Module 2 Introduction To Quantitative Chemistry
IQ 1 What Happens In Chemical Reactions
IQ 2 Moles and Measurement
IQ 3 Concentration and Standard Solutions/Dilution
IQ 4 Gas Laws
Measuring of mass before and after reaction
Increase or decrease
Why
Gain or loss of substance
Examples
Evaporation
New substance formed with unweighed substance
How
Measure mass of original substance
Measure volume of gas
Stochiometry
Law of Conservation of Mass
Matter can not be created or destroyed only transformed
Therefore all chemical equations must be balanced
LHS = RHS
Change Coefficents
Add States of matter
Moles
And all mass produced must = Mass reacted with
A unit of measurement representing the amount of particles in a sample
Moles x Avogadro's Constant = Number of Particles
Avogadro's Constant
6.022x10^23
Mass of Sample / Molecular or Molar Mass
Molar mass
The atomic mass of all atoms in a substance
Limiting Reagents
The Limiting reagent is the reactant that (depending on moler ratios) is fully reacted with first in a reactions
Example: If a 5 moles of Cyanide requires 2 moles of Water to Fully be reacted with but 2 Moles of water 5 Moles of Cyanide to be fully reacted with. Creating 3 Moles of Toxic Solution and 2 Moles of Panadol
A: If there is 1.5 Moles only of water avaliable
B : If there is only 3 moles of Cyanide available
The Limiting Reagent is Cyanide as Cyanide will be used first before the 2 moles of water is used fully
The limiting reagent is water as water will be used first before all 5 moles of Cyanide is reacted fully
Now How many Moles is reacted with
Example A
Example B
So 2 Moles of water reacted with and X moles of Cyanide reacted with.
So 3 Moles of Cyanide reacted with and X moles of water reacted with.
How many Moles is water used
How Many moles of Cyanide used
So to determine how much water is used , the moles of cyanide is divided by Moles needed for 2 moles of water and then multiplied by moles of water
So 3 x 2/5
= 6/5 or 1.2 moles of water reacted
Moles of Water divided by moles needed to react with 5 Moles of Cyanide multiplied by moles of Cyanide
So 1.5 x 5/2
3.75 Moles of Cyanide reacted
Now How many moles of Products
Example A
Example B
First Off, Moler Ratio is 5 : 2 : 3 : 2
so 3.75 moles of Cyanide = 5 so 0.75 moles = 1
So 3 moles of Cyanide = 5 So 3/5 = 1
So Product 1 = 3 x 3/5 = 9/5 moles of Toxic Solution
So Product 1 = 3 x 0.75 = 2.25 moles of Toxic Solution
Product 2 = 2 x 0.75 = 1.5 moles of Panadol
Product 2 = 2 x 3/5 = 6/5 = 1.2 moles
So total moles is 3.75 : 1.5 : 2.25 : 1.5
So total moles is 3 : 1.2 : 1.8 : 1.2
How Moles Determine empirical formula
n = m/mm
When Given mass of sample
Determine moles of each distant substance
Then divided all mole values by lowest mole value
Numbers given decide Empirical Formula
Example
11.36 g of Copper and 28.64g of bromine
So 11.36/mm = 0.178 moles of bromine
So 28.64/77.9 = 0.367 moles of copper
Divided by lowest mole values (0.178) = 2
Divided by lowest mole values (0.178) = 1
Therefore BrCu2
Concentration
c = w/w
c = g/v
c - n/v
Standard Solution
V is in Litres
g is grams
parts per million is just concentration x million
25 g of phosphate 75 g of water
1 g water = 1ml water
c = v/v
25g/75ml = 333g/L = 0.33
2.5 ml of phosphate dissolved in water to make total 50ml solution
2.5/50 x 100 = 5.0%(v/v)
30g NaCl Dissolved in 100g of water
30/130 x 100 = 23.08%(w/w)
a 0.1 ml solution of NaCl of 1L
0.1 = m/mm//1
0.1 = m/mm
0.1 = m/58.44
m = 0.1 x 58.44
m = 5.844g
A solution that the concentration is known accurately
How to prepare one
Measure mass or volume of substance to be added on electronic mass balance or into small beaker from dropper bottle
Add volume to certain amount of water/liquid in volumetric flask using a funnel.
Rinse sides of volumetric flask, funnel and small beaker
Mass add to a minimal amount of water in a small flask (50ml beaker) and stir with stirring rod until dissolved
Pour into volumetric flask using funnel
Rinse 50ml flask, stirring rod, and funnel of with distilled water into volumetric flask
Fill volumetric flask to desiered amount and cap & shake
Put on cap and shake thoroughly for at least 5 seconds
How to dilute one
Using Pipette and take desiered amount to new flask
Seal & Label flask with concentration
Add more water/liquid
done diluted
Charles'
Boyle's
Gay lussac's
Avogadro's
If Temperature Constant
Volume 1 x Pressure 1 = Volume 2 x Pressure 2
Example
Gas in Room at pressure of 1 atm(Measure of pressure) with a volume of 2000L
If Pressure goes to 3 Atm Volume =
P1 x V1/ P2 = V2
Volume = 2000/3 = 666L
Volume 1/Temperature 1 = Volume 2/Temperature 2
Temperature in Kelvin NOT CELCIUS
K = C +273
If Pressure Constant
Example 2L Balloon at temperature of 15 degrees Celcius is increase to 40 degrees celcius
What is new volume
(2/15+273) x 40 = V2
0.2777L
If Volume Constant
Pressure 1/temperature 1 = Pressure 2/Temperature2
Example
Tire has fixed amount of air and on a cool day of 20 degrees celcius has a pressure of 10atm but on a hot day the pressure is 18.5atm
What is the temperature
P2/(P1/T1) = T2
18.5/10/293 = 542K/249C
Combined Gas Law
Pressure 1 x Volume 1/ Temperature 1 = Pressure 2 x Volume 2/ Temperature 2
Example
P1xV1/T2 = P2xV2/T2
40ml x109kPa/298K = 55ml x P2 /383K
What is pressure
P2 = P1xV1xT2/T2xV2
102kPa
1 mole of every gas under same pressure and temperature = Same volume
When under 0C and 100kPa = Standard Temperature and pressure (STP) 22.71L
When Under 25C and 100kPa = Standard laboratory conditions (SLC) Volume = 24.79L
Ideal Gas Law
PV =nRT
Pressure x Volume = Moles x Universal Gas Constant x Temperature
Universal Gas Constant =8.314 J mol K
Example
n = 1.97 P =175 T =306 K
175 x v = 1.97 x 8.314 x 306
V = (1.97 x 8.314 x 306)/175
V= 28.64 Litres
Law used for gas limiting reagents