Although I wasn't here for the lab Timmy, Serena, and Tenchita are nice enough to let me use their data to do a write up about this experiment. They chose to do the experiment involving epsom salt. They put the epsom salt in a test tube and weighed it before any other experimentation. The weight before heating the salt was 4.07 grams. They then heated the salt (telling me that Kandace almost killed them a couple times with matches and the gas). While heating the tablets turned white which they then let cool. After letting the flaming salt cool they weighed it after, noting that it weighed 3.66 grams. They then added water, learning that the crystals turned back to blue.
By subtracting the weight after heating from the weight before heating, we discovered that the crystal only had about 10% water. This is different from the expected percentage because of crystals that got stuck in the tube and crystals that were still blue.
The formula for the crystals is CuSO4 5H20. After multiplying the total CuSO4 mass by 5 parts water, I learned that the total water mass is 90 grams while the total mass was 249.6 grams. After dividing the 90 grams by the 249.6 grams, we discovered that the expected mass was 36%.
Wednesday, January 25, 2012
Sunday, January 22, 2012
Calculations
Moles!
No not the animals, more like the number of moles in an element how the amount of moles needed for a compound.
One mole equals 6.02x10^23 atoms.
So, how do you find how many grams per mole an element has? For example, the element Bromine has a mass of approximately 79.9 so it is 79.9 grams/mol. Now, looking at a compound such as K2O. Potassium has a mass of 39.098 and Oxygen has a mass of 15.9994. Since there are two potassium atoms, we have to multiply the 39.098 by 2 and then add the 15.9994. The final answer is 94.196 K2O. This means there are two potassium atoms for every one oxygen atom, but I'll get into that a little bit later.
Looking at a more complex compound such as C14H18N2O5, it's easier to find the measurement of moles in the compound. First you must calculate the grams per mole, which turns out to be 294. Then you take the mass of the smallest element, which is 225 grams. Divide the 225 by the 294 to get the answer. C14H18N2O5 is .765 of a mole.
Percent of Composition
Let's find the percent of composition of Carbon in an artificial sweetener. The compound is C5H8NO4. To get the percent of composition one has to divide the mass of the total Carbon by the total weight of the compound. Carbon has a 12.0107 mass, but there are 5 atoms of Carbon, so the total mass is 60.055. The total weight of the sweetener is 146.1212. 60.055/146.1212 is approximately 41%. So, Carbon is 41% of the artificial sweetener compound.
The Empirical Formula
The Empirical Formula helps a scientist (or student) determine the compound that is put in front of them based on the given weights. For example, when 2.34 grams of Nitrogen and 5.34 grams of Oxygen are put in front of you, by dividing by the weight we find out that there are .167 moles of Nitrogen and .333 moles of Oxygen. Take the .167 and divide it by .167(the least amount of moles) and do the same for .333 which rounds out to 1 to 2. So, the ratio is one to two. But, that could mean that the compound is NO2 or N2O4. The fact that the compound weighed 93grams was given. Through multiplication, it is found that the compound is N2O4 (Dinitrogren Tetroxide)
No not the animals, more like the number of moles in an element how the amount of moles needed for a compound.
One mole equals 6.02x10^23 atoms.
So, how do you find how many grams per mole an element has? For example, the element Bromine has a mass of approximately 79.9 so it is 79.9 grams/mol. Now, looking at a compound such as K2O. Potassium has a mass of 39.098 and Oxygen has a mass of 15.9994. Since there are two potassium atoms, we have to multiply the 39.098 by 2 and then add the 15.9994. The final answer is 94.196 K2O. This means there are two potassium atoms for every one oxygen atom, but I'll get into that a little bit later.
Looking at a more complex compound such as C14H18N2O5, it's easier to find the measurement of moles in the compound. First you must calculate the grams per mole, which turns out to be 294. Then you take the mass of the smallest element, which is 225 grams. Divide the 225 by the 294 to get the answer. C14H18N2O5 is .765 of a mole.
Percent of Composition
Let's find the percent of composition of Carbon in an artificial sweetener. The compound is C5H8NO4. To get the percent of composition one has to divide the mass of the total Carbon by the total weight of the compound. Carbon has a 12.0107 mass, but there are 5 atoms of Carbon, so the total mass is 60.055. The total weight of the sweetener is 146.1212. 60.055/146.1212 is approximately 41%. So, Carbon is 41% of the artificial sweetener compound.
The Empirical Formula
The Empirical Formula helps a scientist (or student) determine the compound that is put in front of them based on the given weights. For example, when 2.34 grams of Nitrogen and 5.34 grams of Oxygen are put in front of you, by dividing by the weight we find out that there are .167 moles of Nitrogen and .333 moles of Oxygen. Take the .167 and divide it by .167(the least amount of moles) and do the same for .333 which rounds out to 1 to 2. So, the ratio is one to two. But, that could mean that the compound is NO2 or N2O4. The fact that the compound weighed 93grams was given. Through multiplication, it is found that the compound is N2O4 (Dinitrogren Tetroxide)
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