Blog #8

This past week we focused on describing and balancing chemical reactions, and also worked on rearranging atoms. For balancing chemical reactions, we had to put coefficients in front of the reactants and the products to have equal amounts of each element for both of them. We noticed when doing this that after the reaction occurred no mass was added or subtracted in the product. These coefficients tell us how much of that substance you have. The subscripts tell how many atoms you have of the substance like in NH3, there are 1 Nitrogen and 3 Hydrogens. Balancing some equations can be tricky when there are an even and odd match up of an element. An example would be the reaction 2CH4+3O2=2CO+4H2O. How I got the coefficients was by first putting a 2 in front of the H2O to have 4 hydrogens and 2 oxygens. On the product side I had 3 oxygens in total while the reactant side had 2. How you fix that is by putting a 3/2 in front of the O2 (reactant) and multiplying everything else by two, which is how I came to my final answer. Below is a whiteboard of some chemical reactions that have been balanced out, and particle drawings are shows as well for before and after the reaction.

We also described chemical reactions from doing many mini labs testing different things. We did combination reactions where as an example we took a strip of magnesium and ignited it with fire to see a bright flame which is shown below.

We also described chemical reactions from doing many mini labs testing different things. We did combination reactions where as an example we took a strip of magnesium and ignited it with fire to see a bright flame which is shown below.

After we added P-indicator, it showed that there was a presence of a hydroxide ion from the reaction. We then did decomposition reactions where for example solid ammonium carbonate was being heated, and started shaking rapidly because it was starting to form into a gas.

Then we had our single replacement reactions where an example of one was placing a small piece of wire copper in a well plate and adding a few drops of .1M AgNO3 to it, which kept increasingly forming little crystals after a few minutes. This reaction is single replacement because an oxidation reduction chemical reaction occurred where an element moved out of the solutions compound and into the copper forming the crystals.

After this we had double replacement reactions where for example we added a drop of .1M AgNO3 to a transparency sheet and then added .1M NiCl2 to it as well creating this product that was foamy looking, and a mix of both of their colors. It was called double reaction because the cations and anions switched between both reactions to form a new product.

Lastly, we did combustion reactions. We put 10 drops of isopropyl alcohol in a small dish and lit it on fire with a butane lighter. We took a cold watch glass and held it above the fire to see condensation form on the bottom of it. Oxygen was reacted with hydrocarbon to produce energy and light, which was why it was in the combustion category.

In all, I feel that my overall understanding of balancing chemical reactions is a bit shaky, and I usually need help with the harder problems. I need help with knowing exactly what steps to take when I’m given a problem so I don’t start confusing myself, which is what usually happens. I would give myself an overall rating of a 5 or 6 out of 10, but hopefully that will change in the next few days as finals approach!

Blog #7

This past week we focused on naming ionic and molecular compounds, and understanding their differences. With naming ionic compounds, we identified elements that were cations such as: Hydrogen, Lithium, and Magnesium, which are positively charged ions. Also we identified anions such as: Iodine, Chlorine, and Nitrogen, which are negatively charged ions. When looking at ionic compounds, we realized that the metal of the compound is always listed before the non-metal. Also, in an ionic compound both elements have to zero out to become balanced or neutral. For example with the compound Aluminum Oxide, there needs to be two aluminums, and three oxygens since aluminum has a 3+ charge while oxygen has a 2- charge. In this previous example, aluminum oxide is the made up of the regular element name for the metal, but for the non-metal an ide is added on as an ending so instead of saying Aluminum Oxygen, it would be Aluminum Oxide. Other examples would be K3N which is Potassium Nitride, or MgF2, which is Magnesium Fluoride. In some cases certain metals from multiple cations. The metal Copper can have 1 or 2 cations. The way to specify it is by putting the corresponding roman numeral behind the metal in the compound name giving the charge of the element. An example of this would be Cu2O, which is Copper (I) Oxide. Another metal that has this is lead. There can be 2 or 4 cations that correspond with it. Ionic compounds have ionic bonds, and can conduct electricity since its bonds are weaker and allow electrons to pass through.

