Three Weekly Questions, 3/28/14 (extra credit blog)

1. What tasks have you completed recently?
2. What have you learned recently?
3. What are you planning on doing next?

1. I helped clean up some of the nearly countless dead weeds growing in our backyard, I almost 100% completed a video game that I have been playing for about 8 months now, I started building a large library on the game Minecraft which will almost certainly take months to complete, and I otherwise enjoyed my spring break. I also recently read some of a newly refurbished, 38 year-old calculus book, and a geometry book I got from my school’s library two weeks ago, and I’ve helped out at least two people with some kind of homework.

2. I recently learned some facts about ellipses and other geometrical figures, I learned that a square is the rectangle with the largest area for a given perimeter, and I learned many other small and insignificant facts that are too numerous to mention here. For example, one gram of the “activated” carbon that is used in many water filters can have a total surface area greater than 500 m2.

3. I plan on spending the last two days of my spring break reading, watching cartoons, and being outside. I also plan, on Monday, on asking three of my teachers how the report card I was recently sent could show me having a total of nine absences, especially since I haven’t yet missed a single day or period of class this entire year.


Three Weekly questions, 3/14/14

1. What tasks have you completed recently?
2. What have you learned recently?
3. What are you planning on doing next?

1. I recently helped assemble an elliptical machine that took about an hour and a half to set up, completed three lab projects in my Chemistry class including one that dealt with flames and crucibles, and I beat my friend in chess again after he accidentally “gave up” has queen.

2. I recently learned about the very large number of programs that were started in the 1930s to help alleviate the Great Depression in America and I learned how the concept of the mole is used in chemistry. I also learned how to find the gravitational acceleration of an object orbiting the Earth (or really, any other planet or object) if you know the radius of the Earth and the object’s height above the Earth, and I learned a tiny portion of Relativity in my Physics class, and how traveling near the speed of light messes thing up; for example, one year for you on a rocket at 96% the speed of light ages you one year, but would age a twin of yours by 3.6 years.

3. I plan on having some free time this weekend, I foresee not being able to watch some simple cartoons like how I wanted to, and I plan on completing an “extra credit” Glog about “Mole Day”. I also plan on helping some family members dig up some small trees and replanting them somewhere else, being able to read some math and/or geometry books during my free time next week during my school’s period of compulsory testing, and being able to also enjoy Spring Break the week after that.

“Beans in a Pot”

Today, we measured the masses of five different kinds of beans in groups of 50 each, so that we could compare them to the use of the mole in chemistry.  So, similar to the mole, we used the unit of “pots”, which were used to count certain amounts of beans that we had.

My picture

My picture of group data

From the smallest mass in the lab (lentils), we calculated “relative masses”–the masses of each sample of beans divided by the mass of the lentils. This step was included mostly because lentils were used as single “units” and it would show us how many “units” of mass each sample weighed, which would make it easier to solve for the number of beans we would calculate from each sample’s relative mass. The relative masses would also function like molar masses (how many grams per mole of substance), except they would show the masses of one “Pot” of a type of bean. And for every type of bean except Lima, I personally calculated 20 beans per pot, while Lima beans numbered 22 per pot.

The pot is a model for a mole because, like the mole, we can now use it to determine the mass or number of beans in any particular amount or mass of beans. For example, if we had 250 grams of lentils and wanted the number of beans, we could use its “molar mass” of 20 beans per gram, multiplied by 250 grams, to get 5,000 beans. If we had 5 pots of kidney beans, we could use its mass of 9.904g per pot times 5 pots to get 49.52 grams, etc.

And just like how we used relative masses with the beans, chemists use relative masses, calculated by dividing the atomic mass of an element by the mass of one hydrogen atom, which is, like lentils, the element with the smallest mass. I mean, it’s technically divided by 1/12th the mass of a carbon-12 atom, probably because that isotope of carbon is the most abundant and carbon is “so super important” that we had to find some use for it as a standard, but the mass of that 1/12th is essentially 1 amu, which is hydrogen’s mass anyways. And for reading this far, here’s another picture.

My picture

My picture

“The Mole…in All its Glory “

“What have you really learned about it?  What surprised you?  What seems ‘common sense’?”

The concept of the mole and Avogadro’s Number isn’t really new to me and I’ve known about it since the 8th grade, except I haven’t ever had to use it in any class before this current Chemistry class of mine. I think the one major thing I’ve learned about it in this class is that it is used to “standardize” the amounts of the particular elements and/or compounds you have, and so you can use it to determine empirical and molecular formulas of compounds.

One thing that I guess surprised me, although only ever so slightly, is how easy it is to use and work with once you know what you’re doing. It also seems like a versatile concept that can be used in many ways and in many situations to calculate many different things in a chemical composition.

One thing that I think is obvious or ‘common sense’ is that atoms are too small to work with individually. Therefore, it was only natural for the concept of the mole and its companion Avogadro’s Number to be created to make uncountable amounts of atoms easy to work with.

Finally, here’s my own picture of the cat in a bow tie.

