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