MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 8: CHEMICAL COMPOSITION (PART 2)
CLASSNOTES
FORMULA CONVERSIONS
An empirical formula may or may not be the same as a compound’s molecular formula. When the
molar mass (molecular weight) of a compound is known, the following equation is used to determine
the molecular formula.
FORMULA CONVERSION
MF = n(EF)
MF = Molecular Formula
n = whole number
EF = Empirical Formula
Example 1. A compound is discovered with a 58.12 g/mol molar mass. Its empirical formula is C2H5.
What is the molecular formula of this compound?
1A.
(1) C = 12.01 amu
(2) H = 1.01 amu
(3) C2 + H5
(4) C2H5 = 2(12.01 amu) + 5(1.01 amu)
(5) EF = C2H5 = 29.07 g/mol
(6) MF = 58.12 g/mol
(7) MF = n(EF)
(8) n = MF / EF
(9) n = 58.12 / 29.07
(10) n = 2
(11) MF = (2)(C2H5)
(12) MF = C4H10
MOLE TO MOLE CONVERSIONS
Chemical equations are quantitative because they tell us how many reactants and products interact in a
given reaction. In particular, chemical reactions are written in mole to mole ratios.
MOLE TO MOLE CONVERSIONS
For example, 3 H2(g) + N2(g)  2 NH3(g) means that 3 moles of hydrogen gas react with 1 mole of
nitrogen gas to produce 2 moles of ammonia gas.
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
MOLE TO MOLE APPLICATIONS
Example 2. How many moles of oxygen are needed to produce 425 moles of octane in the following
reaction?
__ C8H18(g) + __ O2(g)  __ H2O(l) + __ CO2(g)
2A.
(1) Balance the equation:
2 C8H18(g) + 25 O2(g)  18 H2O(l) + 16 CO2(g)
(2) 425 mol C8H18
(25 mol O2)
----------------- = 5.31 x 103 mol O2
2 mol C8H18
MASS TO MASS CONVERSIONS
Mass to mass conversions are similar to mole to mole conversions. In general, they take the form:
Grams A  Moles A  Moles B  Grams B
Example 3. How many grams HCl (aq) are needed to consume 5.50 grams Mg(OH)2?
__ Mg(OH)2(aq) + __ HCl(aq)  __ H2O(l) + __ MgCl2(aq)
3A.
(1) Balance the equation:
Mg(OH)2(aq) + 2 HCl(aq)  2 H2O(l) + MgCl2(aq)
(2) Mass Mg(OH)2  Moles Mg(OH)2:
(3) Mg = 24.31 amu
(4) 2 x O = 32.00 amu
(5) 2 x H = 2.02 amu
(6) Mg(OH)2 = 58.33 g/mol
(7) 5.50 g Mg(OH)2 (1 mol)
---------- = 9.43 x 10-2 mol Mg(OH)2
58.33 g
(8) Moles Mg(OH)2  Moles HCl:
(9) 9.43 x 10-2 mol Mg(OH)2
(2 mol HCl)
-------------------- = 1.89 x 10-1 mol HCl
1 mol Mg(OH)2
(10) H = 1.01 amu
(11) Cl = 35.45 amu
(12) HCl = 36.46 g/mol
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3A. (continued…)
(13) Moles HCl  Mass HCl:
(14) 1.89 x 10-1 mol HCl
(36.46 g)
------------- = 6.88 g HCl
1 mol HCl
LIMITING REACTANT
Limiting reactants are compounds that are completely consumed in a chemical reaction.
THEORETICAL YIELD
Theoretical yield is the amount of product that can be made in a chemical reaction. It is based on the
amount of limiting reagent.
ACTUAL YIELD
Actual yield is the amount of product actually produced in a chemical reaction.
PERCENTAGE YIELD
Percentage yield is the percentage based on the actual versus theoretical yields:
% Yield = (Actual Yield) / (Theoretical)
Examples 4 - 6. 185 grams Fe2O3 and 95.3 grams CO produce 87.4 grams Fe(s) in a laboratory.
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
4. What is the limiting reactant?
4A.
Mass Fe2O3  Moles Fe2O3:
(1) 2 x Fe = 111.7 amu
(2) 3 x O = 48.00 amu
(3) Fe2O3 = 159.7 g/mol
(4) 185 g Fe2O3 (1 mol)
---------- = 1.16 mol Fe2O3
159.7 g
Moles Fe2O3  Moles Fe:
(5) 1.16 mol Fe2O3
(2 mol Fe)
---------------- = 2.32 mol Fe
1 mol Fe2O3
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
4A. (continued…)
From Fe2O3: 2.32 mol Fe
Mass CO  Moles CO:
(6) C = 12.01 amu
(7) O = 16.00 amu
(8) CO = 28.01 g/mol
(9) 95.3 g CO
(1 mol)
----------- = 3.40 mol CO
28.01 g
Moles CO  Moles Fe:
(10) 3.40 mol CO (2 mol Fe) = 2.27 mol Fe
------------3 mol CO
(11) From CO: 2.27 mol Fe
(12) 2.27 mol Fe < 2.32 mol Fe
(13) Limiting reactant = CO
5. What is the theoretical yield of solid iron?
5A.
(1) Theoretical yield = product from limiting reactant
(2) 2.27 mol Fe (55.75 g)
------------ = 126.4 g Fe
1 mol Fe
(3) Theoretical yield = 126.4 g Fe
6. What is the percent yield for this reaction?
6A.
(1) % Yield = Actual Yield / Theoretical Yield
(2) Actual yield = 87.4 g Fe
(3) Theoretical yield = 126.4 g Fe
(4) % Yield = 87.4 g Fe / 126.4 g Fe
(5) Yield = 69.1%
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