Friday, April 8, 2011
Wednesday, March 16, 2011
Silver Copper Replacement Lab? Wait...what?
Fact: Copper wire reacts with aqueous silver nitrate.
Fact: We did a lab about this 5 weeks ago.
Fact: I should not slack on my blog posts.
So obviously we did another lab, the purpose of this one was to demonstrate a replacement reaction, and show us mole ratios in action, in our case we were taking silver nitrate and copper, trading out the silver for copper, and creating copper nitrate and pure silver. We did this by suspending a copper wire in an aqueous solution of silver nitrate and letting it sit for a day, and then observing the reaction. To make the aqueous solution we mixed silver nitrate (AgNO3) with distilled water until it had dissolved. We then placed a coil of copper wire in the solution, covered the tube with wax, and let it sit in the fume hood until day two. During the time we were waiting for the reaction to take place we sat down and did some math. Using the equation 2AgNO3 + Cu ----> Cu(NO3)2 + 2Ag and we formed predictions for how much silver should be formed and how much copper became Cu(NO3)2 in the reaction. I won't bore you with the details but, our predictions said that we should expect to form about .4513g (.0071 mol) of Ag and loose about .2269g (.0036 mol) of Cu in the reaction otherwise known as a 1:2 ratio of replacement. To explain how the experiment works in a more descriptive manner I will refer to one of my lab partners description: <In the experiment, copper changed from its elemental form, Cu, to its blue aqueous ion form, Cu2+(aq). At the same time, silver ions (Ag+(aq)) were removed from solution and deposited on the wire in the elemental Ag metallic form.> On day two we observed that a crystal structure had formed on the copper wire in the tube (fig. 1). Sadly, we had to remove the structure to measure our results. We then emptied the test tube, and filtered the silver particles out of the copper nitrate solution, set the particles and the wire in the fume hood to dry overnight, and disposed of the copper nitrate solution. We are able to tell how much copper was used by measuring the new mass of the wire and comparing it to the mass from before the experiment. On day three we weighed the copper coil and silver nitrate, and then got to work figuring out how much Ag was formed and how much Cu was lost. In other words, more math...
Our predictions said that we should expect to form about .7703g of Ag and loose about .2269g of copper. The first step was changing the predictions we had from grams to mols, when we did this we got .0103 mol.of Ag, and .0036 mol of copper. In other words a 1 to 3 ratio of Cu to Ag. When this information was compared to our predictions (which said we would have a 1 to 2 ratio of Cu to Ag) we got 144% yield of silver and a 100% yield of copper. The inconstancy most likely stems from the paper we weighed our Ag on was still slightly wet and added weight to our readings. All in all, this was a pretty fun lab.
Fact: We did a lab about this 5 weeks ago.
Fact: I should not slack on my blog posts.
So obviously we did another lab, the purpose of this one was to demonstrate a replacement reaction, and show us mole ratios in action, in our case we were taking silver nitrate and copper, trading out the silver for copper, and creating copper nitrate and pure silver. We did this by suspending a copper wire in an aqueous solution of silver nitrate and letting it sit for a day, and then observing the reaction. To make the aqueous solution we mixed silver nitrate (AgNO3) with distilled water until it had dissolved. We then placed a coil of copper wire in the solution, covered the tube with wax, and let it sit in the fume hood until day two. During the time we were waiting for the reaction to take place we sat down and did some math. Using the equation 2AgNO3 + Cu ----> Cu(NO3)2 + 2Ag and we formed predictions for how much silver should be formed and how much copper became Cu(NO3)2 in the reaction. I won't bore you with the details but, our predictions said that we should expect to form about .4513g (.0071 mol) of Ag and loose about .2269g (.0036 mol) of Cu in the reaction otherwise known as a 1:2 ratio of replacement. To explain how the experiment works in a more descriptive manner I will refer to one of my lab partners description: <In the experiment, copper changed from its elemental form, Cu, to its blue aqueous ion form, Cu2+(aq). At the same time, silver ions (Ag+(aq)) were removed from solution and deposited on the wire in the elemental Ag metallic form.> On day two we observed that a crystal structure had formed on the copper wire in the tube (fig. 1). Sadly, we had to remove the structure to measure our results. We then emptied the test tube, and filtered the silver particles out of the copper nitrate solution, set the particles and the wire in the fume hood to dry overnight, and disposed of the copper nitrate solution. We are able to tell how much copper was used by measuring the new mass of the wire and comparing it to the mass from before the experiment. On day three we weighed the copper coil and silver nitrate, and then got to work figuring out how much Ag was formed and how much Cu was lost. In other words, more math...
