Blog #5

This past week we did a lot of review having to do with molar mass. On Monday, we reviewed unit 5 worksheet 2 which was our homework from the weekend. It went over how to find how many moles there are in a substance if you are given the mass of it. An example of this type of problem would be if you have a penny weighing 3.3g, then how many moles of copper would be in the penny. How I would set up the problem would be take the 3.3g of Cu and multiply it by one mole over 63.546g of Cu (the atomic mass of the element), to cancel out the grams unit, so I am left with .05 moles of Cu. Below is a picture of how I solved this problem on paper (#1).

The other problems below are similar to problem one in finding how many moles are in the given substance. On the backside of the page we were asked to find atoms as well as moles. To do this I would take the amount of moles there are in the substance, and multiply it by 6.022 x 10^23 over 1 mole to get my answer. Doing this process refreshes what we have learned about Avogadro's theorem. We then took a quiz over all these basic concepts we had covered so far.

After getting a good understanding of moles, we used this knowledge to find empirical and molecular formulas on unit 5 worksheet 3. We were given examples where we needed to use our previous knowledge of finding moles of substances to figure out the empirical formula. An example of this was when we were told that there was 32.0g of bromine and 4.9g of magnesium. After finding that there were .4005 moles of Br and .2016 moles of Mg, I divided both by the smaller number which in this case was .2016. After doing this I got 1 for Mg and 1.986 which is about 2 for Br. These numbers were then put respectively after their element coming out to be MgBr2. In another problem, we were asked to do the same thing except now we would find the molecular formula. The empirical formula was based off of data, but now the molecular formula is the simplest formula for the compound. An example of this was a problem where we were given .59g of H, 9.40g of O, and a molar mass of 34.0g/mol. I found the empirical formula which was OH and had a mass of about 17g. I then would divide the molar mass of 34 by the empirical mass of 17 to find the factor I will be multiplying my chemical formula by. This number turned out to be 2, so my molecular formula was H2O2.  

Lastly, we calculated percentages of certain elements in compounds. An example of this was we were given 17.6g of iron and 10.3g of sulfur, with a total mass of 27.9g. The percentage of iron in the compound was found by dividing its mass by the total mass which came to 63.1% and the same thing was done with sulfur and its mass to come to 36.9%.

In all, we came to really understand how to do these many different calculations by doing many example problems. I feel really good about calculating moles, number of atoms, element percentages, relative mass, and figuring out empirical and molecular formulas. All of these things help us figure count the things we don’t see easily, like atoms. I think I really participated in the learning this week, and was able to help others in my class get to the level I was at. I actually really enjoyed the unit since it was mainly just a lot of easy math!