March 4-8

This week, I learned more about the complexities of finding the number of moles. The simple part I learned was the part where I could find the number of moles for 5g of a substances with the given: for every mole of this substances, it contains 10g. So, then I would proportionally equate them and find that for every 1/2 mole, there are 5g of this substance.

However, I still wondered how did scientists find the empirical formula of water to be H2O? This I was curious to find out. I wanted to learn the process of how to get there since I already know the answer. But how do I get there?

Due to the ACT/MME days this week interfering with class time, I could only write about how to find 1) how to find the empirical formula of a compound 2) how to add moles together to find the number of moles in a compound.

But, the complex part of molar mass was to find the empirical formula of a compound with the given: the percentages of the elements chemically combined in the compound. The first step to doing so is to divide the mass of the substance in 1 mole of the compound by the mass of 1 mole of the compound–this is to find the percentage of the element in the compound.

For example, let's find the percentage of H2 in H2O. Volume-wise, we know that H2 is 2/3 of the volume of water, but is it the same thing for the mass? Before I learned about atomic mass and molar mass, I figured that it would be different because mass and volume are two different things, and I had a feeling that hydrogen would have less mass than oxygen. When I found out how much H2 was in water, this supported my hypothesis. The formula to find the percentage of an element in a compound is:

mass of the element in 1 mole of the compound
------------------------------------------------------------   *  100
mass of 1 mole of the compound

I divided the molar mass of diatomic hydrogen (2g) by the total molar mass: 2(1) + 16=18g/mol. I found out that the percentage of H2 in water in terms of mass is 11%. Oxygen would then obviously consist 89% of water's mass.

Then, I wondered how I could use this to write the empirical formula of a compound. I then thought to myself that the best way to do so is to find the percentages of the elements first, one by one. To do so, I would divide the percentages of the elements by their molar masses to find the percentage of the mass of moles.

Then, I would take those answers and divide them by the smallest answer. These answers are the ratios of the elements in the compound, which will be used to determine their empirical formula. With a ratio of 1 to approximately 30.65 (to be exact 1,134/37). If I said the empirical formula for this compound would be O^1H^30.65, this wouldn't make sense. Instead, it should be O^37H^1134. This makes sense because one of the most important rules in writing an empirical formula is the subscripts must be whole numbers, not decimals or fractions.

While learning about this, I realized that if H2O were written as H^1.2O^0.6 or H^3.2O^1.6, the ratio always remains the same. This then reminds me that even if water's empirical formula were written either way (whether the correct or incorrect way), it would still be water. I then thought to myself that, that would be true for chemical currents. Through the electrolysis lab, I realized that the electrical current, which passed through NaCl water solution, split 1 particle of water to many. This would explain why hydrogen and oxygen would be separated into two different tubes based on the anobe and the cathobe plug-ins.

Now I have an idea of what would have happened if we did the electrolysis lab differently (e.g. turning the handle counter-clockwise instead of clockwise and plugging the anobe and cathobe differentlu). If that were the case, the result would be the same, but the process would be different. The electrical current would have gone in a different direction and splitting the particles through this, causing the hydrogen and oxygen to go into the other test tube. I am not sure this is correct, but this a rough idea of what I think happened at the particle level. But, one thing I know for sure is that the actual particles themselves don't change. There would still be H2 and O particles. Hence, through this lab, nothing entered or left the system to change the outcome of the results.

This week, I also learned about adding moles together.

To add moles together, consider whether they are diatomic or not. This is important. Here's an example of one:
2 mol H2 + 1 mol O2 (arrow) 2 mol H2O

But, why would it be 2 mol, not three as we learned in kindergarten that 2+1=3? Well, this isn't the case. Consider that in 2 mol. H, there are 4 H particles, and 1 mol. O2 has 2 O particles. Since we know that hydrogen is diatomic, it can combine with a single oxygen as well as a diatomic one. TIP: One diatomic element as well as a single element can be one mole. This is the case here. To make this problem easier, think of the number of particles involved and how they should be combined. With H+H+O and H+H+O, you get 2 mol of 2H2O (1 mol water=1 H2O). The overall trend I have noticed when combining moles is that the first compound being combined with another one determines the number of moles. (e.g. 4 mol Na and 2 Cl2=4 NaCl, hence 4 moles of NaCl) Na+Cl, Na+Cl, Na+Cl, and Na+Cl=4 pairs. Another trend I have noticed is that if there is a diatomic element, then there are twice as many particles as the same single element (e.g. H2 vs. H). Depending on the ratio of the elements, I've also noticed that determines the number of the moles of the elements combined together. (e.g. H2O: H to O is a 2-1 ratio, so 2 mol H2 + 1 mol O2=2 mol H2O).