In contrast to ionic compounds, molecular compounds are only made of non-metals, and have covalent bonds. They do not conduct electricity since there are no metals present, and their bonds are much stronger than ionic bonds, meaning that electrons can’t slip through. With these type of compounds, prefixes and suffixes are added to the element names depending on how many of each element there are. An example would be with the molecular formula BCl3 the name would be Boron trichloride, since there are 3 chlorides and 1 boron. The prefix mono is only used when there is one of the second element, such as ClF. With that compound it would be named Chlorine monofluoride. There are different rules with the spelling and use of certain prefixes and suffixes such as use tetra when having 4 of an element when there isn’t a vowel that would come after, and use tetr if a vowel comes after it. An example of this would be N2O4, which is Dinitrogen tetroxide since an o was the first letter after the prefix. As you can see these compounds are not only different from ionic compounds in that they are made of non-metals, but they also don’t zero eachother out to become neutral. The only time elements need to have equal charge is with ionic compounds.

With naming ionic compounds, we also drew out their ions and formula unit. Below is a few white boarded examples of some different compounds.

How we get the number of ions is simply by counting how many ions of each element is needed to equal each other in charge. An example would be in CaBr2 there would be 3 ions needed, one ion of Ca and 2 of Br.

Overall I have a good understanding of the main differences between Ionic compounds and Molecular compounds and how each work with classifying them. I would give myself an overall rating of a 9 because from last week to this week I really locked the concept of how to name these compounds and how to make sure that the ionic compounds have a balanced charge. One question I still have is what are Polyatomic ions, and how they compare to molecular and ionic compounds.

Blog #6

This past week we started understanding what positive and negative charges looked like in atoms. We did a sticky tape lab where we charged pieces of tape to see what they attracted. On the hanging dowel we had a piece of foil, paper, and two pieces of charged tape (one that was on top and one on bottom). We then took two other pieces of tape, stuck them together and then ripped them apart to give them a charge. We took the top piece labeled T and put it near the samples on the dowel. We noticed that it attracted the foil, paper, and bottom tape, but repelled the top tape. We then took our bottom charged tape and saw that it attracted the foil, paper, and top tape, but repelled the bottom tape. From this we made some conclusions. There are three rules: 1. The same charged substances repel each other. 2. Opposite charges attract. 3. Neutral charged substances like foil and paper do nothing when they are put next to each other. We found this last rule by putting a piece of paper closer to the hanging foil and other piece of paper on the dowel. When doing this nothing happened, showing that they are neutral. From this lab, we concurred when charging the two pieces of tape together, the top piece has a positive charge and the bottom piece has a negative charge. When we put our hand close the the two tapes before they become charged, they don’t attract or repel because our hands are neutral, and the tape stuck together is neutral as well. However when we put our hand closer to the charge tapes after they have been taken apart they attract each other, since opposites attract! Below is a picture of us conducting the experiment.

Along with this experiment we looked at compounds that combine metal and nonmetal particles that do not conduct electricity as solids. We came up with formulas for many different compounds such as CaO, Al2S3, BeS, and Na2s. After writing them out, w e drew particle drawings for each. After this we looked at a portion of the periodic table that covered all of the elements we used. We wrote all of the chemical formulas we came up with in their corresponding element boxes. For example for the chemical formula CaO, we wrote in in the Ca box and the O box. This helped us find a pattern that wasn’t so easy to see at first. What we found is that the elements in the first row on the table have a +1 positive charge, then the next row was +2, then +3, +4, and then switched to -3, -2, -1, and finally 0. I saw that in a formula such as Li20, both elements have to equal out, so if there are +2 Li’s then there has to be -2 O’s, so they cancel each other out. Below is a picture of what the periodic table looked like with all of the formulas in it. You can take any formula and figure out what charge the elements have by making sure they equal each other out.

In all, my understanding of how charge works if definitely a lot better than when we started this unit. By understanding the basic rules of charge, I can predicted and hypothesize if two things will repel, attract, or do nothing. I would rate my overall understanding at a 7 probably, because there still are some aspects I don’t completely understand: like how many atoms have to be in particles that are oppositely attracted to each other.

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!

