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Unit 9Stoichiometry
Chemistry IMr. Patel
SWHS
Topic Outline• MUST have a scientific calculator (not graphing)!!!• Stoichiometry (12.1)• Mole to Mole Stoichiometry (12.2)• Mass to Mole Stoichiometry (12.2)• Mass to Mass Stoichiometry (12.2)• Gas Volume Stoichiometry (12.2)• Percent Yield (12.3) • Solution Concentration (16.2)
PART I:STOICHIOMETRY
Consider a recipe• 2 eggs (E) + 5 cups of flour (F) + 2 cups oil (O) + 4
cups water (W) produce 1 cake (E2F5O2W4 ):
2 E + 5 F + 2 O + 4 W 1 E2F5O2W4
• A recipe can be considered a chemical equation• How many eggs are needed to make 5 cakes?– 10 eggs
• How many cups of water are needed to make 5 cakes?– 20 cups of water
Balanced Equations
• We can obtain that same type of information from a balanced chemical reaction
• Thus it is important that a chemical equation is always balanced before doing any stoichiometry.
• Just as you learned previously, balance equations by checking one element at a time starting with the leftmost element– Goal: The atoms before and after the arrow are the
same!
Ex: Balance the following equation.
___N2 + ___ H2 ___ NH3
(This reaction is utilized in the Haber Process.)
___N2 + ___ H2 ___ NH3
There are two N on the left and one on the right…so we put a 2 in front of NH3.
There are two H on the left and six (2x3) on the right…so we put a 3 in front of H2.
23
Ex: Balance the following equation.
___ CaCl2 + ___ Pb(NO3)3 ___ Ca(NO3)2 + ___ PbCl3
There is one Ca on the left and one on the right…so we do not need to add a coefficient.
There are two Cl on the left and three on the right…so we put a three in front of CaCl2 and two in front of the PbCl3 (2 and 3 both go into 6).
23___ CaCl2 + ___ Pb(NO3)3 ___ Ca(NO3)2 + ___ PbCl3
There is one Pb on the left and two on the right…so we put a two in front of Pb(NO3)3.
2
There are six NO3 on the left and two on the right…so we put a three in front of Ca(NO3)2.
3
Stoichiometry
• Stoichiometry – calculation of quantities in a chemical reaction– Greek “measure elements”– Quantities = reactants and/or products– Predict how much reactant needed and product
produced– Allows for efficient use of resources (save money)– Aids in safety precautions (volume expansion)
Stoichiometry
• Atoms and mass are always the same before and after a reaction (conservation of mass)
1 N2 + 3 H2 2 NH3
2 atom + 6 atoms 8 atoms
1 molec + 3 molec 2 molec
1x(6.02x1023) + 3x(6.02x1023) 2x(6.02x1023)
1 mole + 3 mole 2 mole
Atoms:
Molecules:
Molecules:
Mole Ratio:
Stoichiometry• Stoichiometry allows for the conversion within
a balanced chemical equation using the coefficients as the mole ratios!
Conversion Factor 5: Chemical Equation (mole to mole)
1 N2 + 3 H2 2 NH3
3 mol H2
1 mol N2
2 mol NH3
3 mol H2
1 mol N2
2 mol NH3Mole Ratios:
LITERS
GRAMS
MOLES PARTICLESAvogadroNumber
1 mole
1 mole Molar MassPeriodic Table
1 mole
22.4 L
ChemicalEquation
Stoichiometry
• Three types of stoichiometry problems:– Mole to Mole (1 step)– Mass to Mole (2 steps)– Mass to Mass (3 steps)
• These are all conversion problems (use factor label method). All the same rules still apply in setting up and solving factor label problems!
PART II:MOLE TO MOLE
STOICHIOMETRY
Ex: How many mol of NH3 are produced from 0.600 mol N2?
___N2 + ___ H2 ___ NH3
mol N2
20.600 mol N2= 1.20 mol NH3
Math: (0.600) x (2) / (1) = 1.20
This is Mole ratiofrom chemical equation
1 mol N2 = 2 mol NH3
mol NH3
1
31 2
Ex: How many mol of Al are produced from 3.70 mol Al2O3?
___Al + ___ O2 ___ Al2O3
mol Al2O3
43.70 mol Al2O3= 7.40 mol Al
Math: (3.70) x (4) / (2) = 7.40
This is Mole ratiofrom chemical equation2 mol Al2O3 = 4 mol Al
mol Al
2
34 2
Ex: How many mol of Al are required to consume 6.00 mol O2?
