Moles and Stoichiometry
http://www.skanschools.org/webpages/rallen/files/notes%20-%20unit%205%20-%20moles%20&%20stoich_2012_key.pdf
Elements
Monoatomic one atom of an element that’s stable enough to stand on its own (very rare)- not bonded to anything
Diatomic (diatoms) elements whose atoms always travel in pairs (N2, O2, F2 etc.) bonded to another atom of the same element
Example
What would be the mass of one molecule of oxygen (O2)?
O2 subscript tells you the total # of atoms in molecule/compound
▪ O2= 2 x 16 amu = 32 amu
Calculating Formula Mass & Gram Formula Mass of Compounds
Formula Mass the mass of an atom, molecule or compound in atomic mass units (amu)
Gram Formula Mass the mass of one mole of an atom, molecule or compound in grams
Mole 6.02 x 1023 units of a substance
Calculating Percent Composition
Step 1: calculate the GFM for the compound
Ex. CaCl2 Ca = 1 x 40.08 = 40.08 Cl= 2 x 35. 453 = 70.906
Formula % composition by mass = mass of part x 100
Mass of whole
“Parts”
* Find the percent composition to the nearest 0.1%
Crystal Hydrates
A hydrate is a crystalline compound in which ions are attached to one or more water molecules
Ex. Na2CO3 10H2O Notice how water molecules are built into the
formula Substances without water built into the formula
are called anhydrates
Application of the Moles
Converting from Grams to Moles
Example
How many moles are in 4.75 g of sodium hydroxide (NaOH)?
Na 1 x = Total
O 1 x =
H 1 x =
22.98 22.98
15.99 15.99
1.007 1.007
GFM = 39.977 g/mol
Example
Plug in the given value and the GFM into the “mole calculations” formula and solve for moles
# of moles = 4.75 g = 39.977 g/mol
.119 mol
Converting from Moles to Grams
Example
What is the mass of 4.5 moles of KOH?
Mass KOH = 4.5 moles x 56.087 g/mol
mass KOH = 252.39 g
Chemical Equations
A chemical equation is a set of symbols that state the products and reactants in a chemical reaction
Reactants the starting substances in a chemical reaction (left side of arrow)
Products a substance produced by a chemical reaction (right side of arrow)
Chemical Equations
Ex. 2Na + 2H2O 2NaOH + H2
Chemical equations must be balanced
Law of conservation of mass: mass can neither be created nor destroyed in a chemical reaction
Balancing Chemical Equations
The number of moles of each element on the reactants side must be the same as the number of moles of each element on the products side
Coefficients and subscripts tell us how many moles of each element we have
Balancing Chemical Equations
Ex. Balanced Equation
C + O2 CO2
1 mol of Carbon 2 mol of Oxygen This means the equation is balanced
Each side of the arrow
Balancing Chemical Equations
Ex. Unbalanced Equation H2 + O2 H2O
Coefficient= integer in front of an element or compound which indicates the # of moles present
Subscript = the integer to the lower right of an element which indicates # of atoms present
Species= the individual reactants and products in a chemical reaction
Balancing Chemical Equations
Ex. Unbalanced Equation H2 + O2 H2O
What do we use to balance equations?
Coefficients * we never change the subscripts in a
formula
Balanced equation:
2H2 + O2 2H2O
Method for Balancing Equations
Method for Balancing Equations
Balancing Chemical Equations
Ex.
