January 26th Chemistry 11

We continued our lesson on Stoichiometry learning how to convert from mass to mass

Example: Lead (IV) nitrate reacts with 5.0 g of potassium iodied. how many grams of Lead(IV) Nitrate are required?
Pb(NO3) + 4 KI = 4KO3 + PbI4
5g x 1 mol KI/ 165.9g x 1 Pb(NO3)4/4KI x 455.2 / 1mol Pb(NO3)4 = 3.4g

when converting mass to mass it requires one extra you just have to convert the number of moles into grams.

Example: given the reaction 4CH3No2 + 3 O2 = 4 Co2 + 6H2O + 2N2 what mass of H2o is produced when .150g of CH3NO2 is burned?
.150g Ch3No2 x 1 mol Ch3NO2/ 61.0g Ch3NO2 x 6mol H2O / 4mol CH3NO2 x 18.0 g H2O/ 1mol H2O = .0662g

after learning all of this Mr.Doktor taught us Percent Yield
-theoretical yield of a reaction is the quantity or products expected
-the amount produced in an experiment is the actual yield

percent yield formula = Actualy mass of product / Predicted mass of product x 100
you want to get the number closes to 100.

Example:
the production of Urea CO(NH2)2 is given by:
2NH3 + CO2 = CO(NH2)2 + H2O
If 47.7g of Urea are produced determine the theoretical yield of 1 mol of CO2 reacts
1mol CO2 x 1mol CO(NH2)2 / 1mol CO2 = 1mol CO(NH2)2
1mol CO(NH2) x 60.1g / 1mol = 60.1g

b) actual yield
47.7g

c) percent yield calculation

47.7g / 60.1g x 100 = 79%

Here is a video on Perecent Yield:

January 22nd Chemistry 11

In chemistry class today we just learned about more stoichimetry we learned how to convert from mass to moles using volume at STP.

example: When 1.5g of oxygen reacts with nitrogen monoxide how many moles of nitrogen dioxide are produced?
1.5 x 1 mol O2 / 32g x 2 mol NO2/ 1 mol O2 = 0.0094mol of NO2

2)If 2g of O2 reacts with nitrogen monoxide what volume of nitrogen dioxide is present at STP?
2.0g x 1mol O2 / 32g x 2 mol NO2/ 1 mol O2 x 22.4L / 1mol NO2 = 7.8L of NO2

3) in the formation of Copper (II) oxide 3.5g of copper react. How many moles of copper oxide are produced.
3.5g x 1mol Cu/ 63.5g x 2mol CuO/ 2mol Cu = .55mol CuO

January 20 Chemistry 11

Today in chemistry 11 we learned about stoichiometry.
coefficients in balanced equations represents the moles and they are also conversion factors.

remember that when you are solving a question it is always what you need over what you have.

Example: if a .15mole sample of methane reacts with oxygen how many moles of each product is produced?

The balanced equation is : CH4 + 2O2 = CO2 + 2H2O you will always need a balanced equation.

.15 mol CH4 x 1 mol CO2/1 mol CH4 = .15mol CH4

you use the coefficients for the number of moles just remember what you need/what you have

.15mol CH4 x 2mol H2O / 1 mol CH4 = .30 mol of H2O

For further help and problems refer to this website: http://www.shodor.org/unchem/basic/stoic/index.html

January 18 Chemistry 11

Today in chemistry we did a lab on heat of combustion of candle wax. we were to find the heat released when one mole of paraffin wax was burnt.
The materials we used were :
Candle
match
Weighing scale
lab stand
wire triangle
150ml of cold water
Pop can
thermometer

The procedure was:
set up the lab stand and the wired triangle
measure 150ml of water and pour it into the pop can
record the temperature
place the candle under the pop can and light it
let the candle burn for 10 minutes then get the new temperature
clean up after your done

Observations:
volume of water used
initial temperature of water
initial mass of candle
final mass of candle
final temperature of water

with that information we can calculate the mass of the paraffin wax that was burnt during the experiment, the moles of wax that was reacted, the number of joules of energy the water gained and the molar heat of combustion of the paraffin wax.

It was a fun lab that we did today in class just remember safety comes first always keep on your safety goggles and aprons or else you and your group members will lose 10%!

Here is a link similar to the lab we did today in class:
http://www.chalkbored.com/lessons/chemistry-12/calorimetry-candle-lab.pdf

January 14 Chemistry 11

In todays class we learned about calorimetry and molar enthalpy.

