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

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!

Chemical Reaction Lab

Last week during chemistry class we performed a total of seven different chemical reaction labs. For this blog I will describe each lab step so that if you want you can do it yourself, and the reaction we observed for each and the chemical formula for the lab. My group for this lab was Cynthia, and Damien.

Reaction Lab 1

1. Take a test tube and fill it with 1/2 full with 3M HCL(aq)-hydrochloric acid.
1. try not to get any on your hands, it will probably burn later for a while and don't forget to wash your hands after the lab. Be sure to wear googles!
2. Place the test tube in a rack
3. Obtain a piece of magnesium metal strip 
1. polish the magnesium stip before hand
4. Place a second test tube into a a test tube holder
5. Have your partner place the magnesium strip into the test tube of acid
6. Immediately hold the second test tube on top of the test tube with the magnesium strip and hydrochloric acid.
1. make sure your fast cause once you stick the magnesium strip in the hydrochloric acid it'll start to bubble up fast
7. Hold the test tube on top until the chemical reaction has stopped
1. in other words until the bubbling has stopped
8. write down any observations you see
1. for this part me and my group saw the chemicals become a clear liquid after a lot of bubbling
9. Test the gas in the top test tube
10. rotate the test tube away from anyone
11. Light a match and and bring it to the end of the test tube
1. you should here a loud popping sound to show the gases reaction to the flame. If you don't get that reaction then you didn't do it right
1. This is the reaction of the gas igniting from the flame
12. Record what happened (the popping noise)
13. When your done pour the liquid down the drain and rinse the tube out with water
1. careful when you pick up the tube it will be hot from the solution reaction.

For this lab as well as for the others we were told by Mr.Ludwig t write the chemical formula showing the reaction and the product.

Chemical Formula: Mg + 2HCl------>MgCl2 + H2 (Magnesium + Hydrochloric Acid------>Magnesium Chloride + Hydrogen) 

Reaction Lab 2

1. Light a Busen Burner
1. This lab was really cool
2. Take a small strip of magnesium and hold it in the flames using the tongs
1. be careful to keep the flames away from your hands don't want to get burned
3. Write down any Observations of anything that was produced including 
1. For this me and my group saw the magnesium strip burn with a really bright white light, after staring at it like an idiot my eyes started to hurt. When I blinked after i saw spots. After the strip finished burning it was black
1. This is a result in my opinion of the magnesium reacting to the oxygen
4. Place the product on a watch glass and allow it to cool. then trow it in the trash.

Chemical Formula: 2Mg + O2-----> 2MgO (Magnesium + Oxygen-------> Magnesium Oxide)

Reaction Lab 3

1. Take a copper wire and shine it using a piece of sand paper
2. Using the tongs, hold the wire in the flame of the Bunsen burner for several minutes
3. Write down observations
1. for this nothing really interesting happened aside from the wire change colors-black, red, and the flame had a green tint to it.
1. I think this is because of the copper reacting to the oxygen in the fire
4. put out the copper wire

Reaction Lab 4

1. Add small amount of ammonium carbonate, (NH4)2CO3 into the test tube
2. Using you test holder, gently heat the test tube
1. careful not to burn yourself
3. Waft the gas coming from the test tube towards your nose
1. don't just stick your nose in it
4. Write down what the gas smells like and any other observations
1. for this we smelled alcohol the smell irritated my nose, by the time we took it off the burner there was nothing in the tube
1. im guessing that evaporation happened

Chemical Formula: (NH4)2CO3-----heat----> NH3 + CO 2 + H2O

Reaction Lab 5:

1. Fill a test tube 1/2 full of hydrogen peroxide, H2O2
2. Add a small amount of magnesium dioxide, MnO2 to the test tube
1. should start to bubble
3. Using a test tube holder, immediately hold an inverted test tube over the first test tube
4. When the reaction is complete, hold the test tube upside down
1. The top test tube
5. Light a match
6. Blow out the flame
7. Immediately put the glowing tip into the inverted test tube
1. if it doesn't work light another match, blow it out and put it in the black solution 
8. write down observations
1. the tip should light back up again
2. I believe it reacted this way because of the carbon dioxide in the solution
9. pour the solution down the sink and rinse with water

Chemical Formula: 2H2O2(aq) → O2(g) + 2 H2O(l)

Reaction Lab 6:

1. Pour a small amount of Potassium iodide, KI into a test tube 
2. Pour a small amount of lead(II) nitrate, Pb(NO3)2 into a test tube 
3. Record the colors of both solutions 
1. KI should be clear
2. Pb(NO3)2 should be clear
4. Pour the chemicals into one test tube
1. should turn the color yellow
5. Record any observations
1. obviously the color change
2. this was one of the more boring experiments
3. reaction happened from chemical bonding
6. Pour the the contents

Chemical Formula:Pb(NO3)2+2 KI--->PbI2+ 2KNO3

Reaction Lab 7:

1. Place a small amount of copper(II) carbonate, CuCO3 into a test tube 
2. Use one test tube holder to heat the test tube while using a second inverted test tube and test tube holder to collect any gas given off
3. Light a match and place it into the test tube that was on top
4. Record any observations
1. burned up
2. smoke moved down the tube
3. another boring one
5. Add water to the test tubes and pour the contents down the drain

Chemical Formula:CuCO3----heat--->CO2 + CuO

Conductivity Lab

Yesterday in Chemistry we did a lab where we had to find the conductivity of different solutions.

For this experiment I chose the solutions Sodium, Corn Syrup, Calcium Acetate, Citric Acid, Molasses, Methyl Cellulose, and Corn Starch. The graph below show the results.

Periodic Table Scavenger Hunt

https://docs.google.com/document/d/1Fssyk1T8S0et53TWsNm-K2fTyBlx_wRA0gLInc491HE/edit

LIne Spectra

Yesterday in Chemistry class we performed a lab were we got to look through a kaleidoscope at  spectra tube. Taking the kaleidoscope we were to look at the light and draw the colors we saw. This was to help us learn the difference between a continuous, emission, and absorption spectra.




Looking at regular light we saw what a continuous spectra was. It was basically all the colors blending into one another as you looked across it.




Next we looked at the Absorptions Spectra. Which was the same as the continuous spectra but some colors like orange had black over it because the energy was absorbed.

Last we looked at the Emission Spectra. This was were we were only able to see a few lines of colors
We were shown Hydrogen, Neon,  and Mercury.

Hydrogen

Neon

                                    Mercury

PIctures

http://intro.chem.okstate.edu/AP/2004Norman/Chapter7/Emission.html

http://astro.unl.edu/naap/hr/hr_background1.html

http://www.astronomyknowhow.com/hydrogen-alpha.htm

http://www.jce.divched.org/JCESoft/Issues/Series_C/6C1/prog3-6C1.html

http://www.jce.divched.org/JCESoft/CCA/CCA8/SampleUseOfMovies/CJM/8_01_9_5_4_.html

http://www1.assumption.edu/users/bniece/Spectra/Complete.html