1. Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations
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2. Law of Conservation of Mass
“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.”
--Antoine Lavoisier, 1789
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3. Chemical Equations
Chemical equations are concise representations of chemical reactions.
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4. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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5. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.
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6. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.
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7. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in parentheses to the right of each compound.
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8. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.
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9. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule.
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10. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule
Coefficients tell the number of molecules.
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11. Reaction Types
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12. Combination Reactions
In this type of reaction two or more substances react to form one product.
Examples:
2 Mg (s) + O2 (g)  2 MgO (s)
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
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13. Decomposition Reactions
In a decomposition one substance breaks down into two or more substances.
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
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14. Combustion Reactions
These are generally rapid reactions that produce a flame.
Most often involve hydrocarbons reacting with oxygen in the air.
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
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15. Formula Weights
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16. Formula Weight (FW)
A formula weight is the sum of the atomic weights for the atoms in a chemical formula.
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
Formula weights are generally reported for ionic compounds.
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17. Molecular Weight (MW)
A molecular weight is the sum of the atomic weights of the atoms in a molecule.
For the molecule ethane, C2H6, the molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
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18. Percent Composition
One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100
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19. Percent Composition So the percentage of carbon in ethane is…
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%
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20. Moles
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21. Avogadro’s Number 6.02 x 1023 1 mole of 12C has a mass of 12 g.
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22. Molar Mass
By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).
The molar mass of an element is the mass number for the element that we find on the periodic table.
The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).
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23. Using Moles
Moles provide a bridge from the molecular scale to the real-world scale.
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24. Mole Relationships
One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.
One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.
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25. Finding Empirical Formulas
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26. Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition.
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27. Rhyme
Rhyme to remember order of steps to convert % composition  empirical formula:
Percent to mass
Mass to mole
Divide by small
Times till whole
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28. Calculating Empirical Formulas
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass.
What is the empirical formula of vitamin C?
(C3H4O3)
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29. Calculating Empirical Formulas
Assuming g of ascorbic acid,
C: g x = mol C
H: g x = 4.53 mol H
O: g x = mol O
1 mol
12.01 g
16.00 g
1.01 g
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30. Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C: = 1.000
H: = 1.33
O: = 1.000
Multiply all by 3: 3 C:4 H:3 O
3.407 mol
3.406 mol
4.53 mol
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31. Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C3H4O3
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32. Practice Exercise 3.13
A g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain g of carbon, g of hydrogen, and g of oxygen.
What is the empirical formula of this substance?
(C4H4O)
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33. Molecular Formula from Empirical Formula
The empirical formula (relative ratio of elements in the molecule) may not be the molecular formula (actual ratio of elements in the molecule).
e.g. ascorbic acid (vitamin C) has empirical formula C3H4O3.
The molecular formula is C6H8O6.
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34. Molecular Formula from Empirical Formula
To get the molecular formula from the empirical formula, we need to know the molecular weight, MW.
(“molecular molar mass”)
The ratio of molecular weight (MW) to formula weight (FW) of the empirical formula must be a whole number.
(“formula molar mass”)
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35. Sample Exercise (p. 96)
Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C3H4. The experimentally determined molecular weight of this substance is 121 amu.
What is the molecular formula of mesitylene?
(Hint: you do not need to figure out the empirical formula first)
(C9H12)
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36. Practice Exercise 13.14
Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its (molecular) molar mass is 62.1 g/mol.
a) What is the empirical formula of ethylene glycol? (CH3O)
b) What is its molecular formula? (C2H6O2)
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37. Combustion Analysis
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this
C is determined from the mass of CO2 produced
H is determined from the mass of H2O produced
O is determined by difference after the C and H have been determined

38. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products

39. Stoichiometric Calculations
From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

40. Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams

41. Sample Exercise 3.15 (p. 97)
Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g CO2 and g H2O. Determine the empirical formula of isopropyl alcohol.
(C3H8O)
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42. Practice Exercise 3.15
Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a g sample of this compound produces g CO2 and g H2O. What is the empirical formula of caproic acid?
(C3H6O)
Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?
(C6H12O2)
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43. Hydrates Anhydrous = dry, does not contain water
Hydrate = a compound formed by the combination of water with some other substance in which the water retains its molecular state as H2O
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44. Determining Hydrate Formulas
1. Find the formula of a hydrate from experimental mass data:
a) Calculate the number of moles of the anhydrous
salt (mass  mole conversion)
b) Calculate the mass of H2O by subtracting mass of
anhydrous salt from mass of hydrated salt
c) Calculate the moles of H2O
(mass  mole conversion)
d) Calculate ratio: # mol H2O:# mol anhydrous salt
e)  correct formula = salt · x H2O (x = mole ratio)
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45. Determining Hydrate Formulas
2. Find the formula from mass % data:
a) Use the method of using 100 g (from 100 %) to
give mass in grams of the salt and water
Use mass  mole conversion to calculate the
# moles of the anhydrous salt and the # moles of
water.
b) Calculate the mole ratio of H2O:anhydrous salt
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46. Determining Hydrate Formulas
Find the % H2O in a hydrate from
its formula: Use mass percent.
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47. Hydrate practice problems
A mass of 2.50 g of blue hydrated copper sulfate (CuSO4 • x H2O) is placed in a crucible and heated. After heating, 1.59 g white anhydrous copper sulfate (CuSO4) remains. What is the formula for the hydrate? Name the hydrate.
(CuSO4 • 5 H2O)
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48. Hydrate practice problems
A hydrate is found to have the following percent composition:
48.8% MgSO4 and 51.2% H2O.
What is the formula and the name for this hydrate?
(MgSO4 • 7 H2O)
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49. Hydrate practice problems
The formula for a hydrate is SnCl2 • 2 H2O.
What is the mass percent of water in this hydrate?
(15.97% H2O)
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50. Limiting Reactants
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51. How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients.
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).
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52. How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.
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53. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount.
In other words, it’s the reactant you’ll run out of first (in this case, the H2).
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54. Limiting Reactants
In the example below, the O2 would be the excess reagent.
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55. Sample Exercise 3.18 (p. 104)
The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):
N2(g) H2(g)  2 NH3(g)
How many moles of NH3 can be formed from 3.0 mol of N2, and 6.0 mol of H2?
(4.0 mol NH3)
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56. Practice Exercise 3.18 Consider the reaction
2 Al(s) Cl2(g)  2 AlCl3(s).
A mixture of 1.50 mol of Al and 3.00 mol of Cl2 are allowed to react.
a) What is the limiting reactant? (Al)
b) How many moles of AlCl3 are formed? (1.50 mol)
c) How many moles of the excess reactant
remain at the end of the reaction?
(0.75 mol Cl2)
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57. Sample Exercise 3.19 (p. 104)
Consider the following reaction that occurs in a fuel cell:
2 H2(g) + O2(g)  2 H2O(l)
This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two significant figures). How many grams of water can be formed?
(1400 g H2O)
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58. Practice Exercise 3.19
A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:
Zn(s) AgNO3(aq)  2 Ag(s) + Zn(NO3)2(aq)
a) Which reactant is limiting? (AgNO3)
b) How many grams of Ag will form? (1.59 g)
c) How many grams of Zn(NO3)2 will form? (1.39 g)
d) How many grams of the excess reactant will be
left at the end of the reaction? (1.52 g Zn)
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59. Theoretical Yield
The theoretical yield is the maximum amount of product that can be made.
In other words it’s the amount of product possible as calculated through the stoichiometry problem.
This is different from the actual yield, which is the amount one actually produces and measures.
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60. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).
Actual Yield
Theoretical Yield
Percent Yield = x 100
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61. Sample Exercise 3.20 (p. 106)
Adipic acid, H2C6H8O4, is used to produce nylon. It is made commercially by a controlled reaction between cyclohexane (C6H12) and O2:
2 C6H12(l) O2(g)  2 H2C6H8O4(l) H2O(g)
a) Assume that you carry out this reaction starting with 25.0 g
of cyclohexane, and that cyclohexane is the limiting
reactant. What is the theoretical yield of adipic acid?
(43.5 g H2C6H8O4)
b) If you obtain 33.5 g of adipic acid from your reaction, what
is the percent yield of adipic acid?
(77.0%)
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