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# What Does Avogadro’s Number Mean Essay

What does Avogadro’s number mean? It is the theory that’s given by the Italian chemist Amedeo Avogadro which states that “equal volumes of gases at the same temperature and pressure contain the same number of molecules regardless of their chemical nature and physical properties”. And the number of molecules or atoms in exactly 1 mole of a substance is 6.022 X1023. [1] What is Density and how does it help us?

Density is defined as the ratio of mass to the volume of a substance: [2] Density = mass/volume [2] Density is an intensive property this means that it is independent of the amount of substance. For example the density of gold is the same for all gold substances regardless of their masses. This is a good way to distinguish and identify different substances because it only depends on the type of substance. [2] In this lab report I will be discussing how I used Avogadro’s number to find the cost of a single aluminum atom and also the different methods I used to calculate the density of aluminum and each one’s implications.

Procedure: Part 1: Cost of an Atom of Aluminum Firstly I will list the instruments used to find the cost (dollars) of one atom of Aluminum: •Ruler •Measuring Balance •Standard household foil and the cost of it (dollars)

Method: 1.Find the area of the entire household foil roll (given in the package in meter2). 2.Take a piece of aluminum foil from the household foil and measure length and width using a ruler. 3.Calculate the area of the piece of aluminum using the formula length X width. 4.Find the mass of the piece of aluminum foil.

5.Divide the mass of aluminum foil (grams) over its area (meter2), to get mass/area. 6.Now get the mass of the whole household foil roll by multiplying what you got in step 5 with the whole area of the household foil (step 1). 7.Find the number of moles of aluminum in the household foil using the formulae moles = mass/atomic mass. 8.Find the number of atoms in aluminum by multiplying moles of aluminum by Avogadro’s number. 9.As we have the number of atoms of aluminum in the standard household foil, we can get the cost of an atom of aluminum by dividing the cost of household foil by the number of atoms (step 8). Part 2: Density of Aluminum

Method 1: List of Instruments used: •10 ml Measuring cylinder Procedure: 1.Fill the measuring cylinder with water and record the initial volume (ml). 2.Take the piece of aluminum in part 1 of the experiment and make it in a shape that fits the hole of the measuring cylinder. 3.Insert the piece of aluminum in the measuring cylinder filled with water and record the final volume. 4.Calculate the volume of the piece of aluminum by subtracting initial volume of the measuring cylinder from the final volume. (final volume – initial volume = volume of piece of aluminum) 5.Calculate the density by using the mass of the piece of aluminum obtained in part 1 of the experiment and the volume obtained in step 4 in the density formulae. Density = mass/volume

Method 2: List of instruments used: •Vernier Calipers Procedure: 1.Measure the thickness of a piece of aluminum foil from the standard household foil in different sides using the Vernier calipers and take the average. 2.Using the length and width of the piece of aluminum in part 1 of the experiment and the thickness calculated in step 1 of this method. You can find the volume (centimeter3) Length X Width X Thickness = Volume

3.Using the mass in part 1 of the experiment and volume in step 2 of this method. You can find the density (grams/centemeter3) Density = mass/volume.

Method 3: Procedure: 1.Use the Internet to find the thickness of a standard household foil. 2.Using the length and width from the experiment in part 1 and thickness found online. Calculate the volume. Volume = Length X Width X Thickness 3.Calculate the density, using the mass of the piece of foil and the volume calculated in the previous step. Density = mass/volume. Method 4 (not tested):

List of instruments used: •Tiny pellets of aluminum. •100 ml Measuring cylinder. •Measuring Balance. Procedure: 1.Take some tiny pellets of aluminum and put it on top of the measuring balance and record the mass (grams). 2.Add water to the measuring cylinder and record the initial volume. 3.Insert the tiny pellets of aluminum which you recorded the mass in step 1 of this procedure in the measuring cylinder containing water and record the final volume. 4.Calculate the volume of the tiny pellet by using this formula: Volume of tine pellet = Final volume (step 3) – Initial volume (step 2) 5.Calculate the density (grams/ml) of tiny pellets of aluminum using formula: Density = mass (step 1)/Volume (step 4)

Results and Discussion: Below I will be discussing the results of each method and its accuracy and the errors and assumptions associated with each method. Part 1: Cost of an atom of aluminum Below is a table of the results for this experiment: Standard Household foilPiece of Aluminum Length (m)0.193 Width (m)0.0920 Area (m2)6.960.0178 Mass (grams)2790.713 Mass/area (g/m2)40.1 Moles 10.3 Cost (\$)3.56 Number of atoms6.20 X 1024 Cost of atoms (\$)5.74 X 10-25

•As you can see we have found the cost of one atom of aluminum to be \$5.74 X 10-25. •To find the moles we used the atomic mass of aluminum and that’s 26.98 g/mole. •To find the number of atoms we used Avogadro’s number which is 6.022 X 1023. Part 2: Density of Aluminum

I will be discussing for the results for each method to find the density. Method 1: Initial Volume (ml)3.80 Final Volume (ml)4.25 Volume of the piece of aluminum (ml)0.45 Mass of the piece of aluminum (g)0.713 Density (g/ml)1.6

•The density of aluminum is 2.70 g/ml[3] •To calculate the percentage error, we should use this formula %Percent error = [(actual yield – calculated yield)/actual yield] X 100 •Percent error = [(2.7-1.6)/2.7] X 100 = 41% •So the percent error is quite considerable since in the experiment above, as we assumed there is no trapped air when we inserted the piece of aluminum in the measuring cylinder, so that contributed to an error. •This made the volume of piece of aluminum higher than what it should be so the density was lower.

Method 2: Piece of Aluminum Mass (g)0.713 Length (cm)19.3 Width (cm)9.20 Thickness (cm)0.00200 Volume (cm3)0.355 Density (g/cm3)2.01

•1 cm3 = 1ml •So 2.70 g/ml = 2.70 g/cm3 [3] (Density of aluminum) •Percent error = 26% (according to the formula) •The percent error here is less than the previous method, this means that this method is more accurate but they are errors though. •One assumption made is that the thickness of the standard household foil is the same everywhere. This would result in an error in the volume which led the density to be less. •Another error could be the accuracy of the Vernier calipers used to measure the thickness. It only measures the thickness to 3 decimal places in centimeters. Method 3:

Piece of Aluminum Length (cm)19.3 Width (cm)9.2 Thickness specified by manufacturer (cm)0.0016 Volume (cm3)0.28 Density (g/cm3)2.5

•%Percent error = 7.4% (using density of aluminum as 2.7 g/cm3)[3] •According to the percent error this method is the most accurate as it has the least percent error. •So the equipment used here to calculate the thickness was the most precise as it calculates the thickness to 4 decimal places in centimeter, so I believe the equipment used here is the micrometer screw gauge which is more accurate than the Vernier calipers. Method 4 (not tested):

•Tiny pellets of aluminum was used instead of the aluminum foil. •The volume of tiny pellets were found by displacement of water but since there is no trapped air in between the tiny pellets, the volume will be more accurate than the aluminum foil used. •This means the density will be more accurate.

•The percent error will also be less.

Conclusion: To summarize everything up, this experiment tells us how important it is to find the cost of an atom so we can know how valuable each different atom is. Also it tells us the most precise and accurate way of calculating the density, which is important to find out what element we are dealing with. [2]

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