Calorimetry: to measure heat absorbed and released by water we have to find out three things:
1) temperature change degrees Celsius
2) amount of water g , kg , ml , L
3) heat capacity kJ/kg ° C the heat needed to change 1 degree in ° C is 1kg

ΔH = mCΔt will be the formula we will need to know.
the heat capacity of water is 4.19 kJ/kg° C

Example: a glass of water 250g is at 20 celsius it is placed on a stove and heated until it reaches a final temperature of 85 clesius, how much energy was added to the water?
m= .250kg change it to kg
c = 4.19
t = 85-20 temperature final - temperature initial
(.250)(4.19)(85-20) = 68.1kJ

Molar Enthalpy: heat absorbed/released by one mole (kJ/mol)

Example:
if there is 1g of wax burnt what is the amount of energy released?
C25H52 + 38O2 = 25CO2 + 26H2O + 11ookJ
convert grams to moles 1g = 1mol/352g molar mass of candle
= .00284mol
then we go .00284mol x 1100kJ / 1molC25H52 = 31.2 kJ/mol
Answer: 31.2 kJ/mol of energy is released when 1g of wax is used

Calirometry video:


Molar Enthapyl video:

January 12 Chemistry 11

In todays class Mr Doktor started out by combing ammonium nitrate and water which made the beaker turn cold. This was an endothermic reaction. After he lighted the bunsen burner and burned magnesium the magnesium turned to light and this was an exothermic reaction. He then burned steel wool and there were sparks which was another example of an exothermic reaction.

We then took notes on heat and enthalpy
reactions that release heat are called exothermic
reactions that absorb heat are endothermic
heat is a form of energy all chemicals have energy stored in them stored energy is called enthalpy
the symbol for enthalpy is H

Exothermic reaction and Endothermic reaction:






To write Exothermic reactions it would be:
2C8H18 + 25O2 = 16CO2 + 18H2O + 5076 KJ
2C8H18 + 25O2 = 16CO2 + 18H2O H = -5076Kf exothermic = negative


For Endothermic it would be:
2C + 2H2 + 52.3 KJ = C2H4 52.3KJ
2C + 2H2 = C2H4 H = 52.3 KJ endothermic = positive

Coefficients can also stand for moles of molecules
N2 + 3H2 = 2NH3 + 46.3 kJ
1mol of N2 = 46.3 kJ
3 mol of H2 = 46.3 kJ
2 mol of NH3 = 46.3 kJ

Example: find the amount of heat produced if 5 mol of H2 are consumed when making ammonia.
46.3 kJ / 3mol H2 x 5 mol of H2 = 77kJ

Here is a video on Endothermic and Exothermic reactions:

January 8 Chemistry 11

In today's class we learned about the 6 types of chemical reactions:
Synthesis
Decomposition
Single Replacement
Double Replacement
Neutralization
Combustion

1) Synthesis: two or more substances that are combined
A+b = C
H2 + Cl2 = 2HCL


2)Decomposition: breaking down a compound to simpler elements
AB = A+B
2KClO3 = 2KCL + 3O2


3)Single Replacement: When the compounds have a metal and a non metal and the metals switch places.
A + BC = B + AC
2AGNO3 + CU = CU(NO3)2 + 2AG


4)Double Replacement : exchange of atoms between two different compounds
AB + CD = AD + CB
AgNO3 + KCL = AgCL + KNO3

5)Neutralization: the products will will be water or salt it usually contains an H and OH in the equation balancing out to water
2NaOH + H2CO3 = Na2CO3 + 2H2o

6)Combustion: reacts with oxygen to form an explosion
C3H8 + 5O2 = 3CO2 + 4H2O


Here is a video describing only 5/6 of the chemical reactions. ( it does not describe neutralization)

January 6 Chemistry 11

In chemistry class today we learned about balancing equations with C H and O. The products are always the same CO2 and H2O.

An Example would be:
C6H12O6 + 6O2 = 6CO2 + 6H2O

After we learned how to use words in our balanced equation.
Example:
Aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas.
2AlBr3 + 3Cl2 = 2ALCL3 + 3Br2
You have to find the symbol for each name Aluminum bromide chlorine etc write the the equation and then balance it accordingly to the number of atoms on each side.

Here is a video helping you to balance word equations:


Mr Doktor also gave us a list of acids that we should memorize:
HCL = Hydrochloric Acid
HNO3 = Nitric Acid
H2SO4 = Sulphuric Acid
H3PO4 = Phosphoric Acid
CH3COOH = Acetic Acid

December 17 Chemistry 11

Today in Chemistry class we learned about balancing equations.

Chemical equations show us the reactants used and the products that are produced during a chemical reaction the numbers in front of the symbols are coefficients which refers to the number of moles.

to balance a chemical equation there must be the same numbers of each atom on both sides of the equation.
Examples :
Al + O2 = Al2O3
C2H6 + O2 = 2CO2 + 3H2O
Fe2O3 + 3H2SO4 = Fe2(SO4)3 + 3H2O

Here is a video showing you how to balance chemical equations:

There are the same number of each type of atom on each side of the equation

December 15 Chemistry 11

Today in Chemistry 11 we started our new unit on chemical reactions Mr.Doktor showed us some chemical reactions today. First he started out by by adding bromothemal blue and hydrochloric acid which turned to a yellow color. Then he added ammonia to the acid and it turned blue. After Mr Doktor mixed potassium iodide with iron iodide and it made the solution turn yellow. Mr.Doktor then showed us another chemical reaction by burning a strip of magnesium which made a very bright light and he told us this is how they started fire works. His last chemical reaction was he dipped ap iece of paper in alcohol and lit the paper on fire, the paper did not set on fire at all but instead the alcohol was burning.