ALUMINUM ATOMS

ABSTRACT

In the aluminum atoms lab, the objective was to find out how thick a piece of aluminum foil is and how many atoms are stacked up to make the thickness of the foil. There were many steps done to complete this experiment. My partner and I had to find out the mass of the aluminum bloack, fill a graduated cylinder and record the volume of the aluminum block, and cut out a 15 cm by 15 cm piece of aluminum foil to finish the experiment. We also had to do a lot of math to figure out our answers to our objective, but it worked. The results in our lab was our height, which was 1.3x10 to the -3rd. 5.2 x 10 to the 4th or 52,000 was how many atoms thick the aluminum foil was. Moreover, the volume of the foil was 17.20 grams which equaled 4.74. This was the 6mL multipled by the .79 grams, which equaled the height of 1.3. Every number in this lab was important; you couldn’t move on without knowing every answer. All of the numbers are recorded on the data table.

INTRODUCTION

In 1825, Hans Christian Oersted found the element aluminum. It is Al on the periodic table. Furthermore, aluminum has a melting point of 660.37°C, boiling point of 2467°C, specific gravity of 2.6989 (20°C). It also has a valence of 3. Pure aluminum is a silvery-white metal. It is soft, light, and relatively nontoxic, with a high thermal conductivity. It also has a high corrosion resistance. It can be easily formed, machined, or cast. Ancient Greeks and Romans used alum as an astringent, for medicinal purposes, and as a mordant in dyeing. It is used in the kitchen and thousands of industrial applications. Aluminum is just an important element in our everyday life and we don’t realize it. Aluminum is used in many industries. It is resistant to corrosion, and it is light while being strong. It is used in packaging, construction, and in electrical applications. Aluminum is very can be very bendable and is just always a great use for us. All in all, we use it for many reasons, and it’ll never stop.

ANALYSIS

In this lab, I found my data by at first weight the aluminum block on the scale which had a mass of 17.20 grams. Then, we put the aluminum in a cylinder of water, and got 50mL, but then subtracted the 50mL from the number 56mL, which was the volume of the water in the cylinder. Also, when we cut out aluminum foil, we got the length as 14 cm and the width as 15 cm. We for the volume of the foil by cross multiplying the mass of the block and the mass of the foil, and the volume of the block to get the volume of the foil. To calculate the thickness of the foil, the height was 1.3x10 to the -3rd. The height was needed to find out how many atoms thick the aluminum foil was. I divided 1.3 by 2.5 (the diameter) and got 5.2x10 to the 4th. To make it simpler, I moved the decimal over the right and got 52,000 atoms thick for the aluminum foil. Yes, my data does match the expected results. The answers were pretty close to the correct numbers and when I compared too many peers in my class, our data all was around the same numbers. There are many sources of area that could have occurred in this experiment. An example would be if I messed up on calculating the volume. In the beginning, I forgot to subtract the amount of water