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.