1. Chapter 3 Mass Relationships in Chemical Reactions
Chemistry: McMurry and Fay, 6th Edition
Chapter 3: Mass Relationships in Chemical Reactions
4/20/2017 6:27:53 PM
Chapter 3 Mass Relationships in Chemical Reactions
Copyright © 2011 Pearson Prentice Hall, Inc.

2. Chemical Reaction In a chemical reaction
a chemical change produces one or more new substances.
there is a change in the composition of one or more substances.

3. Chemical Reaction In a chemical reaction
old bonds are broken and new bonds are formed.
atoms in the reactants are rearranged to form one or more different substances.

4. Chemical Reaction In a chemical reaction the reactants are Fe and O2.
the new product Fe2O3 is called rust.

5. Chemical Equations A chemical equation gives
the formulas of the reactants on the left of the arrow.
the formulas of the products on the right of the arrow.
Reactants Product
O2 (g)
CO2 (g)
C(s)

6. Symbols Used in Equations
Symbols in chemical
equations show
the states of the reactants.
the states of the products.
the reaction conditions.
TABLE

7. Chemical Equations are Balanced
In a balanced
chemical reaction
no atoms are lost or gained.
the number of reacting atoms is equal to the number of product atoms.

8. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Balancing Chemical Equations
A balanced chemical equation shows that the law of conservation of mass is adhered to.
In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical.
2NaCl(s)
2Na(s) + Cl2(g)
left side:
2 Na
2 Cl
right side:
2 Na
2 Cl
Remember, a chemical reaction can be thought of as a rearrangement of the atoms to form new compounds.
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9. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Balancing Chemical Equations
Write the unbalanced equation using the correct chemical formula for each reactant and product.
H2O(l)
H2(g) + O2(g)
Find suitable coefficients—the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation.
2H2O(l)
2H2(g) + O2(g)
Copyright © 2008 Pearson Prentice Hall, Inc.

10. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
Balancing Chemical Equations
4/20/2017
Reduce the coefficients to their smallest whole-number values, if necessary, by dividing them all by a common denominator.
4H2O(l)
4H2(g) + 2O2(g)
divide all by 2
Since balancing at this level is often trial-and-error, the coefficients need to be checked to make sure they are the smallest whole-number ones. An alternative is to use fractions but some students do not do well with fractional coefficients at this stage.
2H2O(l)
2H2(g) + O2(g)
Copyright © 2008 Pearson Prentice Hall, Inc.

11. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Balancing Chemical Equations
Check your answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation.
2H2O(l)
2H2(g) + O2(g)
left side:
4 H
2 O
right side:
4 H
2 O
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

12. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Balancing Chemical Equations
Do not change subscripts when you balance a chemical equation. You are only allowed to change the coefficients.
H2O(l)
H2(g) + O2(g)
unbalanced
2H2O(l)
2H2(g) + O2(g)
H2O2(l)
H2(g) + O2(g)
Balancing chemical equations only determines the proper numerical ratios between all of the substances (both reactants and products). It does not change any of the substances.
Balanced properly
Chemical equation changed!
Copyright © 2008 Pearson Prentice Hall, Inc.

13. Examples Balance the coefficients from reactants to products.
__N2(g) __H2(g) __ NH3(g)
B. __Co2O3(s) + __ C(s) __Co(s) + __CO2(g)
Write a balanced equation for the reaction between carbon dioxide and potassium hydroxide to form potassium carbonate and water.

14. Chemical Symbols on Different Levels
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Chemical Symbols on Different Levels
2H2O(l)
2H2(g) + O2(g)
2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water.
microscopic:
0.56 kg of hydrogen gas react with 4.44 kg of oxygen gas to yield 5.00 kg of liquid water.
macroscopic:
How can we relate the two to each other?
The mole relates microscopic to macroscopic.
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15. Avogadro’s Number and the Mole
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Avogadro’s Number and the Mole
Molecular Mass: Sum of atomic masses of all atoms in a molecule.
Formula Mass: Sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.
HCl:
1.0 amu amu = 36.5 amu
C2H4:
2(12.0 amu) + 4(1.0 amu) = 28.0 amu
Copyright © 2008 Pearson Prentice Hall, Inc.

16. Molar Mass of Na2SO4 Calculate the molar mass of Na2SO4. Element
Number of Moles
Atomic Mass
Total Mass in Na2SO4 Na S O

17. Stoichiometry: Chemical Arithmetic
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Stoichiometry: Chemical Arithmetic
Stoichiometry: The relative proportions in which elements form compounds or in which substances react.
aA + bB
cC + dD
Moles of A
Moles of B
Grams of B
Grams of A
Molar Mass of A
Molar Mass of B
Mole Ratio Between A and B (Coefficients)
Definition from the Oxford Dictionary of Chemistry, Oxford University Press, 2000.
Its origin is Greek for “element” “measure.”
“grams-to-moles-to-moles-to-grams”
You need the coefficients from the balanced chemical equation.
You can relate any of the substances in the chemical equation to each other.
Copyright © 2008 Pearson Prentice Hall, Inc.

18. Stoichiometry: Chemical Arithmetic
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Stoichiometry: Chemical Arithmetic
Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas:
2NaOH(aq) + Cl2(g)
NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react with 25.0 g Cl2?
Grams of
Cl2
Moles of
Cl2
Moles of
NaOH
Grams of
NaOH
Molar Mass Cl2
Mole Ratio
Molar Mass NaOH
Copyright © 2008 Pearson Prentice Hall, Inc.