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| (fig.1) |
Intermolecular Forces Lab (With a mix of Data, and Words)
It's time for another blog post!!! It's only been about a month since my last one, so I'm a little ahead of my normal schedule, but this post is a rather good one on Intermolecular Forces (the bonds that hold the elements together). Recently, we did a lab in chemistry where we measured how different chemicals (in this case alkanes) can affect the rate of cooling of a temperature probe in order to understand how molecular weight plays a part in intermolecular forces. Needless to say, the temperature probes and I did not get along very well.
The basic premis of the experiment was to see how much energy was used by the element when it changed from a liquid to a gas, through evaporation in order to determine if molecular weight is a factor in inter molecular forces [in other words, does molecular weight play a part in the amount of energy needed by an element to induce a change of state (liquid to gas)]. We are able to use temperature to determine the amount of energy needed to change the state of matter thanks to the laws of thermodynamics. Since we know that heat flows from the warmer object (the probe) to the cooler substance (the liquid alkanes) until an equilibrium is reached, state changes are determined by the speed of molecules and that the First Law of Thermodynamics states that "Energy cannot be created or destroyed, only changed from on form to another" then we can correctly assume that the liquid alkanes (having been cooled to below zero in order to liquidize) will take the energy (in the form of heat) from the temperature probe until they evaporate (that energy is used to make the molecules move faster, which changes state), the amount of energy (in the form of heat, measured by temperature) used by the substance to change states is then visible as the amount of energy lost by the temperature probe.
In order to determine this we were to two stick temperature probes (fig. 1) into a pair test tubes each containing the liquid form of one of six different alkanes (Ethanol, Methanol, Propanol, Butanol, Pentane, and Hexane) let the probes sit in the tube for 30 seconds to determine their starting temperature, then remove the probe and tape it to the edge of the desk and measure the decrease in temperature over the course of 5 minutes. After recording the results of all six substances in "Logger Pro" we then found the change in temperature for each substance by subtracting the lowest temperature reached from the starting temperature. I then plotted the decrease in temperature (y-axis) vs. the molecular weight (x-axis) for each element on a graph in order to see if there was a correlation between the molecular weight, and the amount of energy needed to change the state of the substance.
The basic premis of the experiment was to see how much energy was used by the element when it changed from a liquid to a gas, through evaporation in order to determine if molecular weight is a factor in inter molecular forces [in other words, does molecular weight play a part in the amount of energy needed by an element to induce a change of state (liquid to gas)]. We are able to use temperature to determine the amount of energy needed to change the state of matter thanks to the laws of thermodynamics. Since we know that heat flows from the warmer object (the probe) to the cooler substance (the liquid alkanes) until an equilibrium is reached, state changes are determined by the speed of molecules and that the First Law of Thermodynamics states that "Energy cannot be created or destroyed, only changed from on form to another" then we can correctly assume that the liquid alkanes (having been cooled to below zero in order to liquidize) will take the energy (in the form of heat) from the temperature probe until they evaporate (that energy is used to make the molecules move faster, which changes state), the amount of energy (in the form of heat, measured by temperature) used by the substance to change states is then visible as the amount of energy lost by the temperature probe.
In order to determine this we were to two stick temperature probes (fig. 1) into a pair test tubes each containing the liquid form of one of six different alkanes (Ethanol, Methanol, Propanol, Butanol, Pentane, and Hexane) let the probes sit in the tube for 30 seconds to determine their starting temperature, then remove the probe and tape it to the edge of the desk and measure the decrease in temperature over the course of 5 minutes. After recording the results of all six substances in "Logger Pro" we then found the change in temperature for each substance by subtracting the lowest temperature reached from the starting temperature. I then plotted the decrease in temperature (y-axis) vs. the molecular weight (x-axis) for each element on a graph in order to see if there was a correlation between the molecular weight, and the amount of energy needed to change the state of the substance.
| (Fig. 1) |
Now, above you can see my afore mentioned graph of the results of the lab. The first thing that you can see is that there is a very obvious correlation between the molecular weight, and the amount of energy used to change the state of matter. With the first four liquids (represented with boxes, I'll talk about why the circles are different in a moment) you can see that as the molecular weight gets bigger, the amount of energy needed needed to change state is lower. This is because as length of hydrocarbon chains increase (Ethanol has 1, Methanol 2, Propanol 3, and Butanol 4) the strength of the hydrogen bonds holding them together decreases, which makes it easier for the liquids to evaporate. Now, the reason that Pentane and hexane are not following the downward trend set by the other four substances is that they are not held together by a weak hydrogen bond (they can't hold more than 4 hydrocarbon chains together) they are held together by a stronger bond that requires more energy to break, but like the hydrogen bond, gets weaker with the each new set of hydrocarbon chains which would make the pattern repeat its self.
