2 = .3165 mole Cl2 2 mole AuCl3

The mole ratio of chemicals

.3165 mole Cl x 70.9 gram Cl2 = 22.4 gram Cl2 1 mole Cl2

3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3

Problem #2: Calculate the mass of AgCl that can be prepared from 200. g of AlCl3 and sufficient AgNO3?200. gram AlCl3 x 1 mole AlCl3 = 1.50 mole

AlCl3 133.33 grams AlCl3 1.50 mole AlCl3 x 3 mole AgCl = 4.5 mole

AgCl 1 mole AlCl3 4.5 mole AgCl x 143.35 gram AgCl = 645. gram Cl2 1 mole AgCl

2 Au + 3 Cl2 ---> 2 AuCl3Problem #4: How many grams of AuCl3 can be made from 100.0 grams of chlorine?

100. gram Cl2x 1 mole Cl2 = 1.41mole Cl2 70.9 grams Cl2 1.41 mole Cl2 x 2 mole AuCl3 = .94mole AuCl3 3 mole Cl2

.94 mole AuCl3 x 303.35 grams AuCl3 = 285 gram AuCl3 1 mole AuCl3

PARTICLE-PARTICLE CALCULATION 2 AuCl3 ---> 2 Au + 3 Cl2

Problem #1: How many particles of chlorine can be liberated from the decomposition of 6.02x1023 particles of AuCl3?

6.02x1023 particles AuCl3 x 1 mole AuCl3 = 1 mole AuCl3 6.02x1023 particles AuCl3

The partial ratio of chemical

The amount of moles of initial chemical

1 mole AuCl3 x 3 mole Cl2 = 1.5mole Cl2 2 mole AuCl3

The mole ratio of chemicals

1.5 mole Cl2 x 6.02x1023 particles Cl2 = 9.03x1023 particles Cl2 1 mole Cl2

2 AuCl3 ---> 2 Au + 3 Cl2Problem #2: How many particles of chlorine can be liberated from the decomposition of 150. particles of AuCl3?

150. particles AuCl3 x 1 mole AuCl3 = 2.492x10-22 mole AuCl3 6.02x1023 particles AuCl3

2.492x10-22 mole AuCl3 x 3 mole Cl2 = 3.738x10-22 mole Cl2 2 mole AuCl3

3.738x10-22 mole Cl2 x 6.02x1023 particles Cl2 = 225 particles Cl2 1 mole Cl2

3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3

Problem #3: Calculate the particles of AgCl that can be prepared from 132particals of AlCl3 and sufficient AgNO3?132. article AlCl3 x 1 mole AlCl3 = 2.19x1024 mole

AlCl3 6.02x1023 particle AlCl3

2.19x1024 mole AlCl3 x 3 mole AgCl = 6.57x1024 mole AgCl 1 mole AlCl3

6.57x1024mole AgCl x 6.02x1023 particle AgCl =396. particle Cl2 1 mole AgCl

VOLUME-VOLUME CALCULATION 2 AuCl3 ---> 2 Au + 3 Cl2

Problem #1: How many particles of chlorine can be liberated from the decomposition of 22.4 liters of AuCl3?

22.4 liters AuCl3 x 1 mole AuCl3 = 1 mole AuCl3 22.4 liters AuCl3

The volume ratioThe amount of moles of initial chemical

1 mole AuCl3 x 3 mole Cl2 = 1.5mole Cl2 2 mole AuCl3

The mole ratio of chemicals

1.5 mole Cl2 x 22.4 liters Cl2 = 33.6 liters Cl2 1 mole Cl2

N2 + 3 H2 ---> 2 NH3

Problem #2: Suppose 5.6 liters of H2 reacted with sufficient nitrogen. How many liters of ammonia would be produced?

5.6 liters H2 x 1 mole H2 = .25 moles H2 22.4 liters H2 .25 mole H2 x 2 mole NH3 = .167mole NH3 3 mole H2 .167 mole NH3x 22.4 liters NH3 = 3.7 liters NH3

1 mole NH3

2 Na + Cl2 ---> 2 NaClProblem #3: How many liters of Na are required to react completely with 75.0 liters of chlorine?

75. liters Cl2 x 1 mole Cl2 = 3.35moles Cl2 22.4 liters Cl23.35 mole Cl2 x 2 mole Na = 6.7 mole Na 1 mole Cl2

6.7 mole Na x 22.4 liters Na = 150. liters Na 1 mole Na

PERCENT COMPOSITION FORMULASMass 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

PERCENT COMPOSITION CALCULATION

Given: Formula Cu2SProblem #1: Percent Composition of Sulfur2 Mol Cu X 63.55 Cu = 127.1 g Cu mol Cu1 mol S X 32.07g S = 32.07 g S

mol SMolar mass of Cu2S = 159.2 g

32.07 g S x 100 = 20.15% S159.2 g Cu2S

Problem #2: Calculate the percent composition of carbon in the following: C6H12O6

6 Mol C x 12.01 g/mole = 72.06 g C12 Mol H x 1.008 g/mole=12.096 g H6 Mol O x 16.00 g/mole =96.00 g O

180.2 g C6H12O6

72.06 g C x 100 = 39.99% C180.2 g C6H12O6

Problem #3: Calculate the percent composition of carbon in the following: CO2

1 Mol C x 12.01 g/mole = 12.01 g C2 Mol O x 16.00 g/mole =32.00 g O

44.01 g CO2

12.01 g C x 100 = 27.29% C44.01 g CO26