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Chemistry Experiments: For Advanced & Honors Programs
By James Signorelli Trafford Publishing
Copyright © 2014 James Signorelli
All rights reserved.
ISBN: 978-1-4907-4643-2
CHAPTER 1
Determination of the density
Purpose:
To determine the purity of four metal samples using the procedure established by Archimedes 2300 years ago.
Diagram:
Draw or attach a photo of the equipment in set-up mode
Procedure #1:
. Weigh the four metal samples to four decimal places (+/- 0.0001 grams)
. Using a Vernier Caliper, determine the diameter and length of each sample.
. Calculate the density using Archimedes formula.
. Calculate the purity of each pure metal sample (Al, Fe, Cu) percent error is related to purity.
. Calculate the percent composition of Zinc and Copper in the Brass sample.
. Calculate the Density and hypothesize the identity of the unknown given in class.
Calculations:
Show all math required to complete the table, in standard format
. Calculate the density of each sample.
. Calculate the number of moles of atoms in the three (pure) elemental samples using N = SM / MM, where SM is the standard density of that metal. (mass of 1 cm3)
. Calculate the factor of change between the densities of Al and Cu
. Calculate the factor of change between the atomic weights of Al and Cu.
. Calculate the percent composition of the brass.
Conclusion: (essay in MLA format)
Explain completely the concepts of density, how we make assumptions about concentration, and how to approximate the atomic volume.
Example Calculations:
Aluminum:
Mass: 17.35 grams rounded to 3 significant digits = 17.4 grams
Diameter: 1.27 cm
Radius: 0.635 cm
Height: 5.07 cm
1 cubic centimeter can be symbolized as a cc or cm3
Volume = π * (r)2 * H
Vol. = 3.14(0.635cm)2 * 5.07cm
Vol. = 6.419 cm3 rounded to 3 significant digits = 6.42 cm3
Density = mass / volume
Dn = 17.4 gm / 6.42 cm3
Dn = 2.71 g/cm3
Percent Error = (abs difference / expected) * 100%
Error = (2.71 – 2.70) / 2.70 * 100%
Error = (0.01/2.70) * 100%
Error = 0.37 %
REPEAT ALL STEPS ABOVE FOR REMAINING SAMPLES
Number of moles in 1 cm3 = SM / MM
Al Fe Cu
2.7 / 27 = 0.10 moles 7.87 / 55.85 = 0.141 moles 8.96 / 63.55 = 0.141 moles
Density Factor of Change:
Change = Dn of Cu / Dn of Al
Change = (8.96 g/cc) / (2.7 g/cc)
Change = 3.31 X density increased by 331%
Molar Mass factor of Change:
Change = MM of Cu / MM of Al
Change = 63.55 MM / 27 MM
Change = 2.35X molar mass increased by 235%
Brass:
If Brass is 50% copper and 50% zinc, it would have an average density of 8.05 g/cm3
** Since the Brass sample had a density of 8.45 g/cm3, it clearly is more than 50% copper.
Cu + Zn = Brass
8.96 (x) + 7.14 (1-x) = 8.45
8.96(x) + 7.14 - (7.14x) = 8.45
8.96x - 7.14x = 8.45 - 7.14
1.82 x = 1.31
x = 1.31 / 1.82
x = 0.72 (decimal equiv of 72%)
Analysis of Brass sample:
Brass Sample: 72 % copper 28% Zinc
Other common alloys:
Bronze: Copper & Tin mixture
Pure gold is 24 karat:
18 K Gold: 75% gold & 25% nickel
10 K Gold: 42% gold & 58% nickel
In this first lab of chemistry, we study one of the discoveries and inventions of Archimedes of Syracuse. He lived from 287 to 212 BCE. His investigations included the proof that (Pi) was a number between 22/7 and 221/71, therefore, it is an unending value. He devised the formula for the volume of a sphere, he invented the "Archimedes Screw" for bringing water from one level to another, and he wrote the formula for we call Density. Density is defined as a relationship between the mass of an object and its volume. In simplicity, "Density is the mass of an imaginary cube having a total volume of 1.00 cm3" (1 cm X 1 cm X 1 cm).
Archimedes has a legend associated with the discovery of density. The legend, whether real or embellished, states that Archimedes determined the purity of a gold crown by comparing the mass and volume of a bag of gold coins to the finished crown. If the mass and volume of the crown matched the mass and volume of the coins used, then the crown was pure gold. Archimedes went on to prove that no two substances of his era could share the same density unless they were made of the same material. Modern technology has greatly improved upon his method, but his idea is still rock solid.