___Al + ___ O2 ___ Al2O3
mol O2
46.00 mol O2= 8.00 mol Al
Math: (6.00) x (4) / (3) = 8.00
This is Mole ratiofrom chemical equation2 mol Al2O3 = 4 mol Al
mol Al
3
34 2
Ex: How many mol of LiC2H3O2 are required to consume 21.5 mol Mg3(PO4)2?
___LiC2H3O2 + ___ Mg3(PO4)2 ___ Li3PO4 + ___Mg(C2H3O2)2
mol Mg3(PO4)2
621.5 mol Mg3(PO4)2 = 129 mol LiC2H3O2
Math: (21.5) x (6) / (1) = 8.00
This is Mole ratiofrom chemical equation
1 mol Mg3(PO4)2 = 6 mol LiC2H3O2
mol LiC2H3O2
1
6 2 31
PART III:MASS TO MOLE
STOICHIOMETRY
g H2
1126.0 g H2 = 41.58 mol NH3
Math: (126.0) x (1) / (2.02) x (2) / (3) = 41.58
This is Molar Massfrom periodic table1 mol H2 = 2.02 g H2
mol H2
2.02 mol H2
2 mol NH3
3
This is Mole ratiofrom chemical equation
3 mol H2 = 2 mol NH3
Ex: How many moles of NH3 are produced from 126.0 g H2?
___N2 + ___ H2 ___ NH331 2
mol AgCl
38.645 mol AgCl= 411.5 g MgCl2
Math: (8.645) x (3) / (6) x (95.20) / (1) = 411.5
This is Molar Massfrom periodic table
1 mol MgCl2 = 95.20 g MgCl2
mol MgCl2
6 mol MgCl2
95.20 g MgCl2
1
This is Mole ratiofrom chemical equation
6 mol AgCl = 3 mol MgCl2
Ex: How many grams of MgCl2 are required to produce 8.645 mol AgCl?
___ MgCl2 + ___ Ag3(PO3) ___ Mg3(PO3)2 + ___ AgCl 3 12 6
mol MgCl2
21.29 mol MgCl2= 346 g Ag3(PO3)
Math: (1.29) x (2) / (3) x (186.87) / (1) = 346
This is Molar Massfrom periodic table
1 mol Ag3(PO3) = 402.58 g Ag3(PO3)
mol Ag3(PO3)
3 mol Ag3(PO3)
186.87 g Ag3(PO3)
1
This is Mole ratiofrom chemical equation
3 mol MgCl2 = 2 mol Ag3(PO3)
Ex: How many grams of Ag3(PO3) are required to consume 1.29 mol MgCl2?
___ MgCl2 + ___ Ag3(PO3) ___ Mg3(PO3)2 + ___ AgCl 3 12 6
g FeCl3
11.15 g FeCl3 = 3.55 x 10-3 mol Fe2O3
Math: (1.15) x (1) / (162.20) x 1) / (2) = 0.00355 = 3.55 x 10-3
This is Molar Massfrom periodic table
1 mol FeCl3 = 162.20 g FeCl3
mol FeCl3
162.20 mol FeCl3
1 mol Fe2O3
2
This is Mole ratiofrom chemical equation
2 mol FeCl3 = 1 mol Fe2O3
Ex: How many moles of Fe2O3 are produced from 1.15 g FeCl3?
___ K2O + ___ FeCl3 ___ KCl + ___ Fe2O3 3 62 1
PART IV:MASS TO MASS
STOICHIOMETRY
g AgCl
16.90 g AgCl =
2.67gCaCl2
Math: (6.90) x (1) / (143.35) x (1) / (2) x (110.98) / (1) = 2.67
This is Mole Ratiofrom chemical equation2 mol AgCl = 1 mol CaCl2
mol AgCl
143.35 mol AgCl
1 mol CaCl2
2
This is Molar Massfrom periodic table
1 mol CaCl2 = 110.98 g CaCl2
g CaCl2110.98
1 mol CaCl2
This is Molar Massfrom periodic table
1 mol AgCl = 143.35 g AgCl
Ex: How many grams of CaCl2 are required to produce 6.90g AgCl?
___ AgNO3 + ___ CaCl2 ___ Ca(NO3)2 + ___ AgCl 2 11 2
g CaCl2
168.10 g CaCl2 = 208.5gAgNO3
Math: (68.10) x (1) / (110.98) x (2) / (1) x (169.91) / (1) = 208.5
This is Mole Ratiofrom chemical equation
1 mol CaCl2 = 2 mol AgNO3
mol CaCl2
110.98 mol CaCl2
2 mol AgNO3
1
This is Molar Massfrom periodic table
1 mol AgNO3 = 169.91 g AgNO3
g AgNO3169.91
1 mol AgNO3
This is Molar Massfrom periodic table
1 mol CaCl2 = 110.98 g CaCl2
Ex: How many grams of AgNO3 are required to consume 68.10g CaCl2?
