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.

Jan. 26-29

What happens when you chemically mix zinc and hydrogen chloride? What will be the reactant (product)?

This is the experiment I did with my group this week. The challenge was to answer the above question.

I hypothesized that since hydrogen and chlorine are already chemically bonded together and can't bond with the zinc, I figured that the hydrogen or the chlorine would escape from the system and combine with the gas in the air while the zinc would stay in the system since it's a metal.

In order to find out what the reactants will be, my group and I first considered the mass of the zinc, the beakers, and the hydrogen chloride. To measure the zinc, we just put it on a measuring scale to calculate its mass. Then, to find out the mass of hydrogen chloride, I figured that in order to do so, we must find out the change in mass when comparing the beaker's mass to the mass of the beaker and the hydrogen chloride in it. Then, I subtracted the mass of the beaker with the total mass, therefore, to find the mass of hydrogen chloride.

By finding the masses of all these variables, we figured this would be the best way to find out the total mass after the zinc and the hydrogen chloride are chemically combined together, that is, by combining the masses of hydrogen chloride and zinc before they were combined. I assumed that since hydrogen or chlorine would escape from the system, the mass would have to decrease.

However, to get an accurate result, we made sure nothing new entered or left the system to increase or decrease the mass. So, we used a trough, which has an extension cord with a cork attach to it. By capping the beaker with the cork, this can help to trap the reactants in the beaker.

So, my group and I thought of an experimental procedure. To find the volume of the reactant, we would fill a different bottle all the way with water while filling the trough with water. Then, mix the zinc and the hydrogen chloride together. But, we had to seal the beaker quickly so that the gas wouldn't escape. Keeping in mind that the tube is attached to the trough, the gas goes through an opening inside the trough. So, to collect the product, you should fill a bottle with water and tip it over right side using a lens to prevent spilling. Remove the lens once done. Then, combine the zinc and hydrogen chloride so that they will chemically react to form a new substance.

Then, gradually wait until the water level drops and then to collect the gas without it escaping, use the lens once again and flip it up.

Now, with the gas in the bottle, my group and I can test it's chemical properties to determine what the gas is. Hydrogen, for example, is flammable (think the Hindenburg), and oxygen is combustible. To determine if the gas had these properties, we lit a match to test for combustibility and flammability. Once we put the match in the container, the flame got brighter. Therefore, the gas was flammable. When the match got in contact with the gas, a sound was made. Therefore, the gas was combustible.

Also, you can find the density of the gas to determine what it is. To do so, keep in mind the beaker's mass. But, measure it again. It should have changed. Subtract the mass of the beaker from the total mass. Then, find the volume of the gas by looking at the bottle. If there is any condensation on the bottle, this indicates the amount of space it takes up. Then, you divide the mass of the reactant by the volume in which it took up the bottle to find the density. (Do this before testing the gas's properties so you can calculate its density correctly).

With these results, we concluded that since the gas was combustible and flammable, it had both oxygen and hydrogen. But, what happened to the zinc and the chloride? And how did the oxygen get in the picture? We conclude that since the air has oxygen, and since it is diatomic, it combined with H2 (hydrogen molecule) to form water. This would then make sense since the inside of the bottle had condensation on it.

Zinc Chloride

But, what happened to the zinc and the chloride? We noticed that those were left in the beaker that was capped. The zinc, I noticed, dissolved in the chlorine, and when that happened, it formed zinc chloride. But, they chemically combined together. Through this experiment, we identified the two separate reactants: H2O and ZnCl.

After this experiment, I had a hunch that I would then learn about writing chemical equations. I then thought that by combining chemical substances, I could identify the chemical composition of the reactants by writing out their chemical equations. To do so, you need to know what the beginning products consist of. Then, what the reactants consist of.

When writing chemical equations, I noticed that it was like doing Algebra. Remember this? 2(XY)=2X2Y. By using this, you use the distributive property, which is when you multiply the outer term by the inner terms. Here is an example of the distributive property in chemistry: 2(H2O)=(2H2)(O2).

I then wondered what the difference was between these: H2O2 and 2 OH. At first glance, it seemed as if H2O2 was a single compound while 2HO was two compounds. Even though they have the same elements and the same number of those elements, they are different chemical compounds. H2O2 is Hydrgen Peroxide while 2 OH are 2 OH molecules.