So in conclusion, this lab was very useful in demonstrating a number of things. First, it shows how molecular weight affects the bonds within the element, which affect its evaporation time. It also helped demonstrate how some of the laws of thermodynamics work, especially how heat flows from one place to another. And lastly, it proved that temperature probes hate me (spending 15 mins trying to figure out which probe is which was wa to long..)
Friday, January 28, 2011
Mol's, Equations, and Popcorn....Oh My!
Before I get into the chemistry aspect of this post, let me begin with (quite possibly) the most important statement of this entire post: I love popcorn. Over the past 2(?) weeks since school started back up, we've been talking about some pretty interesting stuff, more specifically, mol's and how they relate to the measurement of reactants and products. (In other words, its time for a hypothetical discussion)
602,000,000,000,000,000,000,000-And why it matters
*Warning: Hypothetical Discussion Follows*
Lets say you work for a big chemical company (ie: DOW) and you need to mix up a batch of Sodium Benzoate (Na C6 H5 CO2) the first problem that you will encounter is how to you ensure that you get the proper ratio of atoms in the mixture. In a normal recipe one would typically use a cup of this ingredient and a teaspoon of another to gain that ratio. In chemistry though the ratios are specific to the number of molecules needed to complete the reaction. One unit of Sodium Benzoate contains one Na atom, 6 Carbon atoms, 5 hydrogen, and one molecule of CO2. Sadly you can't just measure out on atom of sodium to throw into the mix, so a old dead scientist (Jean Perrin) through a series of complicated math, came up with Avagadro's (attributed to Amedeo Avagadro) number. Avagadros number is approximately 6.02*10^23 or 602,000,000,000,000,000,000,000. What this number ends up meaning is that there are 602,000,000,000,000,000,000,000 atoms of the element in every mol of that element. Now the weight of a mol for each element is different (due to density and other fancy factors) but easily found on any periodic table. The atomic weight of an element is the weight of one mol of that element. Therefore when you mix up a batch of Sodium Benzoate you will mix 1 mol of sodium (22.990g), 6 mol's of Carbon (12.001g per mol), 5 mol's of Hydrogen (1.0097g per mol) and 1 mol of CO2 (44.009g). When you mix all of this together, due to the Law of Conversation of Mass, the total weight of the final mixture should be 144.0535g. Thereby allowing you to correctly calculate the amount of product needed to produce "X" grams of the solution, solving the second problem you would encounter: How much product do I need to purchase to produce the final chemical. By solving that issue you allow the people in accounting to allocate the funds needed to buy the products and by doing this they can calculate the price that they will charge for the final product, which ultimately means....YOU GET PAID (If you don't get why thats important...I would like to take a moment to welcome you to Earth!) And thats why 602,000,000,000,000,000,000,000 (six-hundred and two million billion, for the record) is important.*Warning: Hypothetical Discussion Follows*
Equations: Why They Exist, and Why I Hate them
| Welcome to Chem 111 |
Unfortunately for the non-mathmatically inclined, Chemistry involves equations. Today, these equations will involve going from precent of composition, Mass to Mol's and writing chemical equations. Equations like these are rather useful in chemistry. They can help you determine basically everything you could ever want to know about a mixture. In the above hypothetical situation, we use an equation to determine the ratio of grams of elements. To do this you determine the atomic weight of every element of the equation, multiply those numbers by the number of mol's needed of each element. To determine the ratio of the elements you divide each of the numbers by the smallest. This creates a gram to gram (is that a real term?) ratio. Which eventually tells you that for every gram of element X that you have you need to add 5 grams of element Y for the correct reaction. Another way that equations are used in chemistry is to determine what a solution is (by way of empirical formulas) based on the percent composition of elements. Take the percent composition of 10.04% Carbon, 0.84% Hydrogen, and 89.12% Chlorine. To determine the empirical formula we are first going to assume that we have a 100g sample of the substance. By assuming this it makes it very easy to determine the grams (10.04% of 100g = 10.04g.) Once you have the number of grams of each element you divide the grams by the molecular weight of the element, which gives you the number of moles of each element. This division gives you .846 mol of Carbon, .832 mol of Hydrogen, and 2.51 mol of Chlorine. The next step is to divide each of the number of mols by the smallest number of mols (.832 mol's of H) which will give you the (approximate(error of margin is always an issue to those super picky chem teachers)) ratio of elements. In this case the ratio is 1C : 1H : 3Cl, which translates into a molecular formula of CHCl3, otherwise known as Chloromethane or Methyl Chloride.
Originally I planned to include data from the recent popcorn experiment we did, but since we just finished another much more interesting experiment I will put a post about that sometime soon, especially since I have a video of it! So, until then, in the words of Spock, "Live Long & Prosper."
Originally I planned to include data from the recent popcorn experiment we did, but since we just finished another much more interesting experiment I will put a post about that sometime soon, especially since I have a video of it! So, until then, in the words of Spock, "Live Long & Prosper."
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