Solution Stoichiometry

Solutions

Solutions are homogenous mixtures composed of a solute and a solvent
- Solute is the chemical present in lesser amount (whatever is dissolved)
- Solvent is the chemical present in greater amount (whatever does the dissolving)
Chemicals dissolved dissolved in water are aqueous
- NaCl(aq); H2SO4(aq)

Molarity

Concentration can be expressed in many different ways
- g/L, ml/L, % by volume, % by mass, mol/L
The most common (In Chemistry 11 & 12) is mol/L which is also called Molarity
- mol/L = M
- [HCl] = concentration of HCl

Molarity = Moles
Volume

Example:
- 100 mL of 0.250 M Iron (II) Chloride reacts with excess Copper. How many grams of Iron are produced?

FeCl2 + Cu ---> CuCl2 + Fe

0.250 mol/L x 0.100 L x 1/1 x 55.8g/mol = 1.40g

- How many moles of Copper Chloride are produced?

0.250mol/L x 0.100 L x 1/1 = 0.025 mol

- Determine [CuCl2]

0.0250 L x 1/0.100 L = 0.250 M

Limiting Reactants:

In today's class we learned how to determine the Limiting Reactant (L.R.)

- In chemical reactants, usually one chemical gets used up before the other
- the chemical used up first is called the limiting reactants
- one it is used up the reaction stops
- L.R. Determines the quantity of products formed
-To find the L.R. assume one reactant is used up. Determine how much of the reactant is required

e.g. 2.5 mol of CuSO4 reacts 2.5 mol of NaNO3. Determine the L.R. and calculate how many moles of NaSO4 are formed.

CuSO4+ 2NaNO3 > NaSO4 + Cu(NO3)2

2.5 mol x 2/1 = 5.0 mol NaNO3
soo the L.R> is NaNO3

2.5 mol x 1/2 = 1.25mol Na2SO4

Energy + Percent Yield

In today's class we learned how to calculate energy and also Percent Yield. Percent yield is basically the amount of products that should be formed in a reaction.

- Enthalpy is the energy stored in chemical bonds
- Symbol of Enthalpy is H
- units of Joules (J)
-Change in Enthalpy is ∆H
-In Exothermic reactions, enthalpy decreases
-In Endothermic reactions, enthalpy increases

Calorimetry:
-To experimentally determine the heat released we need to know 3 things:
1. Temperature change (∆T)
2. Mass (m)
3. Specific Heat Capacity (C)

They are all related by the equation:
∆H= mC∆T

e.g. Calculate the heat required to warm a cup of 700g of water (C= 4.18J/g°C) from 40.0° to 60.0°.
ΔH = mCΔT
= (700) (4.18) (20.0)
= 58 520 J / 1000
= 58.5 kJ


80.0 kJ of heat are added to a 700g glass of water initially at 40.0°C. Calculate the final temperature of the water (C= 4.18J/g°C).
ΔT = ΔH / Mc
= (80 000) / (700) (4.18)
= 27. 3°C
= 27. 3°C + 40.0°C
= 67.3 °C


Percent Yield
- The Theoretical yield of a reaction is the amount of products that should be formed
- 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

Percent Yield = (Actual/ Theoretical) x100

e.g. Determine the percent yield for the reaction between 3.74 of Na and excess O² if 5.34 g of Na²O² is recovered

2Na + O2 > Na2O2

3.24 x 1mol/23.0g x 1/2 x 78.0/1mol = 6.34g

% yield = 5.34/6.34 = 84%

Other Conversions

- Volume @STP can be found using the conversion factor 22.4 L/mol
- Heat can be included as a seperate term in chemical reactions (Enthalpy)
i) Rxns that release heat are exothermic
ii) Rxns that absorb heat are endothermic
iii) Both can be used in Stoichiometry

Examples

If 5.0g of Potassium chlorate decomposes according to the reaction below, what volume of Oxygen gas (@STP) is produced?

