Unit 9 Stoichiometry Chemistry I Mr. Patel SWHS. Topic Outline MUST have a scientific calculator (not graphing)!!! Stoichiometry (12.1) Mole to Mole Stoichiometry

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


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


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


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


1 mol Ag3(PO3) = 402.58 g Ag3(PO3)

mol Ag3(PO3)

3 mol Ag3(PO3)

186.87 g Ag3(PO3)

1


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


1 mol FeCl3 = 162.20 g FeCl3

mol FeCl3

162.20 mol FeCl3

1 mol Fe2O3

2


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


1 mol CaCl2 = 110.98 g CaCl2

g CaCl2110.98

1 mol CaCl2


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


1 mol AgNO3 = 169.91 g AgNO3

g AgNO3169.91

1 mol AgNO3


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


4 mol NO = 5 mol O2

mol NO

30.01 mol NO

5 mol O2

4


1 mol O2 = 32.00 g O2

g O232.00

1 mol O2


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


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


1 mol CO = 22.4 L CO

L CO22.4

1 mol CO


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


3 mol CO = 2 mol Fe

mol CO

28.01 mol CO

2 mol Fe

3


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


mol Fe2O3

159.70 mol Fe2O3

3 mol CO2

1


1 mol CO2 = 44.01 g CO2

g CO244.01

1 mol CO2


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


mol Fe2O3

159.70 mol Fe2O3

3 mol CO2

1


1 mol CO2 = 22.4 L CO2

L CO222.4

1 mol CO2


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


1 mol Cu3N = 204.66 g Cu3N

g Cu3N204.66

1 mol Cu3N


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.

Documents

Unit 9 Stoichiometry Chemistry I Mr. Patel SWHS. Topic Outline MUST have a scientific calculator (not graphing)!!! Stoichiometry (12.1) Mole to Mole Stoichiometry