⏱️ August 28, 2020
🎥Watch: AP Chemistry - Atomic Structure and Mass Spectroscopy
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.
Image Courtesy of Professor Bensely
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.
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.
💡Always convert the % to a decimal when calculating the average atomic mass.
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
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 this😊!
Mass Spectroscopy is more of a lesson that is geared towards the laboratory aspect of Chemistry🧪, but nonetheless, here’s an excellent lesson on Khan Academy which reviews mass spectroscopy:
🎥Watch: Khan Academy - Mass Spectroscopy
✍️ Free Response Questions
AP Chemistry Free Response Questions
⚛️ Unit 1: Atomic Structure and Properties
1.1Moles and Molar Mass
1.2Mass Spectroscopy of Elements
1.3Elemental Composition of Pure Substances
1.4Composition of Mixtures
1.5Atomic Structure and Electron Configurations
1.6Photoelectron Spectroscopy & Graph Interp.
🤓 Unit 2: Molecular and Ionic Compound Structures and Properties
2.0Unit 2 Overview: Molecular and Ionic Bonding
2.1Types of Chemical Bonds
2.2Intramolecular Force and Potential Energy
2.3Ionic Bonding and Ionic Solids
2.4Metallic Bonding and Alloys
2.5Lewis Dot Diagrams
2.6Resonance and Formal Charge
🌀 Unit 3: Intermolecular Forces and Properties
3.0Unit 3 Overview: Intermolecular Forces and Properties
3.2Properties of Solids
3.3Solids, Liquids, and Gases
3.4The Ideal Gas Law
3.5The Kinetic Molecular Theory of Gases
3.6Deviations from the Ideal Gas Law
3.7Mixtures and Solutions
3.8Representations of Solutions
3.9Separation of Solids/Mixtures
3.10Solubility and Solubility Rules
3.11Spectroscopy and the Electromagnetic Spectrum
3.12Quantum Mechanics and the Photoelectric Effect
🧪 Unit 4: Chemical Reactions
4.0Unit 4 Overview: Chemical Reactions
4.1Recognizing Chemical Reactions
4.2Net Ionic Equations
4.4Physical vs. Chemical Changes
4.5Stoichiometry & Calculations
4.6Titrations - Intro and Calculations
4.8Intro to Acid-Base Neutralization Reactions
👟 Unit 5: Kinetics
5.0Unit 5 Overview: Kinetics
5.1Defining Rate of Reaction
5.2Introduction to Rate Laws
5.3Rate and Concentration Change
5.4Writing Rate Laws
5.5Collision Model of Kinetics
5.6Reaction Energy and Graphs w/ Energy
5.7Reaction Mechanisms and Elementary Steps
5.8Writing Rate Laws Using Mechanisms
🔥 Unit 6: Thermodynamics
6.0 Unit 6 Overview: Thermochemistry and Reaction Thermodynamics
6.1Endothermic Processes vs. Exothermic Processes
6.2Energy Diagrams of Reactions
6.3Kinetic Energy, Heat Transfer, and Thermal Equilibrium
6.4Heat Capacity and Coffee-Cup Calorimetry
6.5Phase Changes and Energy
6.6Introduction to Enthalpy of Reaction
6.7Bond Enthalpy and Bond Dissociation Energy
6.8Enthalpies of Formation
⚖️ Unit 7: Equilibrium
🍊 Unit 8: Acids and Bases
8.0Unit 8 Overview: Acids and Bases
8.1Introduction to Acids and Bases
Unit 9: Applications of Thermodynamics
🤺 AP Chemistry Essentials
🧐 Multiple Choice Questions
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