CHEMISTRY SHENANIGANS

Definitions
Solution: A homogeneous mixture
Solute: The one present in smaller amount
Solvent: The one present in greater amount
Concentration: Amount of Solute
Amount of Solvent

Some units for concentration
g, g, mg, mg
ml L L ml

The most common (and useful) units are
mol = Molarity = Molar Concentration
L

THE FOLLOWING ARE ONLY FOR AQUEOUS SOLUTIONS & DO NOT APPLY TO GASES
M= mol
L
mol= M(L)

L= mol
M

NOTE: Think triangle of density/mass/volume

and now an informational video..

Today was the last class before mid-term exams. In class, we discussed the empirical formula and went over what would be on the exam.

You need to know:

1. Binary Ionic
2. Multivalent
3. Polyatomic
4. Acids/Bases
5. Hydrates
6. Molecular Compounds
7. Classical Naming System

Mole Conversions

1. Mole Conversion Table
2. Mole to mass and volume (gases atSTP)
3. Density
4. Number of Molecules
5. Atoms

Significant Digits
SI system
Classification
Lab Safety & Labs

Here are some videos to help!

Molecular

• P₄O₁₀
• C₁₀H₂₂
• C₆H₁₈O₃
• C₅H₁₂O
• 
  
  N₂O₄
  

• 
  
  P₂O₅
  
  
  
• 
  
  C₅H₁₁
  
  
  
• 
  
  C₂H_6O
  
  
  
• 
  
  C₅H₁₂O
  
  
  
• 
  
  NO₂
  
  
  

ex. (in-class) A sample of an unknown compound is analyzed and found to contain 8.4g of C and 2.1g of H, and 5.6g of O.
C_8.4H_2.1O_5.6  <--------- this is wrong because it is not in whole numbers!

Element	Mass (g)	Atomic Mass	Moles
C	8.4	12.0 g/mol	8.4g ÷ 12.0 g/mol = 0.7 mol
H	2.1	1.0 g/mol	2.1g ÷ 1.0 g/mol = 2.1 mol
O	5.6	16.0 g/mol	5.6g ÷ 16.0 g/mol = 0.35 mol

ex. A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?

Element	Mass (g)	Atomic Mass	Moles
Ca	13.5	40.1 g/mol	13.5g ÷ 40.1 g/mol = 0.337 mol
O	10.8	16.0 g/mol	10.8g ÷ 16.0 g/mol = 0.269 mol
H	0.675	1.0 g/mol	0.675g ÷ 1.0 g/mol = 0.675 mol

Here's a helpful hint for putting your ratios into whole numbers:
If the ratio ends in... 
~0.5 multiply by 2
~0.33 or ~0.66 multiply by 3
~0.25 or ~0.75 multiply by 4
~0.2, ~ 0.4, ~0.6, ~0.8 multiply by 5

In today's class, we went over the percent mass of elements in compounds and did a few examples.

The percent composition: the percent mass of each element in a compound

How to find the percent composition:
1. Find the total molar mass of the compound.
2. Next, find the molar mass of each element. (Remember, if there is a subscript, you must multiply the element's mass from the Periodic table by that subscript.)
3. Finally, divide each element's molar mass by the toal molar mass.
Note: Do not round right away to a whole number!

ex. (from in-class) Find the percent composition of K₂Cr₂O₇
step 1:  2(39.10) + 2(52.0) + 7(16.0) = 294.2 g/mol (the total molar mass)
step 2 & 3:
2(39.10) ÷ 294.2 g/mol = K 27%
2(52.0) ÷ 294.2 g/mol = Cr 35%
7(16.0) ÷ 294.2 g/mol = O 38%

ex. What is the percent of carbon in glucose, C₆H₁₂O₆?
step 1: 6(12.0) + 12(1.0) + 6(16.0) = 180.0 g/mol (the total molar mass)
step 2 & 3:
6(12.0) ÷ 180.0 g/mol = C 40%

Here's a video to recap! Enjoy!

Today in class, we reviewed the past topics of conversions to and from Moles, Volume, Molar Mass, Density, and Atoms and Molecules. Don't forget about Mr. Doktor's chart!

Density <----> Molar Mass <----> Mole <-- (conversion factor is 6.02 x 10²³)--> Atoms and Molecules (the subscripts)
                                                  
Mole <--(conversion factor is 22.4 L/mol or vice versa)--> Volume                                                   

Below, we have some examples..

Ex. (in class) 1.25 L of an unknown gas has a mass of 3.47 g. What is the molar mass if it is 22.4 L/mol?

1.25 L x 1 mol  =  0.0558 mol
              22.4 L

Molar Mass  =    3.47 g       =  62.2 g/mol
                       0.0558 mol

Ex. (in class) 250 mL of a gas which is known to contain one sulphur atom and an unknown number of fluorides has a mass of 1.63 g at STP. Find the molar mass, then the number of fluoride atoms. (SFx)

a) the Molar Mass
0.25 L x   1 mol   =  0.011 mol
                22.4L

1.63g / 0.011 mol = 146.048 g/mol

b) the number of fluoride atoms
1(32.1) + x(19) = 146
                  19x = 146-32.1
                      x = 6
Ex.Calculate the volume of 11.2 mol of HCN (g) at STP.
11.2 mol x 22.4 L  =  250.88 L = 251 L
                   1 mol

Ex. What is the volume of 1.3 g of NO₂ at STP?
1N + 2O --> 1 (14) + 2(16) = 46 g/mol
1.3 g  x  1 mol   x   22.4 L  = 0.633 L
                46 g       1 mol


Here are two videos to help you review!

-For now, all bases will be aqueous solutions of ionic hydroxides -NaOH
-Ba(OH)
2

-Use the cation name followed by hydroxide
-Sodium Hydroxide
-Barium Hydroxide
Example: Write the name of the following
-HI(aq)- Hydrogen Iodide
-H PO (aq)- Phosphoric Acid
3 4
-H PO(aq)- Phosphorous Acid
3 3
-HNO (aq)- Nitric Acid
3
-Mg(OH) (aq)- Magnesium Hydroxide
2
-HBr(aq)- Hydrobromic Acid

The Following are Oxalic Acids
-HOOCCOOH(aq)
2-
OOCCOO
2-
C O
2 4