Accel. Chem. Blog 4

                As we come to the end of another chapter in Accelerated Chemistry, I can safely say that this class has probably challenged and pushed me more than my other classes this year.  In our latest unit, we have covered: balancing equations, types of reactions, and reactivity.  These all relate to the overall theme of chemical equations, which builds off of the last chapter, which was about charges in atoms and molecules, where we began to look at how elements are bonded together.

            We started the unit with balancing chemical equations.  The reason we do this is to follow the law of conservation of matter.  It is easy enough to write out a reaction, but you may end up with more of an atom on one side of the reaction, and that defies the law, because you either created to destroyed matter.  First, you make sure that you have written the correct formula, paying attention to charges and diatomic particles.  Then, you need coefficients to ensure that there are equal numbers of the same atoms on both sides.  We also learned some tips to help us in balancing: 1) Adjust the coefficient of a single species (ex. O₂) last.      2) Temporary (key word: TEMPORARY!) use of a fraction or decimal is helpful.  3) If there are polyatomic ions on both sides of the reaction arrow, you should balance them as groups.  To help teach idea of balancing equations, we used models of atoms to put them together and learn about the use of coefficients.

            Next, we learned about the types of chemical reactions.  To learn about this, we performed a series of labs.  There are five types of reactions: Synthesis/Combustion, Decomposition, Single Replacement, Double Replacement, and Combustion.  A synthesis reaction will produce one product. (ex. 2Na + S à Na₂S)  A decomposition reaction will begin with one reactant, but will end with two products, so it’s basically the opposite of a synthesis reaction.  Single replacement begins with an element and a compound, and ends with an element and a compound, but the elements are switched. (ex. 2Li + MgCl₂ à 2LiCl + Mg)  The metal has to react with the compound.  Double replacement basically switches around the elements.  It begins and ends with two compounds.  (ex. 2AgNO₃ + CaCl₂ à 2AgCl + Ca(NO₃)₂)  The last reaction is combustion which is just to burn.  You are always going to end up with Carbon Dioxide and Water, and all you have to do is make sure that the reactant side has O₂ and that the equation is balanced in the end.

            Lastly, we learned about reactivity, because not all elements can react.  There are a few rules when it comes to reactivity.  The main rule is that if an element is above another on the list, it will react with it, but not the other way around.  For example, Barium will react with Tin, but Chromium will not react with Calcium.  (ex. Zn + CuSO₄ à ZnSO₄ +Cu)  Another rule is that all metals about Hydrogen displace Hydrogen from HCl or H₂SO₄.  (ex. Mg + H₂SO₄ à MgSO₄ + H₂)  Another is that metals above Magnesium displace Hydrogen from water.  (ex. Fe + H₂O à no reaction)  Finally, metals above Silver on the list combine directly with Oxygen in a synthesis reaction.  (ex. Pt + O₂ à no reaction)

            A “real life example” of balancing equations is in cooking.  If you have a recipe that calls for two eggs for every three cups of flour, if you put in four eggs, you need to put in six cups of flour.  Another real life application is in a combustion reaction, the fact that you need O₂ in the equation.  I knew that you need oxygen to have a fire, but it is interesting to see that shown in what we are doing with chemical equations. 

            Our work with chemical equations originally proved to be difficult for me, but over the course of the unit I have gained deeper understanding and appreciation for the process.  Balancing equations, knowing the type of reaction, and using the activity series to solve these scientific puzzles is a vital part of chemistry and I am glad for what I have learned in this unit. 

Accel. Chem. Blog 3

             Our most recent chapter in chemistry has been about counting particles.  This has included relative mass, Avogadro’s number, molar mass, mole calculations, and most recently, percentage composition.  A lot of what we covered has had to do with calculations, so a calculator was definitely necessary in this unit! 

            The Periodic Table shows us the masses of all of the elements.  These masses are relative masses, though, based off of Carbon 12.  For example, the relative mass of Lithium is 6.99amu’s, the relative mass of Oxygen is 15.99amu’s, and the relative mass of Sulfur is 31.96amu’s.  We began with a lab activity counting popcorn, rice, and bean particles.  Using a scale, we measured the mass of 1 dozen of each of these objects.  We could then use that number to calculate how much other amounts would weigh.  We could also determine how many dozen of each object would be in say, 400,000 grams by dividing that by the mass of a dozen.  (For the rice, it would have been 400,000 ÷ .22= 1,818,181.818 grains of rice.)

            This then ties into Avogadro’s number, which is the number of atoms in 12.00 grams of C-12.  This number is 6.02x10²³, which is also known as the mole.  Mole (mol.) is the word for 6.02x10²³ of anything, so you could have a mole of grass seeds, a mole of M&M’s, or a mole of seconds.  Molar mass is then the mass of 1 mole of atoms/molecules, but expressed in grams.  For example, a mole of Sulfur weighs 32.1 grams.  To help us understand the enormity of the mole, we ran calculations and we will be creating an electronic poster comparing a mole of something to something large like a planet.

            Mole calculations revolve around a “for every…” statement.  6.02x10²³ stands for the number of atoms/molecules and its statement could read like this: For every one mole of Copper, there are 6.02x10²³ atoms of Copper.  Molar mass is really just equal to grams.  Its statement could read: For every one mole of Copper, there are 63.6 grams of Copper.  This knowledge helped us in a lab in which we found the number of water molecules and where we found how many pieces of chalk it would take to contain 5.62x10²³ molecules of chalk.  Using my mouth, we discovered that I could hold 1.91x10²⁴ molecules of water.  We also discovered that it would take 93.45 pieces of chalk to hold the given amount of molecules.  We needed information like the mass and the number of molecules to solve these problems.

 

   

            Lastly, we covered percentage composition.  This can be used to find the amount of a certain element.  There are two approaches.  One is through relative mass data.  To find the percent of Oxygen in 15 grams of Potassium, Chlorine, and Oxygen, you take the total number of grams (15) and subtract the number of grams in K and Cl (9.1) from that.  That leaves 5.9 grams of Oxygen.  Then, to find the percentage, you calculate 5.9g of O divided by 15.0g total, and multiply your answer by 100.  This shows that it’s 39% Oxygen.  The second method is from the formula.  If we have KClO₃, we find the mass on the periodic table of each of them and add those together (122.6g).  To find the percent of Oxygen again, you would find the mass of 3 Oxygens (48.0) and divide that by the total.  Multiply that by 100, and you have your percentage total of 39%.  In the case of a hydrated compound, you do the same thing, but instead of separately adding 2 Hydrogens and an Oxygen for water, you just add the water, which is 18 grams.

            While explaining this all is rather complicated and probably a pain to read, I can truly say that I feel that I’ve improved my skills in this chapter.  The idea of moles is really mindboggling.  Even when you think about the size of a stick of gum and the size of the moon, it’s still a little hard to grasp, because there’s nothing that we are familiar with that comes close.  That’s probably what will stick most in my mind leaving this unit, and I doubt (especially after being inducted into the Mole Patrol) that I will ever forget Avogadro’s number.