19. Stoichiometry: Chemical Arithmetic
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Stoichiometry: Chemical Arithmetic
The commercial production of iron from iron ore involves the reaction of Fe2O3 with CO to yield iron metal plus carbon dioxide:
Fe2O3(s) + 3CO(g) Fe(s) + 3CO2(g)
Predict how many grams of CO will react with moles Fe2O3?
Remind students about significant figures.
Copyright © 2008 Pearson Prentice Hall, Inc.

20. Yields of Chemical Reactions
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Yields of Chemical Reactions
Actual Yield:
Theoretical Yield:
The amount actually formed in a reaction.
The amount predicted by calculations from the
limiting reactant.
actual yield
Percent Yield =
x 100%
theoretical yield
Maximum percent yield is 100%.
Copyright © 2008 Pearson Prentice Hall, Inc.

21. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Reactions with Limiting Amounts of Reactants
Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant.
Excess Reactant: Any of the other reactants still present after determination of the limiting reactant.
Copyright © 2008 Pearson Prentice Hall, Inc.

22. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Reactions with Limiting Amounts of Reactants
At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers:
C2H4O(aq) H2O(l)
C2H6O2(l)
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23. Examples
How many sandwiches can be made from 10 slices of bread and 8 slices of cheese? Which one is the limiting reactant?
The balanced chemical equation is
2Bd + Ch Bd2Ch

24. Limiting reactant and theoretical yield
Consider the reaction A + B  AB Step1: Grams A  moles A  moles AB Grams B  moles B  moles AB Step 2: Determine which reactant (A or B) produced the least amount of product (AB) and label it as limiting reactant Step 3: Calculate the theoretical yield of AB (or least moles of AB  grams AB)

25. Excess reactant
Whichever reactant (A or B) produced the most moles of product (AB) is the excess reactant
most moles of product - least moles of product = moles of product in excess
moles of product in excess --> moles excess reactant --> grams of reactant = left over from reaction

26. Example Ammonia, NH3, can be synthesized by the following reaction:
2 NO(g) H2(g)  2 NH3(g) H2O(g)
Starting with 86.3 g NO and 25.6 g H2
Determine the limiting reactant
Find the theoretical yield of NH3 in grams

27. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Reactions with Limiting Amounts of Reactants
Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation:
Li2O(s) + H2O(g)
2LiOH(s)
If 80.0 g of water are to be removed and 65.0 g of Li2O are available, which reactant is limiting? How many grams of LiOH are produced? How many grams of excess reactant remain?
Copyright © 2008 Pearson Prentice Hall, Inc.

28. Percent Composition and Empirical Formulas
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Percent Composition and Empirical Formulas
Percent Composition: Expressed by identifying the elements present and giving the mass percent of each.
Empirical Formula: It tells only the ratios of the atoms in a compound.
molecular mass
empirical formula mass
multiple =
Copyright © 2008 Pearson Prentice Hall, Inc.

29. Steps in determine the Empirical formula
Step 1: Obtain the mass of each element (in grams)
E.G 100% = 100g therefore mass percent is the same numerical value in grams
Step 2: Determine the numbers moles of each atom present
Step 3: Divide the smallest moles by numbers of each atoms to obtain the closet integer as possible.
Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then multiply with a factor to get the nearest integer as possible.
E.g 1.5 x 2 = 3.0 atoms
1.33 x 3 = = 4 atoms

30. Example
A compound containing nitrogen and oxygen is decomposed in the laboratory and produces 24.5 g nitrogen and 70.0 g oxygen. Calculate the empirical formula of the compound.

31. Example
An unknown sample gives the following mass percent: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula?

32. Examples
Ibuprofen, an aspirin substitute, has the following mass percent composition: C 75.69%, H 8.80% an O 15.51%. What is the empirical formula of ibuprofen?

33. Molecular formula
Molecular Formula: It tells the actual numbers of atoms in a compound. It can be either the empirical formula or a multiple of it.
molecular mass
empirical formula mass
multiple =

34. Example
Butanedione – a main component in the smell and taste of butter and cheese – contains the elements carbon, hydrogen, and oxygen. The empirical formula of butanedione is C2H3O and its molar mass is g/mol. Find its molecular formula

35. Percent Composition and Empirical Formulas
Chapter 3: Formulas, Equations, and Moles
4/20/2017
Percent Composition and Empirical Formulas
A colorless liquid has a composition of 84.1 % carbon and 15.9 % hydrogen by mass. Determine the empirical formula. Also, assuming the molar mass of this compound is g/mol, determine the molecular formula of this compound.
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36. Determining Empirical Formulas: Elemental Analysis
Chapter 3: Formulas, Equations, and Moles
Determining Empirical Formulas: Elemental Analysis
4/20/2017
Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO2 and H2O, which are separated and weighed by an automated instrument called a gas chromatograph.
hydrocarbon + O2(g)
xCO2(g) + yH2O(g)
carbon
hydrogen
If the unknown compound also contains oxygen, then the masses of carbon and hydrogen are determined first. The mass of oxygen is determined by subtracting those masses from that of the unknown compound.
Copyright © 2008 Pearson Prentice Hall, Inc.

37. Determining Empirical Formulas: Elemental Analysis
Step 1: g CO2  moles CO2  moles C  g C
Step 2: g H2O  moles H2O  moles H  g H
Step 3:
g O (or any missing atom) = grams sample – (g C + g H)
Step 4: determine the moles of O or the missing atom
Step 5: Follow all steps from determining the empirical formula

38. Example
A compound contains only carbon, hydrogen, and oxygen. Combustion of mg of compound yields mg CO2 and mg H2O. The molar mass of the compound is g/mol.

39. Examples
Paradichlorobenzene, a common ingredient in solid air fresheners, contains only C, H and Cl and has a molar mass of about 147g/mol. Given that combustion of 1.68 g of this compound produces 3.02 g of CO2 and g H2O, determine its empirical and molecular formulas