This is mine

This is mine

“How to – Percent Composition”

Percent composition is just what it sounds like: it is finding the percentages of elements that make up a chemical compound. For this post, let’s use sulfuric acid, whose chemical formula is H2SO4. Now, it may look like the atoms are arranged in a ratio of 2 hydrogen to 1 sulfur to 4 oxygen, because that is correct. What isn’t correct, though, is the thought that the percent composition of hydrogen here is (2*100)/ 7= 28% of the entire formula. This is because hydrogen, sulfur, and oxygen all have very, very different masses, which must be taken into account, because the more mass of an element you have, the more atoms of it there are that can form sulfuric acid.

To solve this, we go by the total number of atoms there are in a particular mass of an element. This is called a mole, and is similar to a dozen; while a dozen counts 12 objects at a time, a mole counts 6.022*1023 atoms at a time. While that number, named after its inventor Avogadro, may seem like a big and formidable number, it in fact is, but atoms make up for it by being so tiny. We also use what are called molecular masses, which tell us the mass of one mole of an element or compound. The mole mass of an element is determined by its atomic mass; for example, an element with atomic mass of 5.004 means that one mole of element weights 5.004 grams. With compounds, you would use the sum of all the elements’ atomic masses, multiplied by whatever subscript they have. So finally, let’s do some math.

Hydrogen has atomic mass of 1.008, so that is 1.008 grams in one mole of H . For sulfur, you have 32.06 grams per mole. For oxygen, 16.00 g/mol. Multiply by subscripts to get:
H: 2(1.008g/mol) = 2.016g/mol.
S: 1(32.06g/mol) = 32.06g/mol.
O: 4(16.00g/mol) = 64.00g/mol.

These sum to 98.076 g/mol and round to 98.08g/mol. Since the mole counts the same number of atoms, there are 2 moles of H for every 1 mol of S and for every 4 mol of O. Now that their different masses are all taken into account, we can get rid of the /mol part, and we can divide the mass/amount of each element over the mass/amount of the whole compound, times 100, to get percent compositions:

H: (2.016g of H  / 98.076g of H2SO4)*100 = 2.056%, or 2.056g H per 100g H2SO4.
S: (32.06g S / 98.076g H2SO4)*100 = 32.69%, or 32.69g S per 100g H2SO4.
O: (64.00g O / 98.076g H2SO4)*100 = 65.26%, or 65.26g O per 100g H2SO4.

If you would like, there is this page with more examples on finding percent compositions. Thank you for reading.

“Science by the Numbers”

As many people probably don’t know, there is an actual difference between accuracy and precision. In an experiment, to be precise means that all the measurements you collect have almost the exact same value. For example, if you were measuring an amount of water, and you gathered masses of 700.075 grams, 700.076 grams, and 700.075 grams, all your measurements are pretty precise compared to 700.075 grams, 750.0 grams, and 8 grams. High levels of precision indicate that an experiment can be repeated by others and the results can be reproducible, which is often very good.

Accuracy, on the other hand, is determined by how close your measurements were to the true value, which is often a defined number or is given. For example, imagine you had 500 cubic centimeters of water, and measured the volume three times. Being accurate would be having measurements of 500 grams, 499.999 grams, and 500.001 grams, because they are all very close to the true mass of 500cc of water (which is 500 grams, because 1cc of water= 1 gram of water).

Also relating to accuracy and precision is the concept of significant digits. Significant digits are the numbers in a measurement that are useful. They are counted following a strange rule, called the “Atlantic-Pacific” rule by our chemistry teacher, at least. If a decimal is absent, you count from the Atlantic side (right) of the number, the first non-zero number and every number after it. If a decimal is present, you count from the Pacific side (left), the first non-zero number and every number after it. For example, in a speed of 200.5 feet per second, there are four significant digits: the 2, 5, and the two zeros. In 300 oz, there is only one significant digit, the 3, because the two zeros carry uncertainty; they may have been rounded for all we know. In 0.0860 tons, there are three “sig dig’s”: the 8, 6, and the last 0.

Significant digits are important because they are directly related to precision. The more significant digits you have, the more precise you are, and so also the more repeatable your experiments are. For example, if you went about calculating the value of π (pi), then 3.141592653589 would be much more precise (and accurate) than just 3.14, because the first one is over 592-million trillionths closer to π’s true value.

Hydrate Composition Minilab

Today, we messed with Copper (II) sulfate pentahydrate, CuSO4 * 5H2O, a crystalline compound with a blueish color when hydrated.

We gathered some of the hydrate in a dish, boiled away the water, massed it, boiled it again to make sure we got all the water out of the crystal structure, and massed it a final time. Out of 2.29 grams of Hydrate, my partner and I calculated the mass of the water to be .85 grams, giving a percentage of 100(.85)/2.29= 37%, or that there were .37 grams of water for every gram of the hydrate.

My picture

My picture

I can also say that my results were reliable to one-hundredth of a gram, the finest mass that the scales we used could measure. They were also reliable because I followed the rules of using significant digits taught to us by our Chemistry teacher and because we didn’t lose any great quantities of the hydrate, which would’ve messed up our future calculations. For example, if we had shaken the dish we were boiling the hydrate in, then some of the hydrate or the anhydrous solid would’ve fallen out, and even if we tried to put it back into the dish, the amount would be lower than what it should’ve been and the percentage of water would appear much larger that it really was.

