Chapter 4:

Balancing Equations and Stoichiometry

Key Terms & Concepts

 Stoichiometry  Chemical Equations  reactants and products  balancing chemical equations  Chemical Calculations  Limiting Reactant  Theoretical and Percent Yield

Stoichiometry

 Stoichiometry is the study of the quantitative nature of chemical formulas and chemical reactions.

 Stoichiometry is one the the most essential tools in chemistry  It allows to quantify everything from global warming to drug manufacturing

Chemical Equations

 Chemical reactions are represented in a concise manner by chemical equation  For example, when H 2 is formed.

burns in O 2 , H 2 O  The chemical equation for this reaction is: 2 H 2 + O 2  2H 2 O

Chemical Equations

2 H 2 + O 2  2 H 2 O  The compounds on the left of the arrow are called “reactants”  The compounds on the right of the arrow are called “products”  H 2 and O 2 product are reactants, H 2 O is the

Chemical Equations

2 H 2 + O 2  2 H 2 O  Notice that the number of atoms or each element is equal on both sides of the equation  4 H, 2 0  All chemical equations must meet this requirement 

Chemical equations must be balanced!!

 We balance equations by changing coefficients, not chemical formulas

Chemical Equations

Chemical Equations

 Consider the following chemical equation CH 4 + O 2  CO 2 + H 2 O unbalanced  Start with elements that only appear in one compound on either side of the equation  C and H are only in one compound on either side  C is balanced

Chemical Equations

CH 4 + O 2  CO 2 unbalanced + H 2 O  4 H’s in reactants, 2 H’s in products  Put coefficient of 2 in front of H 2 O  4 H’s in reactants, 4 H’s in products CH 4 + O 2  CO 2 + 2 H 2 O unbalanced

Chemical Equations

CH 4 + O 2  CO 2 unbalanced + 2 H 2 O  2 O’s in reactants, 4 O’s in products  Put coefficient of 2 in front of O 2  4 O’s in reactants, 4 O’s in products CH 4 + 2 O 2  CO 2 + 2 H 2 O balanced

Chemical Equations

Chemical Equations

 Consider this equation: C 3 H 8 + O 2  3C + 8H + 2O  CO 2 + H 2 O 1C + 2H + 3O

Chemical Equations

 Balance C and H C 3 H 8 + O 2  3C + 8H + 2O  3 CO 2 3 C + + 8 4 H 2 H + O 10 O

Chemical Equations

 Balance O C 3 H 8 + 5 O 2  3C + 8H + 10 O  3CO 2 + 4H 2 O 3C + 8H + 10O

Chemical Equations

 Example 4.1

Balance the following chemical equations (1) Mg + HCl  MgCl 2 + H 2 (2) K + H 2 O  KOH + H 2 (3) CaCl 2 (4) NaN 3 (5) C 8 H 18 + Na 3 PO 4   Na + N 2 Ca 3 (PO 4 ) 2 + O 2  CO 2 + H + NaCl 2 O

Chemical Equations

 Example 4.1

Balance the following chemical equations (1) Mg + 2 HCl  MgCl 2 + H 2 (2) 2 K + 2 H 2 O  2 KOH + H 2 (3) 3 (4) 2 CaCl 2 NaN 3 + 2 Na 3 PO 4  2  Ca 3 (PO 4 ) 2 Na + 3 N 2 + 6 NaCl (5) 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O

Chemical Equations

 Example 4.2

Write a balanced chemical equation for the following reactions (1) ammonium nitrate decomposes to nitrogen gas, oxygen gas, and water (2) iron reacts with oxygen gas and water to form iron(II) hydroxide (3) ammonia reacts with oxygen gas to produce nitrogen monoxide and water

Chemical Equations

 Example 4.2

Write a balanced chemical equation for the following reactions (1) 2 NH 4 NO 3 (2) 2 Fe + O 2  2 N 2 + O 2 + 2 H 2 O  + 4 H 2 O 2 Fe(OH) 2 (3) 4NH 3 + 5 O 2  4 NO + 6 H 2 O

Chemical Calculations 2 H 2 + O 2  2 H 2 O 2 molecules 1 molecule 2 molecules 2(6.022x10

23 ) molecules 6.022x10

23 molecules 2(6.022x10

23 ) molecules 2 mol 1 mol 2 mol  Stoichiometric coefficients can be interpreted as either number of molecules or number of moles.