___ AgNO3 + ___ CaCl2 ___ Ca(NO3)2 + ___ AgCl 2 11 2
g NO
155.6 g NO = 74.1g O2
Math: (55.6) x (1) / (30.01) x (5) / (4) x (32.00) / (1) = 74.1
This is Mole Ratiofrom chemical equation
4 mol NO = 5 mol O2
mol NO
30.01 mol NO
5 mol O2
4
This is Molar Massfrom periodic table
1 mol O2 = 32.00 g O2
g O232.00
1 mol O2
This is Molar Massfrom periodic table
1 mol NO = 30.01 g NO
Ex: How many grams of O2 are required to produce 55.6g NO?
___ NH3 + ___ O2 ___ NO + ___ H2O 4 45 6
PART V:GAS VOLUME
STOICHIOMETRY
Mole to Volume Conversion
• Gases are often measured in volume rather than grams
• A conversion is available between mole and volume only at specific conditions– Only for gases (ideal) – Standard Temperature and Pressure (STP)– 0oC and 1 atm
Conversion Factor 4: 1 mole = 22.4 L
Ex: Convert 12.5 mol Ar to liters of Ar at STP. Use the factor-label method.
mol Ar
L Ar12.5 mol Ar = 280 L Ar
Math: (12.5) x (22.4) / (1) = 280
22.4
1
16.5 mol O2 = 246 L NH3
Math: (16.5) x (2) / (3) x (22.4) / (1) = 554
This is Mole Ratiofrom chemical equation
3 mol O2 = 2 mol NH3
mol O2
2 mol NH3
3
This is Molar Volumeat STP
1 mol NH3 = 22.4 L NH3
L NH322.4
1 mol NH3
Ex: How many liters of NH3 are produced from 16.5 mol O2 at STP?
___ N2 + ___ O2 ___ NH31 3 2
g Fe2O3
1172.0 g Fe2O3 = 72.38 L CO
Math: (172.0) x (1) / (159.70) x (3) / (1) x (22.4) / (1) = 72.38
This is Mole Ratiofrom chemical equation1 mol Fe2O3 = 3 mol CO
mol Fe2O3
159.70 mol Fe2O3
3 mol CO
1
This is Molar Volumeat STP
1 mol CO = 22.4 L CO
L CO22.4
1 mol CO
This is Molar Massfrom periodic table
1 mol Fe2O3 = 159.70 g Fe2O3
Ex: How many liters of CO will be liberated/produced from 172.0g Fe2O3 at STP?
___ Fe2O3 + ___ C ___ Fe + ___ CO 1 23 3
PART VI:PERCENT YIELD
Percent Yield
• Measures the efficiency of a reaction
• Similar to a grade on an assignment
• How well you scored based upon best score
• Tell how “well” the reaction proceeded– Can determine the applicability of the reaction process: – High Yield = Less Money Wasted = More Money Earned
Percent Yield• Theoretical Yield– maximum amount of product that could have been
formed (calculated)
• Actual Yield– amount of product actually obtained when doing
the reaction
% Yield = Actual YieldTheoretical Yield X 100%
Note: % Yield is never 100% - usually less due to error, heat loss, etc.
g CO
195.1 g CO = 2.26 mol Fe
This is Mole Ratiofrom chemical equation
3 mol CO = 2 mol Fe
mol CO
28.01 mol CO
2 mol Fe
3
This is Molar Massfrom periodic table
1 mol CO = 28.01 g CO
Ex: How many moles of Fe are produced from 95.1g CO.
___ Fe2O3 + ___ CO ___ Fe + ___ CO2 1 23 3
Math: (95.1) x (1) / (28.01) x (2) / (3) = 1.59
g Fe2O3
184.8 g Fe2O3 = 70.1 g CO2
This is Mole Ratiofrom chemical equation1 mol Fe2O3 = 3 mol CO
mol Fe2O3
159.70 mol Fe2O3
3 mol CO2
1
This is Molar Massfrom periodic table
1 mol CO2 = 44.01 g CO2
g CO244.01
1 mol CO2
This is Molar Massfrom periodic table
1 mol Fe2O3 = 159.70 g Fe2O3
Ex: How many grams of CO2 are produced from 84.8g Fe2O3.
___ Fe2O3 + ___ CO ___ Fe + ___ CO2 1 23 3
Math: (84.8) x (1) / (159.70) x (3) / (1) x (44.01) / (1) = 1.59
g Fe2O3
184.8 g Fe2O3 = 35.7 L CO2
This is Mole Ratiofrom chemical equation1 mol Fe2O3 = 3 mol CO
mol Fe2O3
159.70 mol Fe2O3
3 mol CO2
1
This is Molar Volumeat STP
1 mol CO2 = 22.4 L CO2
L CO222.4
1 mol CO2
This is Molar Massfrom periodic table
1 mol Fe2O3 = 159.70 g Fe2O3
Ex: Calculate the percent yield if 27.5L CO2 are actually obtained from 84.8g Fe2O3.
