Tuesday, May 1, 2012

Determining the Concentration of a Solution: Beer's Law

Last week in chemistry we did a lab where we had to find the concentration of a solution. For this lab you will need the following materials:

  1. Power Macintosh or Windows PC
  2. Vernier Computer Interface
  3. Logger Pro
  4. Colorimeter
  5. One Cuvette
  6. Five test tubes
  7. stirring rod
  8. 30ml of 0.40 M NiSO4
  9. 5ml of NiSO4 unknown solution( the concentration they have to find)
  10. two 10ml pipets, pipet pump or pipet bulb
  11. distilled water
  12. test tube rack
  13. two 100-ml beakers

To start follow the below procedure

  1. add about 30ml of 0.40M NiSO4 stock solution to a 100-ml beaker
    1. add about 30ml of distilled water to another 100-ml beaker
      1. you might need more this is just to start off
  2. label four test tubes 1-5 
    1. in test tube one place 2ml of 0.40 M NiSO4
      1. then put 8 ml of H2O in
    2. in test tube two place 4ml of 0.40 M NiSO4
      1. then put 6 ml of H2O in
    3. in test tube three place 6ml of 0.40 M NiSO4
      1. put 4 ml of H2O in
    4. in test tube four place 8ml of 0.40 M NiSO4
      1. then put 2 ml of H2O in
    5. in test tube five place10ml of 0.40 M NiSO4
      1. then put 0 ml of H2O in
  3. Put all this data in a table:
    1. Trial Number
      0.40 M NiSO4 (mL)
      Distilled H2O (mL)
      Concentration (M)
      1
      2
      8
      0.08
      2
      4
      6
      0.16
      3
      6
      4
      0.24
      4
      8
      2
      0.32
      5
      10
      0
      0.40

  4. Once each tube has both the 0.40 M NiSO4 and water stir them each with a stirring rod
  5. next connect the Colorimeter to you computer and open Exp Colorimeter in the experiment 11 folder of Chemistry with Computers. Te vertical axis should read the absorbance scaled from 0 to 0.6mol/L and a horizontal concentration axis from 0  to 0.5mol/L
  6. Now you are ready to calibrate the Colorimeter
    1. you need to fill the cuvette 3/4 full with distilled water
    2. holding the cuvette by the upper edges, place it in the cuvette slot in the Colorimeter and slide the top shut
    3. Now you are at the point where you are going to calibrate the Colorimeter
      1. choose calibrate from the experiment menue and the click Perform No
      2. turn the wavelength knob on the Colorimeter to the 0%T position
      3. then type zero in the edit box
      4. when the display voltage reading for Input 1 stabilizes click keep
    4. Part two 
      1. turn the knob of the Colorimeter to the Red LED position (635 nm)
      2. type 100 in the edit box
      3. when the display voltage reading for Input 1 stabilizes click keep then click ok
  7. Now we are going to collect the absorbance data for the five standard solutions
    1. click collect and then take out the cuvette and dump out the water
    2. take the solution from test tube one and pur it into the cuvette until it is nearly full
    3. place it back in the Colorimeter and shut the lid
    4. wait for the absorbance value to display on the screen then click keep
      1. your data should be shown on a graph
        1. repeat steps 2 through 4 on the remaining test tubes
      2. you should also make a chart of the data you are collecting, since the data is already there i'll show you the table
        1. Unknown absorbance will be different then the one I have down
          1. make sure to so the first five before you do the unknown
        2. Trial
          Concentration (mol/L)
          Absorbance
          1
          0.080
          0.186
          2
          0.16
          0.255
          3
          0.24
          0.511
          4
          0.32
          0.595
          5
          0.40
          0.726
          6
          Unknown
          0.45

        3. Since the unknown solution in my class was different the results will be different, heres my graph
          1. Trial
            Concentration (mol/L)
            Absorbance
            1
            0.080
            0.186
            2
            0.16
            0.255
            3
            0.24
            0.511
            4
            0.32
            0.595
            5
            0.40
            0.726
            6
            Unknown
            0.45
            Concentration of Unknown
            0.23 mol/L

    5. Now using the absorbance level locate the level of concentration of the unknown using the graph.
    6. Once you have figured out the unknown solution the lab is done make sure to properly discard the solutions and clean the test tubes.










































Tuesday, February 14, 2012

Silver and Copper Lab

For the last three weeks in chemistry we had been doing a copper and sliver lab. Where we were seeing how copper wire reacts with silver nitrate to get silver.