2KClO3 ---> 2KCl+ 3O3

5.0g x 1mol/122.6 x 3/2 x 22.4 L/ 1mol = 1.4 L

When Zinc reacts w/Hydrochloric acid exactly 1.00 L of Hydrogen gas is produced @STP. What mass of Zinc was reacted?

Zn + 2HCl ---> H2 + ZnCl2

1.00L x 1mol/22.4 L x 1/1 x 65.4 g/1 mol = 2.82 g

In the formation of Copper (III) Oxide 3.5g of Copper react. How many litres of Oxygen @STP are needed?

2Cu + O2 ---> 2 CuO

3.5g x 1mol/63.5g x 1/2 x 22.4 L/1mol = 0.62 L

Mass to Mass Conversions!

Mr. Doktor explained to us how Mass to Mass conversions worked. It has a similar concept from Mole to Mole conversions but with an additional step:

Grams of A >>> Moles of A >>> Moles of B >>> Grams of B

Example 1:
How many grams of chlorine from the decomposition of 64.0 g. of AuCl3 by this reaction:
2 AuCl3 ---> 2 Au + 3 Cl2

Step 1: Convert Mass A to Moles A: 64.0g x 1mol/303.32g
Step 2: Convert Moles A to Moles B: 0.211mol x 2/3
Step 3: Convert Moles B to Mass B: 0.316mol x 70.906g/1mol = 22.4g

Example 2:
Lead (IV) Nitrate reacts with 5.0g of Potassium Iodide. How many grams of Lead (IV) are required?

Step 1: Write a Balanced Chemical Equation: Pb(NO3)4 + 4KNO3 + PbI4
Step 2: Convert Mass A to Moles A: 5.0g x 1mol/283.3g = 0.0176mol
Step 3: Convert Moles A to Moles B: 0.0176mol x 1/4 = 0.0044mol
Step 4: Convert Moles B to Mass B: 0.0044mol x 327.2g/1mol = 1.4g

Here is a video futher explaining Mass to Mass conversions:

Stoichimetry Investigation: Testing Stoichimetric Method Lab

PROBLEM: Does Stoichimetric accurately predict the mass of products produced in chemical reactions?

Balanced equation for the reaction: Sr(NO3)2 + CuSO4 ---> SrSO4 + Cu(NO3)2

PROCEDURE:

1. Carefully measure about 3.00g of Copper (II) sulphate.
2. Crush the Copper (II) sulphate into a fine powder using a mortar and pestle.
3. Dissolve the Copper (II) sulphate in 50mL of water. 
4. Carefully measure 2.00s of Strontium nitrate and dissolve it in 50mL of water
5. Slowly pour the two solutions together.
6. Stir the mixture to complete the reaction.
7. Write your group name on a piece of filter paper.
8. Find and record the mass of the filter paper.
9. Using a funnel and an Erlenmeyer flask, place the filter paper in funnel. Slowly pour the mixture into the funnel.
10. Pour the filtrate into the waste collection bottle.
11. Place the filter paper in the drying oven and record the mass when its dry.

*Next class we will check and weigh our filter paper

Mole to Mass and Mass to Mole Conversions


Today in chem. class, Mr. Doktor taught us an additional step from last class, which allows us to determine the mass from an amount of moles or vice versa. Following this chart will help you make the right conversions from “A” to “B”.

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


The balanced chemical equation: 2Al2O3 = 4Al + 3O2


3.5mol * (2/4) * (102g/1mol) <- (molar mass) = 178.5 g = 1.8*10^2g


Example: How many grams of water are produced if 0.84 mol of Phosphoric Acid is completely neutralized by Barium Hydroxide?


The balanced chemical equation: 2H3PO4 + 3Ba(OH)2 = Ba3(PO4)2 + 6H2O


0.84mol * (6/2) * (18g/1mol) <- (molar mass) = 45g


There is no way to go from the mole of one substance to the mass of another directly.


YOU ALWAYS NEED TO CONVERT TO MOLES FIRST.