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Moles and Stoichiometry

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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


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


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


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


Ex.


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


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


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


Type 3: Synthesis

Reaction where we take more than one reactant and create one product

Ex. 4Al + 3O2 2Al2O3

2H2 + O2 2H2O


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


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


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


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


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


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


Multiply all subscripts in our empirical formula by 5

Molecular formula (CH3)= C5H15

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