Feb. 11-15

This was the one thing that I was still thinking about. Isn't it possible for atoms of different elements to have different masses? Indeed, this is so. You can look at the Periodic Table of Elements, but the more interesting thing is how did they find the atomic mass of the individual elements?

To answer this question, one must overcome this fallacy: Assuming that elements bond because they have the same mass. This is not true. This was proven in our experiment with oxygen and magnesium. The mass afterwards was 0.33g. If we were to prove they had the same mass, the mass would have doubled when the two were combined–that wasn't the case. When 0.2g of magnesium was burned, it bonded with O2 since magnesium was not a diatomic element. So, 2Mg bonded with O2. Therefore, this formed 2MgO2.

Based on this, I figured that I could find a ratio between the masses of the elements. Considering that Mg+O=0.33g and that Mg=0.2g, the mass of oxygen was 0.13g. Mg to O has a relationship of 1 to 0.65. Then, I considered the Law of Multiple Proportions, which means compounds should have a proportional rate of the amount of elements as the mass of a substance increases definitely.

But, how did it form 2MgO2? Well, I hypothesized that the magnesium combined with the O2 in the atmosphere since the magnesium went through a chemical change when it was burnt, so it had to have changed into a different substance by bonding with another element.

I also then thought of comparing the ratio of Mg to MgO. Knowing that there are 0.2g of Mg and 0.33g of MgO, I set up an Algebraic proportional relationship between the two. Assuming that MgO is the total amount, I can make it equivalent to any number (let's just say 1) while Mg would be equivalent to x. It wouldn't matter, though, which proportional value I equated to MgO since there must be a definite proportion of elements in order to have the same substance. So, by doing this, I found that Mg to Mgo is 1 to 1.65, which means that MgO has 65% more mass than Mg. Therefore, in spite of the number of particles of Mg and O so long as the ratio is equivalent, it is still MgO.

This week, I applied the Law of Multiple Proportions to what I've also learned this week. In class, I did an experiment with my group. The goal was to compare the mass of 1 item to all the rest. The items were an empty bottle, small brown nails, hexnuts, pennies, screws, washers, bolts, and panel nails. To do the experiment, the first item we measured the mass of was the empty bottle. We found out it was 9.5g, and since all the elements had to stay in bottles, we subtracted 9.5g from all the rest of the total masses to find the masses of the small brown nails, hexnuts, pennies, screws, washers, bolts, and panel nails.

However, we had to choose one item to compare to all the other objects. Since the small brown nails had the least mass, we decided to compare it to the other items, and we found the ratios of their masses to the small brown nails. The picture above shows the data. The small brown nails had a mass of 1.3g, the hexnuts 18.9g, the pennies 11g, the screws 10g the washers 23.6g, the bolts 7.1g, and the panel nails 4g. The small nails to hexnuts were 1 to 14.5, small nails to panel nails were 1 to 5.46, the small nails to pennies were 1 to 8.3, the small nails to washers were 1 to 18.15, the small nails to screws were 1 to 7.69, and the small nails to bolts were 1 to 5.46. Therefore, I have concluded that small nails were the smallest, and the washers have the largest mass since the ratio between small nails and washers was the greatest.

I still felt ambivalent, though, on particles. How can I represent them in a collective group? This question was then answered as I learned in class that for every mole, there are 6*10^23 particles. Thus, the word mole is the collective group of particles.

However, I wondered how I could apply this in the real world. I then thought to myself that it is possible to find the number of moles of any object (e.g. 500g of H2O). Knowing that for every mole of water is 18g, how many moles are there in 500g of water? To find this out, I proportionally represented 500g/Xmoles to 18g/mole. I then found out that for every 500g of water, there are 27.78 moles. But, I still want to find out more about how a mole consists of 6*10^23 particles and how to find out the number of moles in any object without relying on current information. But, how many particles are there in 27.78 moles of water? By knowing how many particles are in a mole, I multiplied (6*10^23) by 27.78 moles of water. I found out that it is equivalent to 166.8*10^23. But, is this the correct way to answer this question? No, because I remembered when I learned about scientific notation in 7th grade. If I remembered correctly, when using scientific notation, a number being multiplied by 10^X has to be less than 10, and since this number is more than 10 and is two value places away from 1.668, I think the correct answer is 1.67*10^25 (since I moved two places back, this would mean adding two more place values).

I then realized that through this lab, this was how scientists found the atomic masses of the elements. They compared it to hydrogen, which happens to have the smallest atomic mass, to all the rest. Also, the overall concept I learned from this was that there can be some way of calling 6 pieces of hardware to 1 collective group (e.g. 1 dozen=12 items). 1 dozen would be the collective group. So, we came up with a collective group name for the 6 hardware pieces: Quinn. Therefore, proportionally speaking, for every Quinn, there are 6 hardware pieces and vice versa.

When doing these kinds, I consider what it is that I'm finding to find. For every X amount of Quinn, how much of the hardwire pieces are there? To solve this problem, I recalled what it was like doing Algebra, so it reminded me of it. In order to find the number of hardware pieces, multiply your unknown value by (Q/6g). Then, you multiply 6g by 5Q to get 30g. Therefore, for every 5 Quinn. there are 30g.

For this problem, flip flop Q/6g since this problem asks for every 24 g hardware pieces, how many Quinns are there? Multiply 6g/Q by the unknown value X, which is equivalent to 24g. Then, divide the total mass by the amount for every Quinn. Therefore, I found out that for every 24g of hardware, there are 4 Quinn.