Blog #4

This past week, our main focus of learning had to do with relative mass and moles. We started off with a worksheet that had different tables where we needed to figure out the masses of different samples. The first table gave us different numbers of chicken/quail eggs, and we needed to figure out their masses.We were given the mass of one egg, so we used that knowledge to see how massive up to a million eggs were. The ration between the masses of both types of eggs were 16:1, for every amount of eggs given. This showed that chicken eggs are 16 times heavier than quail eggs. In the next example we looked at how many atoms were in different samples. For oxygen, when there was a one atom sample, the mass was 16.00 amu. But, when there was one mole in the sample, there were 16.00 grams. We saw that a mole contained the same mass as a one atom sample, just with different units. This introduced us to the mole which we then learned by Avogadro’s theorem is 6.022 * 10^23 particles.

Above is a picture of the third model we looked at comparing different elements masses when there is one atom, vs. when there is one mole. This reinforced that one atom has the same number as one mole, just atoms are measured in amu, while moles are measured in grams.

After this worksheet we did the Empirical Formula Lab. In this lab we took zinc and reacted it with hydrochloric acid to produce zinc chloride, and then figured out the empirical formula (formula based on experimental evidence). My group found the mass of the beaker itself to be 31.89g, and then found the mass of the beaker with zinc to be 34.98g. After getting these masses, we placed the beaker on a hot plate until we the next day. The next day we used a bunsen burner to heat the beaker to get the substance bubbling, and then stopped heating it as soon as the content began to smoke. We then let it cool, and noticed how the zinc chloride started to solidify. After this, we found the mass to be 38.155g, and repeated the heating step again to find the mass to be the same the second time. We looked at the class data, and compared the mass we found the zinc and Cl mass to, and also compared their moles. Every table group had different answers, but they were all very close to each other. In the end, each table came to the same empirical formula which was ZnCl2. Below is a picture of the class data.

After briefly touching upon finding the mass and mole of an atoms in worksheets and the empirical experiment, we started to practice more problems that had to do with finding how many atoms and moles are in certain substances. For example we were asked to find how many moles of iron atoms were in 4.42 grams of iron. To figure this out, you take the mass of the substance you are testing which in this case is 4.42g, and multiply it by 1 mol of iron divided by the mass of the element iron itself which is 55.845g. Then you get your answer to be .079 mol of iron. After doing many problems like these in class and for homework, I came to a good understanding of what the mol really is, and how to calculate it. Everything we did this week really tied to the mol of elements, and how they differ for different substances.

Blog #3

The first two days of class was spent focusing on reviewing the concepts we had learned in unit four. We did a review over how mixtures, elements, compounds, and pure substances differed in particle diagrams and characteristics. For instance some things that were covered were that elements can’t be broken down by physical or chemical means. Also, a mixtures physical properties depend on composition, because the elements mixed are keeping their characteristics. Lastly, compounds can be separated by electrolysis and consist of two or more elements with a fixed mass ratio. We were given different particle diagrams that were examples of compounds, mixtures, elements, pure metals, etc. Then we had to draw particle diagrams ourselves of what a mixture and compound of two elements would look like, and describe their difference. The particles in a mixture are not chemically combined while the particles in a compound are. We then took particle diagrams to the next level, by showing how gas A could combine with gas B to form a compound product. If the gases were diatomic, then there would be 2 volumes or 2 products produced. The next problem on the study guide had to do taking two compounds made up of nitrogen and oxygen and determining their value ratios.  We took the mass of N and divided it by the mass of O to form them. The ratio for compound A was 1.75 and the ratio for compound B was .876. When you compare both of the ratios you find that compound A has about two times more Nitrogen than compound B has. We sketched the particle diagram for each and came to the conclusion that the formula for compound a was ON2 and for compound B it was just ON. The last thing on the study guide was making time vs. temperature graphs for different substances with different boiling points. The graph would incline in the positive direction until it hit a temperature that was a boiling point for one of the substances. Then it would incline again, until it finally reached a boiling point for the other substance. Here’s a picture below of the answers to the problems and the graphs we made.

