Chem Class- January 28, 2010

Today we started a lab that involved testing the stoichiometric method. We are to check to see if the mass of the precipitate matches with our prediction.

According to our prediction, 1.74 g of SrSo4 (Strontium sulphate) should be produced.

2.00 g Sr(NO3)2 x 1 mol SrSO4/1 mol Sr (NO3)2 x 183.7 g SrSO4/ 1 mol = 1.74 g SrSO4

Basically, in the experiment, 2.00 g of Strontium nitrate is dissolved in 50 mL of water and then reacted with excess Copper (II) Sulphate (3.00 g). The product (Strontium sulphate and copper (II) Nitrate) is a precipitate. After mixing the solutions, the precipitate will be separated by filtration, dried and then weighed. The last thing we did so far was place the filter paper in the drying oven. Next class, we are to record the mass of precipitate and filter paper along with a number of other observations: Mass of copper (II) Sulphate used, mass of Strontium Nitrate used and mass of precipitate

As part of our Analysis and Conclusion, we are to calculate the number of moles of precipitate that were formed as well as the percent error for our results. While doing the lab, we needed to make sure not to leave any of the precipitate behind after pouring the mixture into a funnel. This could greatly affect our results. While carrying out the procedure of the lab, we wore our lab aprons and lab goggles. Remember, safety first!

The balanced chemical equation for the lab:

Sr(NO3)2 + CuSO4 -----> SrSO4 + CU(NO3)2

Strontium Nitrate

Copper (II) Sulphate

Chem Class- January 26, 2010

Stoichiometry again. Last class we worked from mass to moles and today we worked from mass to mass.

Here's an example that we did in class: Lead (IV) nitrate reactions with 5.0 g of Potassium iodide. How many grams of Lead (IV) Nitrate are required?
Pb(NO3) + 4 KI -----> 4 KO3 + PbI4
5.0 g x 1 mol KI/165.9 g x 1 Pb(NO3)4/4KI x 455.2/ 1 mol Pb(NO3)4 = 3.4 g

All it is is one extra step, and that's converting the number of moles into grams. Here's some examples we put together:

Ex: C15H12 + 8 O2 -----> 5 CO2 + 6 H2O (Reaction when pentane burns). What mass of CO2 is produced when 100.0 g of C5H12 is burned?
100.0 g C5H12 x 1 mol C5H12/72.0 g C5H12 x 5 mol CO2/1 mol C5H12 x 44.0 g CO2/1 mol CO2 = 306 g

Ex: Given the reaction 4 CH3NO2 + 3 O2 -----> 4 CO2 + 6 H2O + 2N2, what mass of H2O is produced when 0.150 g of CH3NO2 is burned?
0.150 g CH3NO2 x 1 mol CH3NO2/61.0 g CH3NO2 x 6 mol H2O/4 mol CH3NO2 x 18.0 g H2O/ 1mol H2O = 0.0662 g

Ex:4NH3 + 5O2 -----> 6 H2O + 4 NO. Given this reaction, what mass of NO is produced when 2.00 mol of NH3 (g) are reacted with excess O2?
2.00 mol NH3 x 4 mol NO/ 4 mol NH3 x 30.0 g NO/ 1 mol NO = 60.0 g

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We also learned about:

PERCENT YIELD

The theoretical yield of a reaction is the expected (calculated) amount

The experimental amount is the actual yield

% yield = actual/theoretical x 100

Here's the example we did in class which required knowing the above information:

Example: The production of urea CO(NH2) is given by: 2NH3 + Co2 -----> CO(NH2) + H2O. If 47.7 g of urea are produced when 1 mole of CO2 reacts, find the actual yield, theoretical yield and percent yield.

We know that the actual yield is 46.7 g. The theoretical yield: 1 mol CO2 x 1 CO(NH2)2/1 CO2 x 60.1 g/ 1 mol = 60.1 g

Percent Yield = 47.7 g/60.1 g x 100 = 79.4%

Here's a video on calculating the percent yield:

Chem Class - January 22, 2010

More Stoichiometry today! We did a few examples on problems that involved converting from mass to moles and involving volume at STP.

Here are the examples we did in class:
1) When 1.5 g of oxygen reacts with nitrogen monoxide, how many moles of nitrogen dioxide are produced?
1.5 g x 1 mol O2/32.0 gx 2 mol NO2/1 mol O2 = 0.094 mol of NO2

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

3) In the formation of Copper (II) oxide, 3.5 g of copper react. How many moles of copper oxide are produced?
3.5 g x 1 mol Cu/63.5 g x 2 mol CuO/ 2 mol Cu = 0.055 mol CuO

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Mr. Doktor also gave us 2 questions to do on our own and when were were finished we were to show him our answers to get our homework in return:

4) What mass of water is produced when 3.5 L of hydrogen is burnt (reacted with O2)?
2H2 + O2 -----> 2H2O
3.5 L of H2 x1 mol/22.4 L x 2 mol H2O/2 mol H2 x 18 g/ 1 mol = 2.8 g of H2O

5) Lead (IV) Nitrate reacts with 2.5 g of potassium iodide. How many molesof Lead (V) Nitrate are needed?
Pb(NO3)4 + 4KI -----> PBI4 + 4KNO3
2.5 gKIx 1 mol/166 g KI x1 mol Pb(NO3)4/4 mol KI = 0.0038 mol of PBI4

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For further help on solving problems involving stoichiometry, there are a lot of youtube videos. Here's one example:

Chem Class- January 20, 2010

We got a new booklet today called "Gravimetric Stoichiometry".

