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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 STP
Problem #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:
NH3
N2 21
2 mole N2 x 2 mole NH3 = 4mole NH3
1 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 NH3
3 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 Cl2Problem #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 Cl2Problem #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 Cl2Problem #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 NaCl
Problem #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 FORMULAS
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
PERCENT COMPOSITION CALCULATION
Given: Formula Cu2S
Problem #1: Percent Composition of Sulfur
2 Mol Cu X 63.55 Cu = 127.1 g Cu
mol Cu
1 mol S X 32.07g S = 32.07 g S
mol S
Molar mass of Cu2S = 159.2 g
32.07 g S x 100 = 20.15% S
159.2 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