is the study of the relative quantities stoichiometry A. Using Mole Ratios Stoichiometry and Quantitative Analysis of reactants and products in a chemical.

is the study of the relative quantitiesstoichiometry

A. Using Mole Ratios Stoichiometry and Quantitative Analysis

of reactants and products in achemical reaction – stoich video

you can use the number of moles for a given reactant or product to find the moles for any other reactant or product

The equation I use is the following:

WR = Wn

GR = Gn

Example

Consider the following chemical reaction:

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)

a) Write the ratio for all components of the reaction.

b) What amount, in moles, of CO2(g) is formed if 2.50 mol of C2H6(g) reacts?

2:7:4:6

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)n = 2.50 mol n = 2.50 mol

= 5.00 mol

42

c) What amount, in moles, of O2(g) is required to react with 10.2 mol of C2H6(g)?

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)n = 10.2 mol n = 10.2

mol = 35.7 mol

72

d) What amount, in moles, of H2O(g) is formed when 100 mmol of CO2(g) is formed?

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)

n = 100 mmol n = 100 mmol

= 150 mmol

64

= 0.150 mol

gravimetric =mass measurements

B. Gravimetric Stoichiometry

Steps

4. Calculate of the wanted species using

3. Find the of the species using

2. Find the of the species using

Write a including the states. Write the information

balanced equationgiven.

moles givenn=m M

moles wantedmole ratio

mass

m=nM

WR = Wn

GR = Gn

Example 1 Iron is produced by the reaction of iron(III) oxide with carbon monoxide to produce iron and carbon dioxide. What mass of iron(III) oxide is required to produce 1000 g of iron?

CO2(g)  m=?

M = 159.70 g/mol Step 1: n = m

M = 1000 g 55.85g/mol = 17.90… mol

Step 3 : m = nM = (8.95… mol)(159.70 g/mol) = 1429.7225 g   = 1430 g

Fe2O3(s) CO(g) Fe(s) ++ 3 2 31

12

Step 2 : n = 17.90… x= 8.95… mol

m = 1000 g M = 55.85 g/mol

w g

Example 2 The decomposition of the mineral malachite, Cu2(CO3)(OH)2(s), yields copper(II) oxide, carbon dioxide and water vapour. What mass of copper(II) oxide is produced from 1.00 g of malachite?

H2O(g)

m=?M = 79.55 g/mol

Step 3: m = nM = (0.00904… mol)(79.55 g/mol) = 0.7194862 g = 0.719 g

Step 1: n = m M = 1.00 g 221.13 g/mol = 0.00452… mol

Cu2(CO3)(OH)2(s) CuO(s) +CO2(g)+2 1 11

2 1

Step 2: n = 0.00452… x

m = 1.00 g M = 221.13 g/mol

= 0.00904… mol

g w

use to perform calculations

concentrationC. Solution Stoichiometry

c = n v

Example 1 What volume of 14.8 mol/L KOH is needed to react completely with 1.50 L of 12.9 mol/L sulphuric acid?

v=?c= 14.8 mol/L

STEP 2: n = 19.35 x 2

1 = 38.70 mol

v = 1.50 L c = 12.9 mol/L

H2O(l)KOH(aq) H2SO4(aq) K2SO4(aq) ++ 1 1 22

Step 1: n = cv = 12.9 mol/L x 1.50 L = 19.35 mol Step 3: v = n

c = 38.70 mol 14.8 mol/L = 2.6148649 L = 2.61 L

gw

you can use the Law of Combining Volumes when the pressure and temperature conditions are constant for both the

D. Law of Combining Volumes

reactants and the products

Example Consider the following reaction:

N2(g) + 2 O2(g) N2O4(g) 

a) What is the mole ratio for O2(g) and N2O4(g)?

b) What is the volume ratio for O2(g) and N2O4(g)?

c) If 1 mol of N2O4(g) is produced, how many moles of O2(g) must be consumed?

2:1

2:1

1 mol of N2O4(g) 2 = 2 mol of O2(g) 1

d) If 1 L of N2O4(g) is produced, what volume of O2(g) must be consumed?

e) If 2.5 L of N2(g) is consumed, what volume of O2(g) must be consumed?

