Silver Nitrate Lab
Physical Description
Before
- water
- clear, no bubbles, colorless
- copper wire
- silver nitrate crystals
- floating in water and in a small pile on the the bottom of the vial
- shimmery, light grey
After
- water
- blue, transparent/not opaque
- copper wire
- reddish-orange
- thinner where in water
- dull
- silver fluff
- spiky and crystalline (like needles) in center, darker- and softer-looking on outside
- growing off of the copper wire
Quantitative Measurements
Before
beaker with vial and water: 107.71g
beaker with vial, water, and silver nitrate: 108.72g
108.72g - 107.71g = 1.01g silver nitrate
copper wire: 3.3g
After
copper wire: 3.11g
3.3g - 3.11g = 0.19g copper used in reaction
---
Moles
silver nitrate:
1.01g x 1mol/169.88g = 0.0059mol
copper used in reaction:
0.19g x 1mol/63.546g = 0.003mol
Particle Diagram
Before
After
Chemical Symbols/Equation
Before
2Ag(NO3)+Cu
After
Cu(NO3)2+Ag
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In our lab, we started with a copper wire and a vial containing water with dissolved silver nitrate, both of which we measured. We stuck the copper wire on a scale to measure its weight of 3.3g. We also measured the vial in a beaker while it only had water in it, then again when it contained the silver nitrate, subtracting the first measurement from the second so that we got the weight of the silver nitrate (which was 1.01g). Then, after the reaction had occurred, there was a silver metallic growth off of the copper wire, the liquid had turned partly blue, and the copper wire had become thinner. We remeasured the copper wire and got 3.11g, and when we subtracted that from the original measurement, we found that 0.19g had been used in the reaction. I found how many moles there were of silver nitrate (assuming we used all of it) using our original measurement of how much there was (1.01g) and multiplying it by its conversion factor (1mol/169.88) to get the number of moles: 0.0059. Then I did the same with the measurement I had of the amount of copper that was used in the experiment, getting 0.003 moles. Since 0.0059 is almost double 0.003, I realized that there was double the amount of silver nitrate molecules in the reaction than there were copper atoms. This caused me to create a chemical equation where 2Ag(NO3)+Cu is the reactant side and add a 2 in front of the silver on the product side to account for the two silver atoms in the reactants, as well as use the fact that copper can make two bonds to add a 2 after the nitrate (a polyatomic ion) to account for the two nitrates in the reactants. This chemical equation allowed me to display the correct molecules in the after section of my particle diagram.
Empirical Formula Lab
Physical Description
Before
- magnesium
- grey
- in little strips
- solid/metal
- oxygen
- colorless
- oderless
- gas
- in the air
After
- magnesium
- white-grey
- partially melted
- powder
- oxygen
- colorless
- oderless
- gas
- in the air
Quantitative Measurements
Before
crucible: 26.21g
crucible with magnesium: 26.48g
26.48g - 26.21g = 0.27g magnesium
After
crucible with contents: 26.62g
26.48g - 26.62g = 0.41g product
0.41g - 0.27g = 0.14g oxygen in product
---
Moles
magnesium:
0.27g x 1mol/24.305g = 0.0111mol
oxygen:
0.14g x 1mol/15.999g = 0.0088mol
Particle Diagram
Before
After
Chemical Symbols/Equation
Before
2Mg + O2
After
2MgO
---
In our lab, we started with magnesium shavings in a crucible, which we measured. We also had measure the crucible without the magnesium, and we subtracted that value from the measurement of the crucible with magnesium to get the weight of the magnesium (0.27g). Then we heated the crucible and caused a reaction that resulted in a powder in the crucible. We remeasured the crucible with its contents and got 26.62g, and when we subtracted the weight of the crucible from that, we found that the product of the reaction was 0.41g. Then we subtracted the weight of the magnesium to get the weight of the other reactant which was identified as oxygen: 0.14g. I found how many moles there were of magnesium (assuming we used all of it) using our original measurement of how much there was (0.27g) and multiplying it by its conversion factor (1mol/24.305) to get the number of moles: 0.0111. Then I did the same with the measurement I had of the oxygen, getting 0.0088 moles. Since 0.0088 is closer to the same amount as 0.0111 than it is too double or half the amount, I realized that there was a 1:1 ratio of magnesium to oxygen in the reaction. Since I know that oxygen bonds to itself in its pure state, which it would be in on the reactant side, I put a 2 after the oxygen on the reactant side and doubled the amount of magnesium on that side by putting a 2 in front of it so that the ratio of atoms remained 1:1. This caused the final product to be two molecules of MO. This chemical equation allowed me to display the correct molecules, with only one magnesium to each oxygen, in the after section of my particle diagram.
