Download electrical energy and capacitance

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.
1|Page
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
2|Page
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
3|Page
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%
4|Page
CHEMISTRY