This physical property of matter (density & specific gravity)) has survived the test of time. Even after 2300 years, we still use it as the first test of identity for an unknown substance. In this lab we collected data on several "unknown" materials. By using the concepts of density, we can determine the identity of each of the samples tested. Beyond that, however, we can also derive additional information, not known to Archimedes 23 centuries ago. We can use this data to deduce that atoms are not the same size. We can also use this data to count the number of atoms in that 1.00 cm3 sample.
In the lab this week, four metals were tested. The results were used to determine the identity and purity of the three elemental samples. The data was also used to determine the percent composition of the brass alloy. An optional fifth sample can be available at the instructors discretion, if time permits. The data collected from the mass and volume determination was processed using Archimedes formula to find the derived term we call density. This value was then compared to the accepted standard found on the periodic table. The difference between our data and the accepted standard is then used to determine the percent purity of the same, as it is a direct function of the percent error. The brass sample, being an alloy, has many different possible combinations. If the brass was made of 50% copper and 50% zinc, then it would have had a density of 8.05 grams/cm3. If the alloy contained more than 50% copper, the heavier metal, then its density would rightfully be greater than 8.05. Since our sample density was determined to be 8.45, it is safe to conclude that the brass sample was much more than 50% copper.
In determining the number of atoms in the sample, we must use the standard volume of the cube being tested. The density value is the mass of a cube which is 1 cm by 1 cm by 1 cm. The number of atoms filling that cube was determined by using the formula of Amedeo Avogadro. This equation was devised in 1856 and states that the number of atoms (moles) in a sample is the mass of the sample divided by the molar mass of that material. Since our density samples were all solids, it can be assumed that the spaces between the atoms were at a minimum. This allows us to assume that most of the volume of the sample is therefore the volume actually taken up by the atoms of that sample material, and not the spaces between those atoms. The mass of this imaginary cube is the density. The density (grams/cm3) divided by the molar mass (grams/ mole) leaves us with moles of atoms per cubic centimeter.
We see in our data that the number of moles of atoms were higher in the copper sample than in the aluminum sample. Since both samples had the same volume, then the logical explanation must be that the more abundant copper atoms are smaller than the aluminum atoms. This allows us to "pack" more atoms into the same space. When one compares the atomic volumes listed on the periodic table to our sample data, we see that there is a confirmation of this assumption. The table confirms that copper atoms do have a smaller atomic radius than aluminum atoms. Further, since the number of copper atoms in one cubic centimeter is the same as the number of iron atoms,we can assume these atoms to be nearly the same diameter.
Another way to come to this conclusion is to look at another piece of data. The molar mass of the aluminum compared to that of the copper. The mass of copper is only 2.3 times heavier than aluminum. The density of copper is more than 3.3 times greater than that of the aluminum. If mass alone was responsible for density, then the copper sample should only be 2.3 times denser than the aluminum. However, if the atomic volume of the copper was less than that of the aluminum, then the density difference would be greater than the atomic mass difference, which it is! This leads us to a conclusion that as the atomic mass increases, the atomic volume decreases. This general trend is supported by the data across the periodic table.
In closing, we can see that more information has been derived from the work of Archimedes than he could have imagined 2300 years ago. Not only do we have a valid tool to identify a material, but using this with the works of other scientists, we also have insight into the size and quantity of atoms making up the materials around us. This lab points out that to an open mind, there is much more to be seen than what appears to be on the surface.
CHAPTER 2
Investigating Avogadro's First Law (N= SM/MM)
Purpose:
Amedeo Avogadro, an Italian Mathematician, determined that 602 Billion-Trillion atoms of a material would have the same mass as the value called the "Atomic Weight" from the periodic table. This lab will investigate that concept.
Diagram:
Draw or attach a photo of the equipment in set-up mode
Procedure:
. Obtain four different types of beans: Lima, Roman, Black-Eyed Peas, and Lentils
. Fill four 50 ml beakers with each of the four bean types. Make sure to top off each beaker.
. Weigh 15 beans and set aside (Set 1)*
. Weigh 15 more beans, and set aside. (Set 2)*
. Weigh 15 more beans and set aside (Set 3)*
. Add the total mass of the three sets and then divide by 45 or 90 to determine the average mass of one bean. * Lentils are to have three sets of 30 beans per set
. Weigh all of the beans from the beaker, including the 45 or 90 already weighed.
. To predict the total number of beans in the 50 ml sample, divide the total mass of all of the beans by the average mass calculated in step 6
. Manually count the total number of beans in the 50 ml sample.
. Compare your results to the prediction. Calculate the percent error.
. Repeat steps 3-10 for the remaining three beans.