Chemical Calculations  Example 4.3

How many moles of water can be produced from 5.25 mol O 2 ?

5.25

mol O 2  2 mol H 2 O 1 mol O 2  10.5

mol H 2 O

Chemical Calculations  Example 4.4

How many moles of oxygen are required to completely react with 8.50 moles of butane, C 4 H 10 ?

2 C 4 H 10  13 O 2  8 CO 2 8.50

mol C 4 H 10  13 mol O 2 2 mol C 4 H 10  10 H 2 O  55.3

mol O 2

Chemical Calculations  We can’t directly measure moles. We can measure mass.

 We can use the stoichiometric coefficients of a reaction to determine the mass relationships.

 However, we must always convert mass to moles.

 We cannot directly compare the masses of reactants and products.

 We can only compare the moles of reactants and products.

Chemical Calculations  The general scheme is:

Chemical Calculations  Example 4.5

Geranyl formate is used as a synthetic rose essence in cosmetics. The compound is prepared from formic acid and geraniol: HCOOH + C 10 H 18 O  C 11 H 18 O 2 + H 2 O A chemist needs to make some geranyl formate for a batch of perfume. How many grams of geranyl formate can a chemist make from 375g of geraniol?

Chemical Calculations C 10 H 18 O  154.2

g/mol C 11 H 18 O 2  182.3

g/mol 375 g  1 mol 154.2

g  1 mol C 11 H 18 O 2 1 mol C 10 H 18 O  182.3

g 1 mol  443 g

Chemical Calculations  Example 4.6

Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by each 1.00 g of lithium hydroxide?

2 LiOH (s) + CO 2 (g)  (l) Li 2 CO 3 (g) + H 2 O

Chemical Calculations  Example 4.6

2 LiOH (s) + CO 2 (g)  Li 2 CO 3 (g) + H 2 O (l) 1.00

g LiOH  1 mol LiOH 23.95

g  1 mol CO 2 2 mol LiOH  44.01

g 1 mol CO 2  0.919

g CO 2

Theoretical and Percent Yield  The amount of product that

can be produced

from a given amount of reactants is the

theoretical yield

.

 However, no reaction goes to actual completion. The amount of products that is

actually produced

from a given amount of reactants is the

actual yield

.

• Some reactants may not react • Reactants may react in an undesired way (side reactions) • May be difficult to remove products from pot

Theoretical and Percent Yield  The extent of the desired reaction is typically reported as the

percent yield

.

Percent Yield  Actual Yield Theoretica l Yield  100%

Theoretical and Percent Yield  Example 4.7

Look back at Example 4.5. If the chemist starts with 375g of geraniol and collects 417g of purified product, what is the percent yield of the synthesis?

417 g  100 % 443 g  94 .

1 %

Theoretical and Percent Yield  Example 4.8

25.0 g of sodium metal is burned in an excess of chlorine gas. What is the theoretical yield of sodium chloride? If 54.8 g of sodium chloride is actually produced, what is the percent yield of the reaction?

2 Na + Cl 2  2 NaCl

Theoretical and Percent Yield 25.0

g Na  1 mol Na 23.0

g  2 mol NaCl 2 mol Na  58.4

g 1 mol NaCl  63.5

g NaCl This is the theoretic al yield

Theoretical and Percent Yield 54.8

g NaCl 63.5

g NaCl  100%  86.3% This is the percent yield

Theoretical and Percent Yield  Example 4.9

Titanium is a strong, lightweight, corrosion resistant metal that is used in aeronautics and bicycle frames. It is prepared by the reaction of titanium (IV) chloride with molten magnesium between 950  C and 1150  C.