___ Fe2O3 + ___ CO ___ Fe + ___ CO2 1 23 3
= 27.5 L35.7 L 77.0 %=X 100%% Yield
g CuCl
112.85 g CuCl = 8.855 g
Cu3N
This is Mole Ratiofrom chemical equation6 mol CuCl = 2 mol Cu3N
mol CuCl
99.00 mol CuCl
2 mol Cu3N
6
This is Molar Volumeat STP
1 mol Cu3N = 204.66 g Cu3N
g Cu3N204.66
1 mol Cu3N
This is Molar Massfrom periodic table
1 mol CuCl = 99.00 g CuCl
Ex: Calculate the percent yield if 5.124g Cu3N are actually obtained from 12.85g CuCl.
___ CuCl + ___ Sr3N2 ___ Cu3N + ___ SrCl2 6 21 3
= 5.124g8.855g 57.87 %=X 100%% Yield
Ex: The theoretical yield for a reaction is 25.0g. When performed, only 21.0g of product was obtained. Calculate the percent yield of the
reaction.
% Yield = Actual YieldTheoretical Yield
= 21.0g25.0g 84.0 %=
X 100%
X 100%
PART VII:SOLUTION
CONCENTRATION
Concentration
• Measure of the amount of solute in a given volume of solvent– Solute = lesser quantity particle– Solvent = greater quantity particle– Dilute = solution with lesser concentration– Concentrated = solution with greater concentration
• These are all qualitative terms
Molarity
• In chemistry, there are many quantitative measurements to describe concentration– Molarity (M)– Molality (m)– Parts per million (ppm)– Parts per billion (ppb)
Molarity
• Molarity (M or [X]) is determined by the following equation:
• Consider: molarity x liters = moles grams• Written as “3.00 M” and read as “3.00 Molar”
molarity = moles soluteliters solution
Ex: What is the concentration of a 3.2 L solution containing 9.3 mol CaCl2?
M = molL = 9.3 mol CaCl2
3.2 L 2.9 M CaCl2=
Ex: What is the molarity when 0.90g NaCl are dissolved to a volume of 0.100 L?
M = molL = 0.015 mol NaCl
0.100 L 0.15 M NaCl=
g NaCl
10.90 g NaCl = 0.015 mol NaCl
mol NaCl
58.50
Ex: How many grams of Al(C2H3O2)3 will I need to dissolve to prepare a 1850mL solution with a
concentration that is 0.75M?
M =mol
L
mol Al(C2H3O2)3
204.111.40 mol Al(C2H3O2)3 = 286 g
Al(C2H3O2)3
g Al(C2H3O2)3
1
mL
11850 mL= 1.85 L
L
1000
Step 1:Convert mL to L
We can only use L
mol = M x L= (0.75M)(1.85L) = 1.40 mol Al(C2H3O2)3
Step 2:Determine #
of moles
Step 3:Convert moles
to grams
Try the following.
1) What is the molarity of a 0.725 L solution containing 0.673g K2C2O4?
1) 5.59 x 10-3 M K2C2O4
or 0.00559 M K2C2O4
Dilutions
• In a lab, chemicals are stored as stock solutions (concentrated)
• We need to dilute these stock solutions with solvent in order to obtain the amount and concentration we desire
• Note: moles do not change• Dilution Equation: – M = Molarity– V = Volume
M1V1 = M2V2
Ex: How many mL of 2.00 M MgSO4 must be diluted to prepare 100.0 mL of 0.400 M MgSO4?
M1V1 = M2V2
(2.00M)(V1) = (0.400M)(100.0mL) V1 = 20.0 mL
Preparation of solution:Need to mix 20.0 mL of the concentrated
MgSO4 with about 80 mL of H2O to get the desired 0.400M MgSO4.
Ex: How many mL of 12.0 M HCl(aq) must be diluted to prepare 250.0 mL of 3.80 M HCl(aq)?
Explain how to prepare this solution.
M1V1 = M2V2
(12.0M)(V1) = (3.80M)(250.0mL) V1 = 79.2 mL
Preparation of solution:Mix 79.2 mL of concentrated HCl(aq) with about (250-79.2)= 170.8 mL H2O to get
the desired 250.0 mL of 3.80M HCl(aq).
Note: Always add acid to water…NEVER add water to acid – very exothermic.