DATA: Lima Roman Black-Eyed peas Lentils
Mass - set1 (15)
Mass - set2 (15)
Mass - set 3 (15)
Mass of three sets (45)
AVERAGE BEAN mass
Total mass of all beans
predicted count
actual count
Percent error
Calculations:
Show all math work in HSPA format.
Conclusion:
Write a brief biography of the Italian Mathematician, Amedeo Avogadro. Then, explain his work on the Mole concept. How is this lab, and the Density lab, related to his work? Explain why we use three sets of beans for our weighing analysis. How is this rooted in Probability & Statistics? The main concept of Avogadro's work is that you can count by weighing, if you know the fundamental mass to divide by.
Calculations:
Lima Beans
Sample #1 ... 16.155 g. (15 beans)
Sample #2 ... 15.495 g. (15 beans)
Sample #3 ... 16.875 g. (15 beans)
Total mass of three samples = 48.525 g (S1+S2+S3)
Average mass of one bean = total mass / total beans
Ave mass = 48.525 g / 45
Ave mass = 1.078 g / bean
Mass of all beans in 50 ml beaker = 54.98 g. (includes the 45 already counted)
Predicted count = total mass / average mass (N = SM / MM)
Predicted count = 54.98 g / 1.078 g/bean
Predicted Count = 51 beans
Actual Count = 52 beans
Percent Error = (Absolute difference / actual count) * 100%
Error = (52 – 51) / 52 * 100%
Error = 1.92 %
Repeat these calculations for the remaining bean samples.
Avogadro's 1st law: N = SM / MM Moles = sample mass / molar mass
Amedeo Avogadro was an Italian mathematician in the middle 1800's. He devised a way of measuring the concentration of particles in a sample. Whether the sample is composed of atoms or molecules, he determined that the number of particles having a mass number matching the "atomic weight" was always 6.02 X 1023. In other words, if you had a sample of carbon with a mass of 12.011 grams, then you would have an Avogadro number of atoms. If you had a sample of water with a mass of 18.00 grams, then you would have his number of molecules in that water sample. Avogadro gave us the ability to count particles (concentration) by simply weighing the sample. This made balancing equations and the application of chemistry a predictable science. Avogadro's number has become the standard in chemistry. When we look at the periodic table, and read the "mass", we are really getting the mass of six hundred and two, billion-trillion atoms.
In this lab, we take large lima beans and determine their average mass per bean. As no two lima beans are the exact same size, then they can't be expected to have the exact same mass. To negate these differences in the beans of our sample, we randomly gather and weigh three groups of beans. Each groups mass is added together. The simple math of dividing the total mass of all of the beans in the three groups by the number of beans which were weighed produces the average mass of one bean. This would be the equivalent of what is now called the molar mass. Now one needs only to weigh a sample of that bean and then divide by this determined average mass, to predict the total number of beans in the sample being investigated. This procedure is repeated for the other three types of beans and a pattern emerges.
A fifty milliliter beaker was used to create a fixed sample volume. In this experiment, we are trying to determine the number of beans which fit into a 50 ml sample volume. So now, for each of the four bean types tested, a fifty milliliter sample was weighed, and then the number of beans in that sample was predicted. This number is based on the average mass of that bean type. As the data shows, the method is very accurate. As the beans become smaller and lighter, they also have become more numerous. A greater number of smaller beans fits into the fifty milliliter beaker than larger beans. If we were to extend this concept down to the size of an atom, then one can see how particle counts in the trillions could be expected. Avogadro gave us the ability to draw relationships between particle size, particle mass, and particle concentration, contained in a fixed volume of a sample.
To achieve an accurate test of these relationships, a fixed volume for atoms and molecules had to be devised. If we compare the density of a sample, which is the mass of a fixed volume (one cubic centimeter) to the atomic weight (molar mass), then we have our link between Archimedes and Avogadro. How many atoms of aluminum fit into one cubic centimeter of volume? How many atoms of copper fit into that same one cubic centimeter of volume? The math shows that more atoms of copper fit into this common volume associated with density. If copper atoms are smaller, then more will fit into the 1 cc space. Looking at the periodic table, and comparing atomic radii, we see that copper atoms are definitely smaller than aluminum atoms.
In this simple lab exercise with four different bean types, we have linked atomic weight to molar mass. We have related molar mass to particle size and also particle concentration. Thus we have found a relationship between Archimedes and Avogadro across 2200 years of science.
(Continues...)
Excerpted from Chemistry Experiments: For Advanced & Honors Programs by James Signorelli. Copyright © 2014 James Signorelli. Excerpted by permission of Trafford Publishing.
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