Day 1

  1. Obtain a 30 cm length of bare copper wire. Using a plastic sanding pad, clean off the copper wire. Coil the wire around a pencil, forming a loose coil or "spring" on on end. Stretch the coil to reach from the bottom to the top of the tube. The other end should reach to the top of your test tube and be uncoiled.
  2. Weigh the coil(copper wire) as accurately as possible with the balance. place the copper wire in the test tube to make sure it fits. Then take it out and put it aside for later.
    1. the weight of the copper wire we got was 3.408g
  3. Measure out 1 gram of silver nitrate
    1. Before that weigh the filter paper by its self and record the the amount of grams
    2. Then measure the silver nitrate on the filter paper together record the measurement.
      1. For the filter paper we got a measurement of  1420g   , and for the silver nitrate we measured    1.726g   .
  4. After recording the measurements pour the silver nitrate in the test tube and pour distilled water until the water is about 2 inches from the top.
  5. Cover put something on top of the tube and shake to dissolve the silver nitrate in the distilled water. Then place the copper wire inside the test tube and seal it with parafilm( parafilm is so fun to play)
  6. Set aside until the following day.
Day 2
  1. This time you will take the test tube that contains the copper wire and silver nitrate that you prepared the pervious day and take off the parafilm. 
  2. Carefully use distilled water to wash off any silver that is stuck to the copper wire.
  3. Once you are sure all the silver is in the test tube and not on the copper wire 
  4. Lay aside the copper wire as you dont need it anymore and carefully pour the silver into the filter paper.
    1. make sure that the filter paper is in a cone and the cone is over a flask so that the liquid drains into the flask and you are left with only the silver.
  5. Allow the silver and copper wire to dry over night
Day 3
  1. For the final part of the lab all you are doing is taking measurements.
  2. First measure the copper wire now that it is dry to find out how much of it reacted
    1. We got a measure of 3.193g we then take the measurements from before the copper reacted and subtract 3.193g from 3.408g. to get the amount of copper left 0.215g
  3. Next gently take out the filter paper and silver and measure to find the mass.
    1. For this step we got a measurement of 1.726g Then we took the mass of both the silver and filter paper 1.726g and subtracted the mass of the filter paper 1.420g.
    2. We then got the measurement of silver that was produced 0.396g.
  4. Once you have all the data you then answer the analysis questions
Silver and Copper Lab Analysis Questions
  1. Find the number of moles of silver produced
    1. .0028molAg
  2. Find the number of moles of copper consumed during the reaction
    1. .0033molCu
  3. Find the mole ratio of copper to silver in this reaction (round o the nearest whole number)
    1. 1:1
  4. Write a balanced equation for the reaction using the mole ratio derived above as the coefficients
    1. 14:10:108
  5. During the middle ages, a group of people called alchemists searched unsuccessfully for a way to "transmute" or change one element to another, e.g. lead to gold. Was copper "transmuted" into silver in this experiment
  6. The correct equation for this chemical reaction is 
    1. 2AgNO3 + Cu -----> Cu(NO3)2 + 2Ag
  7. Using your starting amount of silver nitrate, how much Ag should be formed in grams
    1. .65gAg
  8. Using your starting amount of silver nitrate how much Cu became Cu(NO3)2 in the reaction in grams
    1. .77gCu

Friday, January 20, 2012

Ch.11 Questions 42,42,44(1-3)

These are questions that I did for chemistry to improve me being able to find out the percent of an element in a compound.


  1. Calcium Chloride
    1. first i find the atomic mass of each compound(Ca and Cl)
      1. Ca=40.08
      2. Cl=35.45
    2. Now since there are two Cl we 35.35x2=70.9
    3. we than add 70.9+40.08=110.98
    4. CaCl2 has a mass of 110.98
      1. to find out the percent of Ca and Cl we simply divide the individual mass by the total mass
        1. 40.08/110.98=36.11%
        2. 70.9/110.98=63.88%
  2. Sodium Sulfate
    1. find the total mass of Sodium Sulfate
      1. Na2So4
        1. Na=22.990x2=45.980
        2. S=32.066
        3. O=15.999x4=63.996
    2. add all the individual masses together
      1. Na2SO4=142.042
    3. divide each individual mass by the total
      1. Na(45.980/142.042=32.37%)
      2. S(32.066/142.042=22.58%)
      3. O(63.996/142.042=45.05%)
  3. Which has a larger % of Sulfur H2SO3 or H2S2O8
    1. find the total mass of 
      1. H2SO3
        1. H=1x2=2
        2. S=32.066
        3. O=15.999x3=47.997
    2. add all the individual masses together
      1. H2SO3=82.063
    3. divide each individual mass by the total
      1. S(32.066/82.063=39%)
    4. find the total mass of
      1. H2S2O8
        1. H=1x2=2
        2. S=32.066x2=64.132
        3. O=15.999x8=127.992
    5. add all the individual masses together
      1. H2S2O8=194.124
    6. divide each individual mass by the total
      1. S(64.132/194.124=33.03%)
H2SO3 has a larger sulfur percent.