On top of this review sheet, we had another one that focused on describing the substance objectives. This covered very similar topics that the other review guide had, but instead the questions required more writing, and less drawing particle diagrams. We talked about how the Law of Definite Proportion states that compounds contain the same proportion of elements by mass, and that the Law of Multiple Proportions states that when two elements combine, the ration of the masses of one element that combines with the fixed mass of the other are whole numbers. Lastly, Avogadro's hypothesis was discussed. His hypothesis was that gases that have the same volume temp, and pressure, contain the same amount of molecules. The laws and hypothesis these scientists had are used to this day, and show a great deal of importance in the science world. Below is a picture of one of my classmates review sheet white boarded.

After review, we started to focus on using the mass of objects to determine how many objects there are in a given quantity. In class, we had a bag filled with packing peanuts, and were told to guess how many there were. Everyone made their guess, and then we were asked to figure out the real amount. How we did this was we took the mass of three or four packing peanuts, and found the average mass of them. The we massed the entire bag of peanuts and divided it by the average we found. I don’t remember the exact amount that was found, but my classmates predictions, and even my prediction for the most part was very off. This activity was similar to what we had to do on our Relative Mass worksheet, and our Unit 5, Worksheet 1. We found the mass of one unit, and used that to find the mass of containers filled with many more units. This can be a very useful skill to use in the real world for many things. If you have a container filled with many of the same chapter books, and just don’t have time to count each one, why not just find the mass of one book, and take the mass of all the books (minus the container it’s in) to find you answer!

Blog #2

September 21-25, 2015

The main ideas that we learned this past week were how theories were formed by different philosophers like Democritus and Dalton, how iodine was discovered, deriving formulas for different masses of elements, and comparing different element levels of the same or different substances.
Democritus reasoned that if you take a stone and cut it into smaller and smaller pieces, at some point it will be too small to cut anymore. He called these very small pieces atomos, which means indivisible, and said that can’t be destroyed. This theory sparked interest in other philosophers and helped explain the physical aspect of the world. Later John Dalton discovered that water could be in a gas and solid. The gas could mix with the air but, ice could not. His theory was composed of four basic concepts that are, all matter has atoms, all atoms in a given element are the same, chemical reactions combine atoms not destroy them, and elements react in whole number ratios when forming compounds. These different theories and concepts were used to inspire and question other scientists to find more about the world they didn’t know.
The next topic was how Iodine was discovered. Nicolas Clement and Charles Bernard Desormes reported a new discovery that was found in seaweed. It gave off a purple vapor, created crystals, and was in a solid form. They thought that he had many similarities to the substance chlorine. A man named Joseph-Louis Gay-Lussac reviewed the scientists experiments, and names then substance “iode” which comes from an ancient Greek word for purple. A scientist with the name Andre-Marie Ampere had a hold of the substance from Clement and Bernard and showed the famous scientist Davy it. Davy studied the substance and ended up calling it its English name Iodine, which is now an element on our periodic table to this day.
We then worked on a worksheet that had us derive formulas for different masses of elements. We were given two compounds, Compound A and Compound B, and were told to find the ratio of it the masses of two elements that made the compound. We compared the ratios to see what the difference was between them, and expressed them as improper fractions. We were given different hypothesis to draw particle diagrams for. Hypothesis one dealt with us using the improper fractions we had previously derived, which helped us decide how many particles of each element should be in it. In hypothesis two we had to take in consideration that the atoms of the element oxygen would be heavier than carbon by the ratio of compound A. Below is what this problem looked like.

Lastly, we did a worksheet on looking at the elements in the substance sucrose. We were given the masses of each element that is in it and found the percent mass each element took up of the substance. Then there was another chemical analysis done that consisted of a heavier sample of pure sugar cane with the same elements in it, just consisting more mass. We found the percentages again which came out to be the exact same as the sucrose percentages were. We found that both of these were sucrose.

This helped explain the Law of Definite proportions. When you have the same proportion just using different masses, it can still be the same substance, since the percentages and ratios would still be the same.

 All of the theories  that were derived in history are what we use today. The basic compound, elements, and mixtures, that we  use everyday for science, is what helped further our knowledge in even more new discoveries. I think I learned a lot of new interesting material this week that I had no clue about. Reading about Iodine and the different theories, opens my eyes and shows me that these laws we take into consideration when exploring science were derived from scratch by some very talented scientists.