My picture

My picture

The 37% should be a constant that can be used to show how much of any mass of the hydrate is water alone. Using units, the percentage would be .37gWater / 1gHydrate, and 6.0 grams of hydrate would cancel out and leave you with about 2.2 grams of water.


Today, we calculated the molecular mass of copper sulfate pentahydrate. The formula weight of the compound is derived from the atomic weights of the elements in it; copper gives 63.5; sulfur, 32.1; oxygen, 4*16.0 (because there are four oxygen atoms in the compound); hydrogen, 5*1.0; and the second amount of oxygen gives 80 amu. the total mass is 249.6 amu’s, and using the mass of water, 90.0 amu’s, the percentage of water in any amount of the compound should be (90*100)/249.6= 36.05% by mass. This means that from the 2.29 grams my group used, we should’ve calculated (2.29*90)/249.6= .826 grams of water.

However, my group got .85 grams of water, so there must be some percent error. That error is 100(.85- .826)/.826= 2.9% more mass due to water than there should’ve been. To remedy this, we must realize that in calculating the mass of water lost from the difference between the hydrate before and the anhydrous after boiling, any bit of anhydrous solid lost during the process, such as being left as residue on our stirring rods or having been splattered out of the dish, would be calculated as the mass of the lost water. however, the lost mass equates to about two-hundredths of a gram, so I think I can say my group’s results are still pretty reliable.

Three Weekly Questions, 2/28/14

1. What tasks have you completed recently?
2. What have you learned recently?
3. What are you planning on doing next?

1. Recently, I have helped multiple people with math work/homework, I calculated the thickness of a piece of aluminum foil, and I cooked some delicious steak. I also recently won a game of chess against the leader of my school’s chess club.

2. I have recently learned how to determine the number of significant digits an piece of numerical data has. I also learned about the economic growth of America during the 1920s, and I learned how to calculate the force of friction on two surfaces and the force required to surpass it and start moving .

3. I plan on not having homework this weekend, on watching some television, and on playing some video games. I know my family is also planning to go up to a lake for some fishing and such sometime soon, which we haven’t done since about July.

Thickness of Aluminum Foil

Another recent minilab of ours strove to find the thickness of a piece of aluminum foil. I’ve concluded that the piece my partner and I used was about .046 millimeters, or 46 microns.

I figured that out by using the density of aluminum, the measured mass, the area of our piece, and by solving for the thickness. The density we used was 2.7g/cm3, my partner and I settled on a square that was 7cm by 7cm, though we ultimately had to decide it was 6.98cm by 6.95cm, which gave us an area of 48.511cm2, which through converting to mm and rounding was ultimately made 4,850 mm2. The mass we found the square to be was .603 grams.

If density= mass/volume, then v= m/D, and v= l*w*h, and l*w=4,851.1mm2, then 4,851.1mm2*h= .603g/(2.7g/cm2). Converting units a bit more, 4,851.1mm2*h= 603mm3/2.7. Therefore, h= 603mm3/(2.7*4,851.1mm2), h= 603mm/(2.7*4,851.1), and finally, H= .046037mm, which rounds to .046mm, as seen above in the first paragraph.

My result is reliable up to two significant digits, because our density of 2.7 had the least number of significant digits at two. We also calculated our masses and lengths to the best of our ability, so I would say that the calculation of my aluminum foil’s thickness is very accurate and reliable.

What’s in a BB?

Today, we performed a minilab in which we tried to find the composition of small BB’s in a container. We took the mass of a graduated cylinder filled with 20 ml of water, and then massed the same cylinder with a volume of BB’s equal to 5 ml, 10 ml, and then finally, 15 ml. We then subtracted the mass of the cylinder and water to get the masses of the BB’s. Here are the masses my group had, along with the calculated masses of the BB’s by themselves:
1. Grad. Cyl.+20mL water: 66.12g
2. G.C.+20mL+ 5mL BB’s: 107.29g
3. G.C.+20mL+10mL BB’s: 146.24g
4. G.C.+20mL+15mL BB’s: 184.49g

1. 41.17g of BB’s.
2. 80.12g of BB’s.
3. 118.37g of BB’s.

My picture

My picture

It was here when we were subsequently told of how imprecise measurements are, especially with equipment in which you must eye the measurement like with tripe-beam balances, how a number such as 146.24 is not equal to and is less precise than 146.240, and that a greater number of zeros at the end of a number with a decimal means greater precision.

Anyways, from my data, I concluded that the BB’s are most likely composed of a copper and zinc alloy in an approximately 50:50 proportion. I include my group’s average density of the BB’s, 8.012 g/cm3, and the density of copper and zinc at about a 50:50 ratio, which is about 8.03 g/cm3, as evidence. Our data is reliable to the nearest hundredth of a gram, which is the most precise decimal point our balances could go to. Additionally, one rule of significant-digit calculation requires the produce to have at most the number of significant digits as the number with the least digits, which I have followed. Therefore, I can conclude that my data and my calculations are reliable and valid.