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๐Ÿงชย ap chem

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โš›๏ธย Unit 1

ย ย โ€ขย ย โฑ๏ธ4 min read

1.2 Mass Spectroscopy of Elements

Jeremy Kiggundu

jeremy kiggundu

Dalia Savy

dalia savy

Dylan Black

dylan black


โฑ๏ธ August 28, 2020


๐ŸŽฅWatch: AP Chemistry - Atomic Structure and Mass Spectroscopy

Mass Spectroscopy and Isotopes

An important takeaway from this unit is that by looking at the mass spectrum of an element, you can identify different isotopes of that element. For instance, looking at the mass spectrum of the element Carbon, we can identify the isotopes of Carbon such as Carbon-12 and Carbon-13.ย

Image Courtesy of Professor Bensely

What are Isotopes?

Let's take a look at Carbon on the Periodic Table:

Here, you might notice that instead of a nice "12" or "13" under Carbon, there's a really messy decimal of 12.01. Similarly, oxygen will most likely be 15.999 and even hydrogen will be 1.001. This is because the atomic masses that you're given on a periodic table are the average atomic masses of these elements. This can be described as a weighted average of the abundance of each isotope.

Isotopes are variants of an element. They have the same number of protons and electrons, but a different number of neutrons. Note: Carbon has 6 electrons and protons. The mass number (12/13) minus the number of protons equals the number of neutrons.

Carbon-12 has 6 neutrons (mass number of 12 - 6 protons) whereas Carbon-13 has 7 neutrons (mass number of 13 - 6 protons). This variation of isotopes is summed up on the periodic table using the average atomic mass, which again, is calculated using the mass number and natural abundance, whether it be in decimal form or % form.

Average Atomic Mass Example

So how did the different number of neutrons eventually lead to an average atomic mass of 12.01? Let's look at this. Carbon's 2 main isotopes, Carbon-12 and Carbon-13, account for roughly 100% of the Carbon in nature. There is a tiny sliver of Carbon-14, but these are so small that it's somewhere in the thousandths of a percent. Let's suppose a chemist finds that 98.9% of all carbon in nature is carbon-12, and thus the other 1.1% is carbon-13. To calculate the average atomic mass of carbon from this data:

AAM = 0.989(12) + 0.011(13) = 12.01, which is the same as the number on the periodic table.

Let's see what would happen if we changed the numbers. Let's say now the chemist finds that only 75% of the carbon in nature was carbon-12 and 25% was carbon-13. Now, the AAM is:

AAM = 0.75(12) + 0.25(13) = 12.25

As you can see, the abundance of isotopes can have a significant effect on the average atomic mass of an element. If you are only given the mass numbers of 12 and 13, as well as the average atomic mass, you can easily tell which isotope exists in greater amounts in nature. Since 12.01 is much closer to 12 than 13 is, Carbon-12 is clearly more abundant.

๐Ÿ’กAlways convert the % to a decimal when calculating the average atomic mass.

Another Practice Question

It is also good to know how to identify an element just from its mass spectrum. Let's try with this example:

The peaks below show a mass spectrum of element X. What is element X based on this spectrum? What is its average atomic mass?

Image Courtesy of Kenyaplex

Without a calculator, you can easily tell that the average atomic mass is going to be somewhere between 24 and 25 since X-24 has the greatest % natural abundance. However, let's do some work! Here is what you should do for this problem:

AAM = 0.828(24) + 0.081(25) + 0.091(26) = 24.263.

When looking at the periodic table, we can identify this element as Magnesium.

Image Courtesy of AP Chemistry Reference table

Sample AP Question - 2007 Exam

Part a of question 2 on the AP Chemistry 2007 Exam (Form B) includes basic stoichiometry and advanced mass spectrum calculations. Here it is:

(i)Instead of giving you the graph and asking you to solve for the average atomic mass, they are giving you the average atomic mass and asking you to solve for the percent abundance.

Don't worry! As long as you remember the basic formula, you can solve this problem:

AAM = (natural abundance)(mass) + ... + (natural abundance)(mass)

Now, just plug in what you know!

20.18 = (natural abundance)(19.99) + (natural abundance)(21.99)

One crucial piece of information that you have to remember for this question is that the percent abundances add up to 100%. This allows us to have x as one natural abundance and 1-x as the other, like you have probably done in previous algebra classes. Therefore, the setup for this problem is:

20.18 = (x)(19.99) + (1-x)(21.99)

Once you solve for x, you would get x = 0.905. Therefore, the percent abundance for Ne-20 is 90.5% and the percent abundance for Ne-22 is 9.5% (100%-90.5%).

(ii) To recall from the last guide, you always want to put the value you know first when doing dimensional analysis. Well, how do you even know you have to do stoichiometry? It is asking you to convert from grams to atoms which usually means performing dimensional analysis. Try this problem on your own first but here is the set up:

In the first conversion step, you are simply using the molar mass of Neon on the periodic table. In the second conversion step, you are using the ratio of Ne to Ne-22, which is the percent abundance you found in (i). Then, you use Avogadro's number to convert to atoms and you get the answer of 3.558 x 10^22 Ne-22 atoms.

You got this๐Ÿ˜Š!

Mass Spectroscopy as a Laboratory Procedure

Mass Spectroscopy is more of a lesson that is geared towards the laboratory aspect of Chemistry๐Ÿงช, but nonetheless, hereโ€™s an excellent lesson on Khan Academy which reviews mass spectroscopy:

๐ŸŽฅWatch: Khan Academy - Mass Spectroscopy

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ap chem study guides

โš›๏ธย  Unit 1: Atomic Structure and Properties

๐ŸŒ€ย  Unit 3: Intermolecular Forces and Properties

๐Ÿงชย  Unit 4: Chemical Reactions

๐Ÿ‘Ÿย  Unit 5: Kinetics

๐Ÿ”ฅย  Unit 6: Thermodynamics

โš–๏ธย  Unit 7: Equilibrium

ย  Unit 9: Applications of Thermodynamics

๐Ÿคบย  AP Chemistry Essentials

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