Then, we learned the following points that have to do with Stoichiometry:

• Coefficients in balanced equations represent moles
• Coefficients are also conversion factors

When doing any problem solving questions that involve stoichiometry, use the following:

What you need/What you have

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Here's some examples:

ex: Given the reaction N2 + 3H2 -----> 3NH3, how many moles of NH3 are produced when 18 mol of H2 are reacted?
- 18 mol H2 X 2 mol NH3/3 mol H2 = 12 mol

ex: How many moles of Fe are required to produce 16 mol of H2 in the reaction 3Fe + 4H2O ----->Fe3O4+ 4H2?
-# of moles of Fe = 16 mol H2 x 3 mol Fe/4 mol H2 = 12 mol

ex: According to the eqyatuon 2H2 + O2 ----> 2H2O, how many moles of H2O are produced when 9.6 mol of O2 react?
- 9.6 mol O2 x 2 mol H2O/1 mol O2 = 19 mol

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Converting to mass requires one additional step

ex: In the reaction 4NH3 + 5 O2 -----> 6H2O + 4NO, what mass of NO is produced when 2.00 mol of NH3 are reacted with excess O2?

- 2.00 mol NH3 x 4 mol NO/4mol NH3 x 30.0 g NO/ 1 mol NO = 60.0 g

ex: C5H12 (Pentane) burns according to the reaction C5H12 + 8 O2 -----> 5 CO2 + 6 H2O. What mass of O2 is required to produce 60.0 g of H2O?

- 60.0 g H2O x 1 mol H2O/18.0 g H2O x 8 mol O2/6 mol H2O x 32.0 g O2/1 mol O2 = 142 g

Chem Class- January 18, 2010

Today we did a lab in class on 'The Heat of Combustion of Candle (Paraffin) Wax. We were to find the heat that is released when one mole of Paraffin wax is burnt.

Materials:
candle

match

electronic weigh scale
lab stand
wire mesh or triangle

Thermometer

150 mL of cold water

A metal can

                                                                                            

Procedure:

1. Set up the lab the stand and wire mesh
2. Measure 150 mL of water and pour it into the metal can
3. Record the temperature of water and the mass of the candle before lighting it
4. Place the candle under the can and carefully light it
5. Let the candle burn for approximately five minutes and then record the new temperature of the water as well as the mass of the candle after the experiment
6. Clean up your lab station

                                                                              

We were also to make the following observations:

• The volume of water used
• The initial temperature of the water
• The initial mass of candle
• The final temperature of water
• The final mass of candle

                                                                                             

- With that information, we calculate the mass of Paraffin wax that was burn during the experiment (final mass-initial mass), the # of moles of Paraffin wax that was reacted during the experiment (number of grams x molar mass/ C25H52), the # of joules of energy that the water gained and determine the molar heat of combustion  (Joules/Mole) of Paraffin Wax.

- In addition, knowing that the accepted value for molar heat of combustion of Paraffin wax is approximately -14500 kJ/mol, we determine the percent error (Remember that the percent error is measure value- accepted value/ accepted value). We need to also keep in mind that there is a possibility of error such as miscalculations and changing ambient conditions

It was a simple and fun lab to do and it had a lot to do with what we were learning (Calorimetry and Enthalpy). Try it out yourself, and see if you come close to the accepted value for the molar heat of combustion of Paraffin Wax. Remember to wear your safety goggles and lab apron!

Chem Class- January 14, 2010

New topics today:

Calorimetry
To measure heat absorbed/ released bywater, we need to know three things:

• Temperature change (° C)
• Amount of Water (g, kg, mL, L)
• Specific Heat Capacity (kJ/kg ° C)- the heat needed to change 1 degree C in 1 kg

ΔH = mCΔt
Read as: Change in enthalpy is equal to the mass of water times the specific heat capacity times the change in temperature.

The specific heat capacity of water is 4.19 kJ/kg ° C

Ex: -calculate the amount of heat (in joules) required to warm 400g of water from 10°C to 35°C.(Heat Capacity (C) for H₂O is 4.18 kJ/ kg·°C) ***mr. doktor said 4.19 but it doesn't matter whether it's off by 1...- refer to Hand in Assignment #9 - energy in chemical reactions #7***

-When looking for a final temperature change (and other similar−) the equation can be (and is) rearranged using Algebra.
ΔH = mC(Tfinal - Tinitial)
Tfinal = (ΔH ÷ mC) + Tinitial
the answer gives you the final change in temperature.