Aluminum and Blue Stuff Lab
Physical Description
Before
After
- liquid
- aluminum
- shiny
- silver-colored
- smaller
- red-stuff
- sponge-like
- red-brown/rust colored
- some on bottom of vial, some clinging to aluminum
Particle Diagram
Before
After
Chemical Symbols/Equation
Before
3CuCl2+2Al
After
2AlCl3+3Cu
---
In our lab, we started with aluminum foil in a ball and a blue liquid in a vial. We took no quantitative measurements, but we were given the elements used in the reaction. The blue liquid was copper chloride, and the ball of metal was, as stated, aluminum. Using tests, we figured out that the red sponge material that resulted from the reaction was copper, and that the liquid (which had turned colorless) was aluminum chloride. Using the periodic table, we figured out that aluminum had three bonds, so it needed to have three chlorine atoms attached (since they only had one bond). This meant that there would have to be three of the copper chloride molecules in the reactants so that there would be six chlorine atoms, which could be used to form two aluminum chloride molecules. In order to have the two aluminum necessary for the two molecules in the product, I put a 2 in front of the aluminum in the reactants, and a 3 in front of the copper in the product to account for the three copper used in the molecules in the reactants. This allowed me to have an accurate representation and number of molecules in my particle diagram by specifying the number of molecules minimum needed to represent the reaction and teaching me how many of what atoms were bonded to what.
Water Investigation
Physical Description
Before
- water
- half-way filling beaker, filling vials almost all the way full
- clear, no bubbles
- soapy/slippery feel
- vial
- two
- upside down
- very small amount of air in the top of vials, otherwise filled with water
- wires
- one black, one red
- go a little bit up into the bottom of each vial
- attached to a large battery
After
- water
- filling less of vial
- clear, were bubbles
- still soapy
- vial
- two
- upside down
- more gas at top of each vial, the red wire with double the gas of the black wire
- gasses
- gas in red wire tube:
- when exposed to flame, was a popping sound
- twice the amount of black wire
- gas in black wire tube:
- when exposed to flame, flame flared
- half the amount of red wire
Quantitative Measurements
Before
amount of gas in black-wire vial: 0cm
amount of gas in red-wire vial: 0cm
battery: 43.75g
After
amount of gas in black-wire vial: 1.5cm
amount of gas in red-wire vial: 3cm
battery: 43.75g
Particle Diagram
Before
After
Chemical Symbols/Equation
Before
2H2O
After
2H2+O2
---
In our experiment, we ran electricity from a battery through two wires (one red, one black), that were submerged in water. These wires each had an upside-down vial filled with water covering their ends, which started fizzling after electricity started running through them. After some time, gas had accumulated in the top of each vial, the red vial containing double the amount as the black vial. We discovered that the gas in the red vial created a popping sound when exposed to fire, denoting it as hydrogen, while the black vial gas flared the flame, meaning it was oxygen. Because our measurement showed that there was twice the amount of hydrogen as oxygen (3cm of hydrogen: 1.5cm of oxygen), we knew that whatever reactant the product came from must have had two hydrogen atoms for every one oxygen atom. This lead us to the conclusion that the reactant was the molecule H2O. Since we knew that H2O is water, we proposed that the electricity was splitting the water in a decomposition reaction, and that informed my choice on having the hydrogen and oxygen in my particle diagram come from preexisting water molecules. In the chemical equation, having merely one water molecule on the reactant side was incorrect, however, because oxygen on its own bonds to itself, and thus I would need two oxygen atoms on the reactant side in order to have the correct diatomic oxygen molecule on the product side. This made me add the 2 in front of water molecule so that there would be two oxygen atoms, which in turn caused me to add a 2 in front of the diatomic hydrogen molecule on the product side to account for the doubled amount of hydrogen atoms on the reactant side.
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