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.

Mole to Mole Conversions

In today's class we learned how to do from mole to mole conversions. We used the coefficients in front of balance equations. The coefficients tell us the number of moles reacted or produced. These formulas can also be used as conversion factors.

e.g. CH4 + 2O2 > CO2 + 2H2O
0.15 mol x 1/1 = 0.15 mol (what you need over what you have)
* leave in the correct ratio even it is 4/4

e.g. 2.1 x 10^-2 of Aluminum Hydroxide + Sulphuric Acid
2Al(OH)3 + 3 H2SO4 > AL2 (SO4)3 +6H2O

2.1 x 10^-2 x 1/2 = 0.011 mol
= 1.1 x 10^-2 mol

Stiochiometry: Quantitative Chemistry

Todays class we learned that Stiochiometry 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 6 types of chemical reactions is the foundation of stiochiometry.

6 Types of Reactions:
1) Synthesis: A + B ---> AB / Usually elements ---> compounds
ex. 2Al + 3F2 ---> 2AlF3
4K + O2 ---> 4K2O

2) Decomposition: AB ---> A + B (The reverse of Synthesis)
* Always assume the compounds decompose into elements during decomposition
ex. 4H3PO4 ---> 6H2 + 1P4 + 8O2
Mn(C2O4)2 ---> Mn + 4C + 4O2

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

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

5) Neutralization: Reaction between Acid + Base
ex. H2SO4 + 2KOH ---> 2HOH + K2SO4
3Ca(OH)2 + 2H3PO4 ---> 6HOH + Ca3(PO4)2

6) Combustion: Reactions w/something (usually hydrogen) with air.
* Hydrocarbon combustion always produces CO2 & H2O
ex. CH4 + 2O2 ---> CO2 + 2H2O
C8H18 + 25/2 O2 ---> 8CO2 + 9H2O

Molecular Formulas

In today's class we learned how to make or get a molecular formula

Its basically if you know howto make a empirical formula you can find the molecular formula you need the molar mass
e.g. the empirical formula for a substance is CH20 and its molar mass is 60.0g/mol
Determine the molecular formula

Empirical
CH2O
30.0g/mol

Molecular
C2H4O2
60.0g/mol

Percent Composition!

Today in chemistry class, Mr. Doktor taught us about percent composition by mass. We learned how to figure out the percent by mass of each element in a compound.



* To relate the mass of each element in a compound to the entire mass of the compound we use percent composition.



Example. H20--------- (2) molar mass of hydrogen in the compound / (18) total molar


mass of the compound = 0.111 = 11.11% of hydrogen


(16) molar mass of oxygen in the compound / (18) total molar mass of the compound


= 0.889 = 88.89% of oxygen


* To check your answer, add up all the percents and if the sum is 100 you did it right.


* 11.11%

+

* 88.89%


= 100%

Empirical Formulas:

- Empirical Formulas are the simplest formula of a compound
- They only show the simplest ratio, not the actual atoms
- Molecular Formulas are the actua number of atoms
- To determine the empirical formula we need the ratio of each element
- to determine the ratio fill in the table below for each problem

e.g. Find the percentage composition of a compound that contains 17.6g of iron and 10.3g of sulfur. The total of the compound is 27.9

Fe 17.6g/27.9= 63.1 %
S 10.3/27.9=36.9 %

e.g. A compound was analyzed and was found to contain 9.8 g of nitrogen, 0.7 g of hydrogen, and 33.6 g of oxygen. What is the empirical formula of the compound?

N = 9.8g x 1mol/14.0g = 0.7 mol = 1
H = 0.7g x 1mol/1.0g = 0.7 mol = 1
O = 33.6g x 1mol/16.0g = 3 mol = 3

empirical formula: NHO3