Percent Yield

-The theoretical yield of a reaction is the amount of products that SHOULD be formed.
-The actual amount depends on the experiment. 
-The percent yield is like a measure of success  

     - How close is the actual amount to the predicted amount?

     % Yield       =  Actual

                           Theoretical   x 100

Evergy & Percent Yield

-Enthalpy is the energy stored in chemical bonds
-Symbol of Enthalpy is H
      - units of joules (J)
- Change in Enthalpy is ΔH
- In exothermic rxns enthalpy decreases
-In endothermic rxns enthalpy increases

Calorimetry
- To experimentally determine the heat released we need to know 3 things
   1) temperature change ( ΔT)
   2) Mass (m)
   3) specific Heat capacity (c)
There are related by the equation: ΔH = mC ΔT

ex: calculate the heat required to warm a cup of 400g of water (C= 4.18J/g °C) from 20.0 °C to 50.0 °C.
 ΔH=mCΔT
 ΔH= (400)(4.18)(50.0-20.0)
 ΔH=50160 J = 50.2 kJ


-Paul Dinh
 

Other Conversions : Volume & Heat

-Volume at STP can be found using this conversion factor 22.4 L/mol
-Heat can be included as a separate terms in chemical reactions (This is called Enthalpy)
-Rxns that release heat are exothermic
-Rxns that absorb heat are endothermic
-Both can be used in stoichiometry

ex: If 5.0 g of Potassium chlorate decompose according to the reaction below, what volume of Oxygen gas (at STP) is produced?

2KClO3 > 2KCL + 3O2
5.0 g x 1mol/122.6 g x 3/2 x 22.4 L/ 1 mol = 1.4 L of Oxygen


ex2: When Zinc reacts with hydrochloric acid exactly 1.00 L of hydrogen gas is produced (at STP). What mass of Zinc was reacted?


Zn +2HCl > ZnCl2 + H2
1.00 L x 1mol/22.4 x 1/1 x 65.4 g / 1 mol = 2.92 g of Zinc


- Paul Dinh

Mass to Mass Conversions

mass to mass problems are additional conversions.

Grams of A                                                                      Grams of B
            |                                                                                     |
Moles of A -  >>Multiplied boy the stoichiometric ratio >>> Moles of B

ex: lead (IV) nitrate reacts with 5.0 g of Pottasium iodide. How many grams of lead (IV) nitrate are required for a complete reaction?

Pb (NO3)4  + 4KI   > 4KNO3  = PbI4
5.0g x (1 mole over 166 g/mol) x 1/4  x (455 over 1 mole) = 3.4 g

ex2: how many grams of O2  are produced ffrom the decomposition of 3.og of Pottasium chlorate?


2KClO3 >> @K + Cl2 + 3O2


3.0 g x (1 mole over 122.6 g  x 3/2 x (32 over 1 mol) = 1.2 g


-Paul Dinh

Mole to Mass & Mass to Mole

ex How many grams of Bauxite (Al2O3) are required to produce 3.5mol of pure aluminum?

Step 1 - Write a balanced equation (2Al2O3 - > 4Al + 3O2)

Step 2 - Change moles of aluminum to moles of bauxite using "what you need over what you have" (2/4)

Step 3 - Multiply the moles of bauxite by the molar mass of bauxite (1.75mol * 102g/mol)

=1.8*10^2g of bauxite

How many moles of Lead(II)Nitrate are consumed when 4.5g of sodium sulphide completely react?

Step 1 - Write a balanced equation (Pb(NO3)2 + Na2S -> PbS + 2NaNO3)

Step 2 - Change grams of sodium sulphide to moles using molar mass ( 4.5/78.1)

Step 3 - Change moles of sodium sulphide to moles of Lead(II)Nitrate using "what you need over what you have" (1/1)

= 0.058mol of Lead(II)Nitrate


- JanCarlo Paysan

Mole to Mole Conversions

- Coefficients in balanced equations tell us the number of moles reacted or produced
- They can also be used as conversion factors
ex 3X + Y -> 2Z    ratios :  x to y 3:1, x to z 3:2, y to z 1:2

- What you need over what you have (stoichiometric ratio)
ex

if 0.15mol of methane are consumed in a combustion reaction how many moles of CO2 are produced?
CH4 + 2O2 -> CO2 + 2H2O
0.15 * 1/1 = 0.15 mol of CO2


When 2.1*10^-2 mol of Aluminum Hydroxide react with Sulphuric acid how many moles of Aluminum sulphate are produced?
2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + 6H2O
2.1*10^-2 * 1/2 = 1.1*10^-2 moles of aluminum sulphate 