1 L of N2O4(g) 2 = 2 L of O2(g) 1

2.5 L of N2(g) 2 = 5.0 L of O2(g) 1

if the other information is given for the chemicals in the reaction (eg. mass), use to perform calculations

E. Gas Stoichiometry

ideal gas law

PV = nRT

Example 1 If 300 g of propane burns in a gas barbeque, what volume of oxygen at SATP is required for the reaction?

v=? P = 100.000 kPaT = 298.15 KR = 8.314 kPaL/molK

m = 300 g M = 44.11 g/mol

n = 6.80… mol x 5 1

= 34.0… mol

H2O(g) C3H8(g) O2(g) CO2(g)++ 5 3 41

PV = nRT(100.000kPa)V = (34.0… mol)(8.314)(298.15K) V = 842.94…L = 843 L

n = m M = 300 g 44.11 g/mol = 6.80… mol

g w

limiting reagent = the that determines in a reaction

let’s make double burgers…

1 bun + 2 meat patties 1 double burger2 buns + 4 meat patties 2 buns + 2 meat patties

2 buns + one million meat patties

excess reagent = the that is present in than necessary

21

2excess

excess

limiting

limiting

Chapter 8: Applications of Stoichiometry 8.1 Limiting and Excess Reagents

double burgersdouble burger

double burgers

reactanthow much product can be formed

reactantlarger quantities

Steps

1.

2.

3.

Write the including states.

Calculate the of using

Use to calculate the answer.

balanced chemical equation,

number of moles each reactantn=m or C = n/v M

limiting reagent moles

Do step 2 for both reactants. WR = Wn

GR = Gn

- Use the lower n for step 3. (it’s the limiting reagent)

Example 1  When 80.0 g copper and 25.0 g of sulphur react, which reactant is limiting and what is the maximum amount of copper(I) sulphide that can be produced?

16 Cu(s) + 1 S8(s) 8Cu2S(s)

m=? M = 159.17 g/mol

m = 80.0 g M = 63.55 g/mol

n =

m = 25.0 g M = 256.56 g/mol

n/16 = 0.0786…mol n/1 = 0.0974… mol\ limiting \ excess

1.25…mol

= 0.629… mol

n = 80.0 g 63.55g/mol = 1.25… mol

n = 25.0 g 256.56g/mol = 0.0974… mol

8/16

m = (0.629…mol ) (159.17 g/mol) = 100.17… g  = 100 g

g w

Example 2You are supplied with 9.00 g of KCl and 6.50 g of AgNO3. What is the mass of the precipitate formed when these two chemicals react?

m = ?M = 143.32 g/mol

m = 9.00 g M = 74.55 g/mol

n =

m = 6.50 g M = 169.88 g/mol

n/1 = 0.120…mol n/1 = 0.0382… mol\ excess \ limiting

0.0382… mol

n = 9.00 g 74.55g/mol =0.120… mol

n = 6.50 g 169.88g/mol = 0.0382… mol

1/1

1 KCl(aq) + 1 AgNO3(aq) ® 1 KNO3(aq) + 1 AgCl(s)

m = (0.0382…mol ) (143.32 g/mol) = 5.48 g

Example 3A 200 mL sample of a 0.221 mol/L mercury (II) chloride solution reacts with 100.0 mL of a 0.500 mol/L solution of sodium sulphide. What is the mass of the precipitate formed?

m = ? M = 232.66 g/mol

C = 0.221 mol/L V = 0.200 L

n =

C = 0.500 mol/L V = 0.1000 L

n/1 = 0.0442…mol n/1 = 0.0500… mol\ limiting

0.0442…mol

m = (0.0442…mol ) (232.66 g/mol) =10.3 g

n = CV=(0.221 mol/L) (0.200 L) =0.0442… mol

1/1

1 HgCl2(aq) + 1 Na2S(aq) ® 2 NaCl(aq) + 1 HgS(s)

n = CV=(0.500 mol/L) (0.1000 L) =0.0500… mol

the is called the

the quantity of the product is called the

it is for the predicted and experimental yield to be

predicted or theoretical yieldexpected amount of product

experimental or actual yieldactually obtained

the sameextremely rare

8.2 Predicted and Experimental Yield

factors affecting the experimental yield include: 1. two chemicals can react to give