Jan. 22-25

This week, I learned about the concepts of Dalton's theory.

Dalton's Playhouse Visual Learning

On Monday, I did a simulation on Dalton's theory in the computer game. The first part of the game consisted of burning the calx (Priestley). As 7.39g of the 100g calx was burned, only 92.61g remained. This time, I tried 200g of calx. Using my understanding of the Law of Conservation of Mass, I predicted that twice the amount of mass from the first trial would burn. It turns out that this was true. The change in mass was 14.78g was subtracted from the 200g calx. Therefore, depending on the mass of a substance burning, it loses mass at a proportional when comparing it to the same substance with a different mass.

However, I considered this: What about the volume of gas that was in the experiment? Well, I figured that the more mass was burned, the more volume the calx would be surrounded in since mass and volume have an indirect relationship. Using the 100g, the volume of gas came out to be 5.171L, and then using the 200g, the volume of gas came out to be 10.34L. Therefore, since there is only half of the mass of calx left, the volume doubled. Also, the volume of gas changed at a proportional rate to the rate at which the mass of calx changed.

So, this then establishes a core Dalton principle: Chemical reactions occur at a proportional rate. This is indeed true as everything else (mass, volume, etc.) change at that same proportional rate.

The next experiment I did in this activity was the Lavoisier stage. This is where the phlogiston and the oxygen were tested at burned at different rates to see how they would change in mass or volume. First, 1/3 of the phlogiston was burned. To begin with, both started out with volumes of 6L and they were both in separate beakers , which had tubes connecting to a center beaker, which is where the gas would go. As the oxygen and phlogiston burned, 5L for both were left.

Next, as 2/3 of the phlogiston were burned, 4 L for both oxygen and phlogiston were left.

So far, one could extrapolate that they would change at the same rate based on the results of these two trials. However, when all the phlogiston was burned, only half of the oxygen was consumed so that 3L of oxygen were left.

Therefore, this concludes that phlogiston burned at a quicker rate than oxygen. But how? And why? I speculate that it has to do with the fact that phlogiston was more flammable, therefore, it would burn at a quicker rate. Then, as soon as I realized that phlogiston was renamed hydrogen, I then thought, "Of course!" And I then thought of the Hindenburg incident where it blew up because the hydrogen that it was filled with was flammable, and so when it came in contact with the flame (presumably from an explosion), the Hindenburg exploded. Therefore, I think that hydrogen is flammable possibly because it may be just able to bond with almost any element since it makes just one bond (hence, any element can make one or more bonds). Overall, hydrogen burned twice as fast as the oxygen in the Dalton simulation.

Both consist of a similar chemical composition and consist
of carbon. Therefore, they change mass and volume at
the same rate and number during a chemical reaction.

Lastly, I worked on the Diamond and Charcoal lab. I started out with 0.20g of charcoal and diamond and kept the mass of oxygen at 1.06g and volume at 0.74L constant (charcoal and diamond were tested individually). The mass of the oxygen decreased from 1.06g to 0.73g as the subtracted amount went to the 0.20g of charcoal, thus increasing the mass of charcoal to 0.53g. Next, 0.40g of charcoal was tested with the same volume of gas (0.74L). Then, the mass of gas dropped from 1.06g to 0g and the volume dropped from 0.74L to 0L. Therefore, with twice the mass, the rate at which volume dropped doubled, hence the rate at which charcoal's mass increased. Hence, chemical reactions occur at a proportional rate and inversely affect each other, and subtracted amount is added somewhere else in a closed system.

Then, 0.20g and 0.40g of diamond were tested. The results were the same. Thus, charcoal and diamond have similar chemical properties (e.g. melting point, boiling point, etc.) so that they can change at the same rate and may consist of the same kind of elements (carbon, for example).

From this lab, I learned that chemical reactions occur at the same rate, whether they are in a closed or open system. I learned that the amount of mass in a closed system would stay the same while in an open system, the total mass of the reactants won't come out of the system. Based on this, volume could be affected as well. If the mass left the system, then the volume would increase. But, if it is in a closed system, the total volume won't change.