⏱️ August 28, 2020
Imagine an atom. Well, most likely you can't even begin to grasp how small an atom even is⚛️. How do scientists then perform laboratory work when it is nearly impossible to count the atoms they are working with🤔? This is where the concept of a mole emerged from. A mole relates the mass of an element to the number of particles there are. Let's begin with the molar mass.
🎥Watch: AP Chemistry - Moles and Molar Mass
The molar mass of a substance is the amount of grams there are in a mole (the units for molar mass are grams/mole). Finding the molar mass of an element or compound is not as hard as it might seem: the only things that you need to know are which elements are involved and how many of them are present. Also, there's one huge important thing that you'll need: a periodic table!
Reference table you will have with you during the AP
For example, let’s say that we have the compound CO2. To calculate the molar mass, you simply have to multiply the atomic mass of each specific element by it’s subscript, and then add it all together. Simple, right?
So we're working with CO2. Carbon has a subscript of 1 and has an atomic mass of 12.01 grams. Oxygen has a subscript of 2 in this element and has an atomic mass of 15.99 grams. Always multiply the subscript by the atomic mass, and when it's done where, it equals 31.98 grams. Finally, we add 31.98 grams to 12.01 grams to get 43.99 grams. Therefore, Carbon Dioxide has a molar mass of 43.99 grams per mole.
One of the most fundamental takeaways from this unit is dimensional analysis! What this means is that you can change the units that a number is represented in. For instance, you can convert from miles per hour to meters per second. We can apply the concept of moles and perform dimensional analysis with Avogadro's number. Avogadro’s principle tells us that 1 mole is equal to 6.022x10^23 atoms of a pure substance. This will be important to know when converting from grams to moles to atoms.
Now that we have learned the basics, let’s try converting a sample of 50.0 grams of CO2 between units.
When doing dimensional analysis, you want to put the number that you know first, which in this case, is 50.0 grams of CO2. Then, you want to multiply 50.0 by the molar mass in order to convert to the moles of CO2. The unit that you have (grams CO2) should always be on the bottom of the next ratio in order for the units to cancel out. Here, the grams of CO2 cancel out and you are left with a measurement in moles.
💡Tip: It is good to memorize that moles = grams/molar mass. The conversion step in this problem is actually using this concept since you are ultimately dividing the number of grams you have by the molar mass to get the number of moles.
Now let's convert 1.13 moles of CO2 into atoms using Avogadro's number.
Here, you are once again taking the number that you have and putting it first. Then, you are putting the unit of measurement that you want over the unit of measurement that you have, making that step the unit conversion. This enables the moles of CO2 to cancel out, leaving you with just 6.80x10^23 atoms of CO2.
Since the subscript on Carbon is 1, the number of atoms of CO2 is equivalent to the number of carbon atoms in CO2. There is nothing to multiply by because of this 1 to 1 ratio, therefore the number of carbon atoms in this 50.0g sample of CO2 is 6.80x10^23.
Unlike carbon, oxygen has a subscript of 2. This makes the ratio of CO2 atoms to oxygen atoms 1:2. Therefore, we have to use dimensional analysis:
Since there are two atoms of O in one atom of CO2, we had to multiply by 2 to get the number of atoms of O.
You got this! Once you practice multiple problems involving dimensional analysis, it'll seem like a piece of cake. Sadly, these problems become more difficult as the course progresses but as always, practice makes perfect.
✍️ 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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