…called

eg) C(s) and O2(g) can react to form CO2(g) or CO(g)

2. reaction is very

3. methods

4. reactant or product

5. reaction doesn’t go to

different productscompeting reactions

slow

collection and transfer

purity

completion

ideally, percent yield should be as close to as possible

100%

percent yield is calculated as follows:

% yield = actual yield 100 predicted yield

we can also calculate our for the experiment, which tells us

the closer to the percent error is, the

% error = actual yield – predicted yield 100 predicted yield

how far we are from the theoretical yield

% error

better the experiment

zero

Example 1 Calculate the % error and % yield for the following:

predicted mass of ppt = 6.20 gactual mass of ppt = 7.12 g

% error = 7.12 g - 6.20 g x 100 6.20 g

= 14.8 %

% yield = 7.12 g x 100 6.20 g = 115 %

% error = 93.5 g - 100 g x 100 100 g

= -6.50 %

% yield = 93.5 g x 100 100 g = 93.5 %

Example 2 Calculate the % error and % yield for the following:

predicted mass of ppt = 100 gactual mass of ppt = 93.5 g

in solution stoichiometry, sometimes you don’t have enough information to solve the problem on paper

A. Titrations

eg) 10 mL of acetic acid reacts with a 0.202 mol/L NaOH solution. What is the concentration of the acetic acid?

CH3COOH(aq) + NaOH(aq) H2O(l) + NaCH3COO(aq)x mol/L

v = 0.0100 L

c = 0.202 mol/Lv = ??? *****

**** you need this volume in order to solve the problem

8.3 Acid-Base Titration

a is a used to find the of substances so you can calculate

is when a solution of concentration, a , is reacted with a solution of concentration

both need to be since their concentrations will

titration procedure

concentration

standardization knownstandard solution

unknown

strong acids and strong basesstandardized

change over time

volume

a solution called the is transferred from a precisely marked tube called a to a containing the and anindicator

sampleflaskburette

titrant

an indicator (eg. methyl orange, bromothymol blue) is used because a sudden change in colourindicates the

completion of the reaction

the endpoint is the point where the titrant reacts completely with the sample

equivalence point is the volume needed to reach the endpoint

you need a minimum of 3 trials within of each other to ensure results are accurate

0.20 mL

Here is a website to view a virtual titration:

Virtual Titration

Practice using this simulation.

ExampleA 10.00 ml sample of HCl(aq) was titrated with a standardized solution of 0.685 mol/L NaOH(aq). Bromothymol blue indicator was used and it changes from yellow to blue at the endpoint. What is the concentration of the HCl(aq)? Note: HCl(aq) “is titrated with” NaOH(aq)

sample in flask

titrant in burette

Trial 

1 overshoo

t

2 

3 

4

Final Volume (mL)

Initial Volume (mL)

Volume NaOH (mL)

Endpoint Colour

Data Table

average volume =

NaCl(aq) x mol/LV = 10.00 mL = 0.01000 L n = C = n V = =

C = 0.685 mol/L V = mL = L n = = mol

( + + )

3= mL

HCl(aq) + NaOH(aq)→ H2O(l) +1 1 1 1

B. Titration Curvesa plot of the

is called a pH titration curve pH vs. the volume of titrant added

titration curves areS-shaped

when a is titrated with a the will always have a pH of 7 (at 25C)

equivalence pointstrong monoprotic base,strong monoprotic acid

the on the curve is always the pH of the sample

first point

the pH changes at first…this is called thebuffer region

very gradually

as the endpoint is approached, the pH changesvery rapidly

the is (goes to infinity/never ends) to the pH of thetitrant

asymptoticovertitration

Strong Acid Titrated with Strong Base

pH

volume of titrant added (mL)

7

0

14

Strong Base Titrated with Strong Acid

pH

volume of titrant added (mL)

7

0

14

C. Indicators for Titrations can be used to carry out a

titration but it is much more convenient to use an

the indicator should change colour

the pH of the endpoint should

indicatorpH meters

immediately after the endpoint is reached

fall within the pH range of the indicator

Review assignment: p. 328 #1-33 (omit 22)