Balancing Chemical Equations
When balancing chemical equations, polyatomic ions may be balanced as a single element rather than as separate elements as long as they stay intact during the reaction
Balancing Chemical Equations
Al2(SO4)3 + Ca(OH)2 Al(OH)3 + CaSO4
Polyatomic ions= sulfate and hydroxide
Polyatomic ions remain intact during the reaction, can be considered one unit
Balance the equation:
Al2(SO4)3 + 3Ca(OH)2 2Al(OH)3 + 3CaSO4
Types of Chemical Reactions
Type 1: Single Replacement
Reaction where one species replaces another (one species alone on one side and combined on the other)
Ex. 3Ag + AuCl3 3AgCl + Au 2Cr + 3H2SO4 Cr2(SO4)3 + 3H2
Types of Chemical Reactions
Type 2: Double Replacement
Reaction where compounds react, switch partners and produce 2 new compounds
Ex. Pb(NO3)2 + 2 NaCl PbCl2 + 2 NaNO3
Na3PO4 + 3 Ag NO3 Ag3PO4 + 3NaNO3
K2CO3 + 2AgNO3 Ag2CO3 + 2KNO3
Types of Chemical Reactions
Type 3: Synthesis
Reaction where we take more than one reactant and create one product
Ex. 4Al + 3O2 2Al2O3
2H2 + O2 2H2O
Types of Chemical Reactions
Type 4: Decomposition
Reaction where we take one reactant and create 2 products
Ex. BaCO3 BaO + CO2 2H2O2 2H2O + O2 2Bi(OH)3 Bi2O3 + 3H2O
Mole-Mole Problems
A chemical equation = recipe for reaction
Coefficients= tell the amount of reactants and products needed
Reactants in an equation react in specific ratios to produce specific amounts of products
Mole-Mole Problems
Method for solving mole-mole problems
Set up a proportion using your known and unknown values
Cross multiply and solve for your unknown
Always check to make the equation is balanced
Determining Empirical Formulas
Empirical formula= the reduced formula; a formula whose subscripts cannot be reduced any further
Molecular formula= the actual formula for a compound; subscripts represent actual quantity of atoms present
Empirical vs. Molecular Formula
Empirical vs. Molecular Formula
Calculating Empirical Formulas from % Mass
Step 1: always assume you have 100 g sample (the total % for the compound must = 100, so we can just change the units from % to g)
Step 2: convert grams to moles
Step 3: divide all mole numbers by the smallest mole number
Calculating Empirical Formulas from % Mass
ex. a compound is 46.2% mass carbon and 53.8% mass nitrogen. What is its empirical formula?
Step 1: assume 100 g sample 46.2 % C = 46.2 g C 53.8 % N= 53.8 g N
Calculating Empirical Formulas from % Mass
Step 2: Convert grams to moles (have g need moles)
46. 2 g C= 3.85 mol C 53.8 g N = 3.84
12 g/mol C 14 g/mol
* must have whole numbers for subscripts
Calculating Empirical Formulas from % Mass
Step 3: divide each mole number by the smallest mole number (we will round in this step to the nearest integer)
3.85 mol C = 1 3.84 mol N = 1
3.84 mol N 3.84 mol N
Empirical formula: CN
Determining Molecular Formula
We know how to: 1. find an empirical formula from %
mass 2. find an empirical formula from
molecular formula
Next how do we find out the molecular formula from an empirical formula?
Finding Molecular Formula
Ex. a compound is 80.0% C and 20.0% H by mass. If its molecular mass is 75.0 g, what is its empirical formula? What is its molecular formula?
First, determine the empirical formula using the 3 step process
Finding molecular formula
Step 1: assume a 100 g sample 80.0 % C= 80.0 g C 20.0 % H = 20.0 g H
Step 2: convert g to moles 80.0 g C = 6.66 mol C 20.0 g
H= 20 mol H 12 g/mol C 1
g/mol
Finding Molecular Formula
Step 3: divide each by the smallest number of moles and round to the nearest whole
C= 6.66 = 1 H= 20.0 = 3.00
6.66 6.66
Empirical formula: CH3
Finding Molecular Formula
Empirical mass (the mass of 1 mol of CH3)= 15 g
Molecular mass = 75 g
Molecular mass is 5 times larger than empirical mass
Molecular formula must be 5 times larger than empirical formula
Finding Molecular Formula
Multiply all subscripts in our empirical formula by 5
Molecular formula (CH3)= C5H15