TiCl 4 (g) + 2 Mg( l )  Ti(s) + 2 MgCl 2 ( l ) In a certain industrial process 3.54x10

are reacted with 1.13x10

7 7 g of TiCl 4 g of Mg. (a)Calculate the theoretical yield of Ti in grams. (b)Calculate the percent yield if 7.91x10

6 g of Ti are actually produced.

Theoretical and Percent Yield 3.54

 10 7 g TiCl 4  1 mol TiCl 189.7

g 4  1 mol Ti 1 mol TiCl 4  47.9

g 1 mol Ti  8.94

 10 6 g Ti 1.13

 10 7 g Mg  1 mol Mg 24.3

g  1 mol Ti 2 mol Mg  47.9

g 1 mol Ti  1.11

 10 7 g Ti TiCl 4 is the limiting reactant and 8.94

 10 6 g Ti is the theoretic al yield

Theoretical and Percent Yield 7.91

 10 6 g Ti 8.94

 10 6 g Ti  100%  88.5% This is the percent yield

Limiting Reactants  Most reactions do not occur with stoichiometric equivalent amounts of each reactant.

• One reactant is used up first • This reactant is the

limiting reactant

because it limits the amount of products that can be formed

Limiting Reactants  Consider the “ham sandwich” example • one sandwich is made from one slice of ham, one slice of cheese and two slices of bread • How many ham sandwiches can be made from six slices of ham, seven slices of cheese and 14 slices of bread?

• What is the limiting reactant?

Limiting Reactants

Limiting Reactants  If a problem gives specific amounts of two or more reactants it is a limiting reactant problem.

 Determine the amount of product that can be formed from each reactant • The reactant which produces the smallest amount of product is the limiting reactant • The remaining reactants are said to be in excess

Limiting Reactants  Example 4.10

How many moles of water can be formed when 10.0 moles of H 2 reacts with 4.50 moles of O 2 ? What is the limiting reactant?

Limiting Reactants  Example 4.10

2 H 2 + O 2  2 H 2 O 10.0

mol H 2  2 mol H 2 O 2 mol H 2  10.0

mol H 2 O 4.50

mol O 2  2 mol H 2 O 1 mol O 2  9.00

mol H 2 O O 2 is the limiting reactant and 9.00

mol of H 2 O can be formed

Limiting Reactants  Example 4.11

Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and aqueous acetic acid. If 15.0 g of sulfuric acid and 15.0 g of lead (II) acetate are mixed, calculate the number of grams of lead (II) sulfate that can be produced. Also calculate the number of grams of the excess reagent remaining after the reaction is completed.

Limiting Reactants H 2 SO 4 + Pb(CH 3 COO) 2  PbSO 4 + 2 CH 15.0

g H 2 SO 4  1 mol H 2 SO 4 98 .

1 g 3 COOH  1 mol PbSO 4 1 mol H 2 SO 4  303.3

g 1 mol PbSO 4  46.4

g PbSO 4 15.0

g Pb(CH 3 COO) 2  1 mol Pb(CH 3 COO) 2 3 25.3

g  1 mol PbSO 4 1 mol Pb(CH 3 COO) 2  303.3

g 1 mol PbSO 4  1 4 .

0 g PbSO 4 Pb(CH 3 COO) 2 is the limiting reactant and 14.0

g of PbSO 4 can be formed

Limiting Reactants 15.0

g Pb(CH 3 COO) 2  1 mol Pb(CH 3 COO) 2 3 25.3

g  1 mol H 2 SO 4 1 mol Pb(CH 3 COO) 2  98.1

g 1 mol H 2 SO 4  4 .

52 g H 2 SO 4 4.52

g of H 2 SO 4 are used in the reaction leaving 15.0

g 4.52

g  10.5

g of H 2 SO 4 in excess.

End-of-Chapter Exercises  Suggested End-of-Chapter Exercises 5, 9, 10, 14, 17, 20, 22, 28