Empirical Formula and Molecular Formula

The empirical formula of a chemical compound is the simplest positive integer ratio of atoms of each element present in a compound. Using the ration you can identify the name of an unknown compound. In this blog i am going to demonstrate my understanding on how to use this formula. Lets start with some problems:



  1. You have 70% of Fe(Iron) and 30% of O(Oxygen) is the compound FeO or Fe2O3.
    1. Assume that one rock of Iron Oxide is 100g.
      1. you can than say that there is 70g of Fe and 30g of O
    2. next is to find the number of moles in 70g of Fe and the number of moles in O
      1. 70gFe=              mol/Fe, 30gO=             mol/O
    3. to figure this out you would look at the atomic mass of each element
      1. Fe has an atomic mass of 56 and O has an atomic mass of 16
      2. taking each atomic mass you divide by the number of grams
        1. 70/56=1.25, 30/16=1.875
      3. the answer will be the number mole in each element.
        1. 70gFe=  1.25  mol/Fe, 30gO=   1.875  mol/O
      4. you than take the lower number in this case 1.25 and divide 1.875 and 1.25 by 1.25
        1. 1.25/1.25=1, 1.875/1.25=1.5
        2. When you get a decimal like 1.5 you will multiply both answers by two
          1. 1.25/1.25=1x2=2, 1.875/1.25=1.5x2=3
      5. you than have your ration for every two iron there are three oxygen
        1. 2:3
This next problem I will not only use the empirical formula but to find the right answer when given to compounds that can work for the ratio I will than use the Molecular Formula.

Empirical Formula part 1
  1. YOu have 2.34g of N(Nitrogen) and 5.34g of O(Oxygen) you need to find out if this compound is No2 or N2O4? Weight is 92g/mol
    1. First find the number of moles in each
      1. 2.34gN=      mol/N, 5.34gO=      mol/O
    2. the atomic mass of N is 14 while the atomic mass of O is 16 divide the grams by the mass
      1. 2.34g/14=0.167, 5.34/16=0.334
    3. 2.34gN= 0.167  mol/N, 5.34gO=  0.334  mol/O
      1. divide by the lowest number(0.167)
        1. 0.167/0.167=1, 0.334/0.167=2
    4. you have a ratio of 1 nitrogen for every 2 oxygen
      1. 1:2
    5. Now find the weight simple add 16x2+14=46g/mol the weight doesn't match what we were given.
  2. Find out if the compound is N2O4 we have to use the molecular formula. You are given the weight of 92g/mol to get the answer you simply do the following.
    1. n x46=92g/mol
      1. 2x47=92g/mol
        1. answer is N2O4

Friday, January 13, 2012

The Mole

The mole is a unit of measure used in chemistry to show the amount of a chemical substance. One mole is always shown as the equation of 6.02X1023 or 602 billion trillion. A mole is abbreviated as mol, g/mol.-grams per mole. The number of a mole is always the same only the mass s different.When you are looking for the number of moles in an element for example 1 mol. of antacid Al(OH)3 would equal 78g/mol. this is because the mass of 1 mol of Al is 27, you than look at the other elements 1 mol of O is 16 but if you look at the compound it has Al(OH)3. You take the 3 and multiply it by the 1 mol. of O which gives you 48. You do the same with H and 1X3 which is 3. You than add all of them together 27+48+3=78.  There are other ways to solve a mole. Here are some of them:


  1. How many moles are in 15 grams of lithium(Li)
    1. to find out you first have to find out how much grams are in one mole of lithium.
      1. 7g/mol.
    2. Once found you than divide the number of grams so 15 by 7 to get the number of moles which is 2.13g/mol.
  2. How many grams are in 2.4 moles of sulfur
    1. now this is different we are looking for grams not moles so you really just to the opposite that you did to find moles.
      1. there is 32g/mol in 1 mole of Sulfur
    2. You than take the 32 and multiply by the number of moles you should get 76.8g/mol.
  3. How many moles are in 22 grams of argon
    1. 1 mol. Ar=40g/mol
      1. 22/40=0.55g/mol
  4. How many grams are in 88.1 moles of magnesium
    1. 1 mol Mg=24
      1. 24*88.1g=2114.4g/mol.
        1. *-multiply
        2. /-divide
  5. Now here are some with compound elements
    1. How many grams are in 4.5 moles of sodium fluoride, NaF
      1. to solve this you first have to figure out how many grams are in one mole of NaF to do this you find the mass and add them together
        1. Na=23g/mol., F=19g/mol.
          1. Na+F=42g/mol.
      2. taking the 42g/mol you would then multiply by 4.5 to equal 189g/mol.
        1. 42*4.5=189g/mol.
    2. How many moles are in 98.3 grams of aluminum hydroxide, Al(OH)3
      1. Al=27g/mol, O=16*3, H=1*3, than add Al+O+H=78g/mol.
      2. divide 98.3 by 78 to equal 1.26g/mol.
        1. 98.3/78=1.26g/mol
Hope this post helped you to understand the mole better bye!