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

Density & Moles!

The formula for Density is:
d=m/v

The conversion between Density and Moles is a two step process.

Example:
How many moles are in a 25.0mL sample of Iron if the density of Iron is 7.87g/mL

Step 1: Convert from Density to Mass using the equation.
25.0mL x 7.87g/1mL = 197g

Step 2: Convert from Mass to Moles using Molar Mass
197g x 1mol/55.8g = 3.53mol

Lab 4B from Heath Chemistry Lab Manuel

In class, our the purpose of our Lab was: To find out how many moles, raio of atoms of formula units are produced in the reaction of iron and copper (II) chloride.

Materials: Beaker, Washbottle, Stirring Rod, Tongs, Scapula, Scale, Drying oven, Safety goggles, Filter, Copper (II) Chloride, 2 Iron nails, 50 mL of distilled water

Procedure:
1. Find the mass of a dry 250 mL beaker, the 2 nails, and the piece of paper you are using to place the copper (II) chloride. Record it's mass using the scale.
2. Add the chloride onto the paper until you have reached 7 grams (Make sure it's below 10g)
3. Add the powder into the beaker
4. Weigh the beaker with the Copper (II) Chloride powder and record it's mass.
5. Add 50 mL of distilled water into the beaker filled with the powder. NOTE: To prevent spillage, use a stirring rod.
6. Add the 2 nails into the beaker filled with the solution.
7. Wait 10-15 minutes.
8. Remove the nails from the beaker one by one, with tongs.
9. Holding the nails above the beaker, spray them with water to remove the copper. NOTE: The less water sprayed, the less water to filter.
10. Once the copper is removed from the nails, set them to dry.
11. As the nails are drying, filter the Copper by decanting.
12. Take the filtered paper filled with copper and set it into the drying oven to dry.
13. After both the nails and filter paper of copper are dry, measure their masses.
14. Clean up all lab stations and equipment.