CHAPTER 3: Stoichiometry

Question 1

BALANCING/ED CHEMICAL EQUATIONS (COEFFICIENTS)

Answer 1

• Trial and error
• Must be WHOLE NUMBERS
• At least ONE coefficient will be an ODD NUMBER
• SMALLEST whole number ratio of coefficients

Question 2

COMBINATION REACTIONS

Answer 2

• When two or more substances react to form one product
  1. Metal (representative) and a nonmetal –> IONIC COMPOUND (binary) –> (PREDICTABLE)

2. Nonmetal + Oxygen –> Nonmetal oxide –> UNPREDICTABLE

Question 3

DECOMPOSITION REACTIONS

Answer 3

• In a decomposition, one substance breaks down into two or more substances.
  Ex: CaCO3 (s) –> CaO (s) + CO2 (g)
• Metal hydrogen carbonates (PREDICTABLE):
  Ni(HCO3)2 (s) –>(heat) NiCO3 (s) + H20 (l) + CO2 (g)
• Metal carbonates (PREDICTABLE):
  CaCO3 (s) –>(heat) CaO (s) + CO2 (g)

Question 4

COMBUSTION REACTIONS

Answer 4

• Typically, these reactions are rapid and produce a flame; most often involve hydrocarbons reacting with oxygen in the air.
• PRODUCTS –> CO2 (g) + H20 (g)

Ex:
CH4 (g) + 2O2 (g) –> CO2 (g) + 2H20 (g)

Question 5

AVOGADRO’S NUMBER

Answer 5

6.022 x 10^23 (atoms)

1 mole = Avogadro’s number = 6.022 x 10^23 atoms

Question 6

FINDING EMPIRICAL FORMULAS

Answer 6

Empirical formula of a compound:
- the simplest whole number ratio of atoms of each element in a molecule

TO DETERMINE THE EF, given mass % or masses of each element in a compound:

Question 7

TO DETERMINE THE EF, given mass % or masses of each element in a compound:

Answer 7

1. Convert masses to moles
2. Find the mole ratio (ie. divide by the lowest number of moles)
3. Verify that it is a whole number ratio; if not, multiply each ratio by the lowest multiple so that they are all whole numbers

Question 8

BALANCED CHEMICAL EQUATIONS: Coefficients

Answer 8

Coefficients in a balanced equation give the relative numbers of molecules –> they do not relate masses!

Question 9

STOICHIOMETRY

Answer 9

The process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction

g of X mol of X mol of Y g of Y
MM of X bal. eq. MM of Y

Question 10

LIMITING REACTANT

Answer 10

In a chemical equation, the LIMITING REACTANT (LR) is the reactant present in the smallest stoichiometric amount, ie. it is used up first

Question 11

1ST METHOD TO FIND LR

Answer 11

Divide the number of moles of each reactant by the respective coefficient to find out which number is the smallest.

The smallest –> LR

Question 12

What to do with the LR?

Answer 12

After identifying the LR,
- use its number of moles to calculate the amount of product obtained and the amount of the other reactants that is used.

• To find out what is left of the reactant in excess:
  Subtract the initial quantity of the excess reactant from the quantity used = quantity remaining

Question 13

ICE TABLE (Table of Amounts in moles)

Answer 13

Substance: balanced chemical equation
Below each compound:
I(nitial)      # mol
C(hange)  - / + (coefficient) x
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
E(nd) # mol - (coefficient) x

Using the LR, which is completely used up, final LR = 0
–> # mol - (coefficient) x = 0; x = # mol / (coefficient)

Question 14

THEORETICAL YIELD

Answer 14

The maximum amount of product that can be produced

- The amount of product produced in (calculated) stoichiometric problems

Question 15

ACTUAL YIELD

Answer 15

The amount ACTUALLY obtained in the lab (usually less than the theoretical)

Question 16

PERCENT YIELD

Answer 16

(ACTUAL YIELD / THEORETICAL YIELD) x 100 = % yield