How many moles of bauxite (Aluminum Oxide) are required to produce 1.8mol of pure aluminum?
2Al2O3 -> 4Al + 3O2
1.8 * 2/4 = 0.90 mol of aluminum


- JanCarlo Paysan

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions
- it is a generalization of mole conversions to chemical reactions
- understanding the types of chemical reactions is the foundation of stoichiometry

6 types of reactions
1) synthesis
2) decomposition
3) single replacement
4) double replacement
5) neutralization
6) combustion

Synthesis
- A + B = AB
- usually elements -> compounds
ex 2Al + 3F2 -> 2AlF3
     4K + O2   -> 2K2O
     CoCl + 6H2O -> CoCl2 * 6H2O
     2SO2 + O2 -> 2SO3


Decomposition
- AB -> A + B
- Reverse of Synthesis
  - Always assume the compounds decompose into elements during decomposition unless told different
ex 4H3PO4 -> 6H2 + P4 + 8O2
     Mn(C2O4)2 - > Mn + 4C + 4O2
     2C12H22O11 - > 24C + 22H2 + 11O2


Single Replacement
- A + BC -> B + AC
ex Ca + 2KCl -> 2K + CaCl2
     3Mg + 2Al(NO3)3 -> 2Al + 3Mg(NO3)2


Double Replacement
- AB + CD -> AD + CB
ex MgCl2 + K2SO4 -> MgSO4 + KCl
     Mn(ClO4)4 + 2CaCO3 -> Mn(CO3)2 + 2Ca(ClO4)2


Neutralization
- Reaction between an acid and a base
ex H2SO4 + 2KOH -> 2H2O + K2SO4
     3Ca(OH)2 + 2H3PO4 -> 6H2O + Ca3(PO4)2


Combustion
- Reactions of something (usually hydrocarbon) with air
- Hydrocarbon combustion always produces CO2 and H2O
ex CH4 + 2O2 -> CO2 + 2H2O
     2C8H18 + 25O2 -> 16CO2 + 18H2O
     2C11H23OH + 33O2 -> 22CO2 + 24H2O

- JanCarlo Paysan
 

Molecular Formula

Molecular formulas give the actual number of atoms
- if you know the empirical formula, to find the molecular formula you need the molar mass

example
empirical formula - CH2O  molar mass - 60.0g/mol  Determine molecular formula
step 1 - find molar mass of CH2O (30.0g/mol)
step 2 - divide molar mass of molecular formula by molar mass of empirical (60.0/30.0 = 2)
step 3 - multiply all subscripts of the empirical formula by the answer (2)
molecular formula = C2H4O2


- JanCarlo Paysan

Empirical Formulas

Empirical formulas are the simplest formula of a compound
- examples
    - empirical    |   molecular
        C4H9      |     C8H18
        NO2        |     N2O4
        CH7O2   |     C3H21O6
           P2O5     |     P4O10
        C5H11   |     C10H22
        C2H6O  |     C6H18O3
        C5H12O|     C5H12O
        NO2       |     N2O4
        H2O       |      H2O
- show only the simplest ratios, not the actual atoms
   - the empirical formula for chlorine gas is Cl
   - dinitrogen tetraoxide is not equal to N2O4
- To determine the empiricaL formula we need to know the ratio of each element
- To determine the ratio fill in the table below

Atom | Mass | Molar Mass| Moles | Mole/Smallest Mole | Ratio

- The simplest ratio may be decimals. For certain decimals you need to multiply everything by a common number
           Decimal           | Multiplying Coefficient
    0.5                         |                2
    0.33 or 0.66           |                3
    0.25 or 0.75           |                4
    0.2, 0.4, 0.6, 0.8    |                5

- JanCarlo Paysan

Density & Moles

Density is a measurement of mass per volume
formula is -> d=m/v

examples
Water has a density of 1.0g/m. Mass of 11.5 mL of water? How many moles?
d=m/v  d * v = m  1.0 * 11.5 = 11.5g
11.5 * 1mol / 18 = 0.639mol of water


Unknown compound with molar mas of 65.0g/mol. If 0.25mol occupies a volume of 50mL, determine the compound density.
0.25mol * 65.0 / 1mol = 16.25g / 50mL = 0.33g/mol


The density of aluminum is 2.70g/mL. A solid piece of aluminum has a volume of 45.0mL. Find the number of atoms.
d = m/v   d*v = m  2.70 * 45.0 = 121.5g * 1mol / 27g * 6.02*10^23 / 1mol = 2.71*10^24


- JanCarlo Paysan