 Molar Enthalpy
- Heat absorbed/released by one mole (kJ/mol)

Example:
C25H52 + 38O₂ → 25CO₂ + 26H₂O +1100kJ
If there are 1.0g of wax burnt, what is the amount of energy (aka heat) released?
-Firstly you need to convert g to mol

1.0g = 1mol / 352g [352g is taken from the total mass of wax (C25H52)]
        = 0.00284mol
- we take the Coefficient and us it like:
0.00284mol × 1100kJ / 1molC25H52 = 31.2 kJ/mol
Therefore: 31.2 kJ/mol of energy are released when 1.0g of wax is burnt (used up)

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Here's a site on caliometry:

http://www.emsb.qc.ca/laurenhill/science/calorimetry.html

Here's a site on caliometry and molar enthalpy:

http://www.emsb.qc.ca/laurenhill/science/calorimetry.html

And here's a video on someone solving a caliometry question:

Chem Class- January 12, 2010

To start the class off, we went over the answers to our "Classifyung Chemical Equations" worksheet. Then Mr. Doktor combined ammonium nitrate and water and the beaker turned cold. This was an example of an endothermic reaction. Then he lighted a bunsen burner nad burned magnesium (Equation: Mg + O2 -----> MgO). The magnesium produced light and was an example of an exothermic reaction. After that, he burned steel wool (Equation: Fe + O2 ----> Fe2O3). It lighted on fire and sparks flew from it. This was another example of an exothermic reaction.

We also took notes:
Heat and Enthalpy

• Reactions that release heat are exothermic
• Reactions that absorb heat are endothermic
• Heat is a form of energy
• All chemicals have energy stored in them. Stored chemical energy is enthalpy (Mr. Doktor dropped a textbook from different heights to demonstrate that enthalpy is sotred chemical energy)
• Enthalpy of gasoline > Enthalpy of water
• Enthalpy symbol is H and change in enthalpy is ΔH

There are two ways to write enthalpy in reactions:
For exothermic: 2C8H18 + 25O2 -----> 16 CO2 + 18H2O + 5076 kJ
                         2C8H18 + 25O2 -----> 16 CO2 + 18H2O     ΔH= - 5076 kJ
For endothermic: 3.2 C + 2H2 + 52.3 kJ -----> C2H4
                           3.2 C + 2H2 +  -----> C2H4  ΔH= - 5076 kJ

Also, when given an equation, the coefficients can stand for moles or molecules.
Example: N2 + 3H2 -----> 2NH3 + 46.3 kJ
So 1 mol of N2 produces 46.3 kJ, 3 mol of H2 produces 46.3 kJ and there are 2 mol of NH3
This gives you the following conversion factors: 46.3 kJ/1 mol N2, 46.3 kJ/ 3 mol H2 and 46.3 kJ/2 mol NH3
So, if you were asked to find the amount of heat released in...say 5.0 mol of H2 that is consumed, here's how it'll work: 4.63 kJ/ 3 mol H2 x 5.0 mol of H2 = 77 kJ

Chem Class - January 8, 2010

Today in class, we corrected our homework on balancing equations and then took some notes of the 6 types of chemical reactions: Sythesis, Decomposition, Single Replacement, Double Replacement, Neutralization, and Combustion.

1) Synthesis

A+B ----> C (Two or more substances combine)
EX: H2 + Cl2 ----> 2HCl

2) Decomposition

AB---->A+B (Breaking down into simpler substances)
EX: 2Ag2O---->4Ag + O2

3) Single Replacemnet

A + BX---->B + AX (Compunds must have a metal and a nonmetal-replacing one atom in a compound by anohter atom)
EX: Cl2 + 2KI ----> I2 + SKCl

4) Double Replacement

AB + XY ----> AY + XB (exchange of atoms between two different compounds)
EX: 2NaCl+ H2SO4 ----> 2 HCl + Na2SO4

5) Neutralization

(Products are water and an ionic salt/ Always between acids and bases)
EX: HCl + NaOH----> NaCL + H2O

6) Combustion

There are two types: Metallic (Can also be known as a synthesis reaction/includes oxygen) and hydro-carbon (includes carbon and oxygen). Also, the productes are always CO2 and H2O

EX: C5H12 + 8 O2 ----> 5CO2 + 6 H2O


Here's a video on five of the six chemical reactions we learned about in class:

Chem Class- January 6, 2010

Today we learned about balancing equations with C, H and O. The products are always the smae (CO2 and H2O). Here's an example:

C3H8 + O2 -----> CO2 + H2O
C3H8 + 5O2 -----> 3CO2 + 4H2O (balanced)

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We then learned how to put words into a balanced equation. It isn't very hard.

 Let's say aluminum reacts with oxygen to form Aluminum Oxide. You just convert the chemical names into the appropriate symbols (Al + O2 -----> Al2O3) and then balance it. You might come up with 2Al + 1.5 O2----->Al2O3. One and a half of a molecule seems pretty silly though simply multiply everything by 2 (look at the last number: .5) to get:

 4Al + 3O2----->2Al2O3

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We also were given a short of list of acids that we should know:
Hydrochloric Acid (HCl)
Nitric Acid (HNO3)
Sulphuric Acid (H2SO4)
Phosphoric Acid (H3PO4)
Acetic Acid (CH3COOH)

Acetic Acid or Vinegar