12/9/13-12/13/13

Previously on this blog, I have stated that there have been some challenging topics. However, those topics pale in their abstrusity when compared to the current subject matter. The unit at hand for the last several weeks in AP Chemistry is Thermodynamics, which covers a wide variety topics including energy, heat, work, the system and surroundings, state functions, enthalpy, entropy, and Gibbs free energy.

To start, energy is the ability to do work or transfer heat. Its SI unit is the joule (J). Heat is energy in transit. It is represented by the variable q. Work is equal to force multiplied by distance and is represented by the variable w. Additionally, work is equivalent to the negative pressure multiplied by change in volume (this will come into play later). Heat energy can be used to do work. Likewise, work can be transformed into heat. When referring to the transfer of energy, one must consider the system and the surroundings. The system is the actual chemical reaction that is taking place. The surroundings are everything outside of the reaction. The energy inside a system is referred to as a system's internal energy. Changes in a system's internal energy can be modeled by the following equation:

∆E = q + w

In other words, the change in internal energy is equivalent to heat plus work.

Enthalpy is related to internal energy. If a reaction takes place at constant pressure and the only work done is pressure-volume work, the heat flow of the process can be accounted for by measuring the enthalpy of the system. Simply put, enthalpy is the internal energy plus the product of pressure and volume. This can be modeled by the following equation:

H = E + PV

When the system changes at constant pressure, the change in enthalpy (∆H) is equivalent to:

∆H = ∆E + P∆V

Since it is known that ∆E = q + w, "q + w" (heat plus work) can be substituted into the equation for the change in enthalpy. It is also known that w = -P∆V, so this can also be substituted into the equation for change in enthalpy. Thus:

∆H = (w + q) - w
∆H = w - w + q 

∆H = q

At constant pressure, the change in enthalpy is equivalent to the heat gained or lost. When a reaction is endothermic (takes in heat) ∆H is positive. When ∆H is negative, the reaction is exothermic (heat leaves the system). There are four methods for determining the change in enthalpy, the first being using average bond energies. To do this, one must calculate the energy needed to break bonds in the reactants and then form bonds in the products. This process is modeled by the equation:

∆H = ∑BE(bonds broken) + ∑BE(bonds formed) 

Where the energy of the bonds broken is always negative

∑BE(bonds broken) is always positive as breaking bonds is an endothermic process, while ∑BE(bonds formed) is always negative as the formation of bonds in exothermic.

∆H can also be found using a calorimetry. Calorimetry is the measurement of heat transfer. Heat flow is measured using a calorimeter, which measures heat flow by measuring the before and after a reaction. Calculating change in enthalpy using this method requires values of heat capacity, the amount of energy required to raise the temperature of a substance by 1 K, as well as specific heat, the heat capacity of a substance per unit mass (side note: this website helped me to differentiate between specific heat and heat capacity). The heat capacity of a substance as well as the mass of the sample are two influential factors in temperature change. Thus, the change in enthalpy can be determined using this equation:

∆H = m x Cp x ∆T

The change in enthalpy is equivalent to the mass of the sample multiplied by the specific heat capacity of the substance multiplied by the change in temperature. However, calorimetry uses the energy gained or lost in the surroundings to calculate the energy gained or lost in the reaction. Due to the first law of thermodynamics (also known as the law of conservation of energy) the energy gained or lost by the system is is equal and opposite to the energy gained or lost by the surroundings.

∆Hreaction = -∆Hsurroundings  

The third method of obtaining the change in enthalpy value is through Hess' Law which states that if a reaction is carried out in a series of steps, the overall change in enthalpy will be equal to the sum of enthalpy for each of the individual steps. Hess' Law is closely related to the concept of state functions. A state function is a property that depends on the present state of the system, not the path taken to arrive at that state. 

An example of a state function. 

The fourth and final method of calculating ∆H is through enthalpies of formation. Enthalpies of formation are hypothetical values that indicate how much heat would be lost or gained during the formation of one mole of a compound from its elements in their standard states (standard states meaning 25˚C and 1 atm). Heat of formation reactions are always written so that all reactants exist as they would under standard conditions and if there was one mole of product. Using the method of enthalpy of formation, ∆H is calculate using the equation:

∆H = ∑n∆H˚f(products) - ∑n∆H˚f(reactants)

The heat gained or lost in the reaction is equal to the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the products minus the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the reactants. Methods of finding ∆H were covered in several thermodynamics lectures as well as in the Thermodynamics I and II worksheets.

Entropy is a statistical measure of the number of most probable distinguishable microstates or degrees of freedom available in a system. When there are a greater number of distinguishable microstates (i.e. greater entropy) the reaction is referred to as favorable or spontaneous. Thus, when the change in entropy, ∆S, is positive, the value is favorable. The change in entropy is equal to the sum of the change in entropy of the products minus the sum of the change in entropy of the reactants.

∆S = ∑S(products) - ∑S(reactants) 

The second law of thermodynamics states that the entropy of the universe is constantly increasing. Reactions may have a reduction in entropy as long as the entropy of the surroundings increase. Entropy is a state function. ∆S is greater than 0 when heat is added (such as in processes like melting and vaporization). ∆S is less than zero in certain cases, such as when substances are dissolved and when solutions of gases and liquids are made. The concept of Entropy was covered in a lecture as well as in the Thermodynamics III worksheet.

Entropy plays a part in determining thermodynamic favorability. A thermodynamically favored process (or a spontaneous process) is a reaction that proceeds without any assistance from the surroundings. In reversible reactions, the thermodynamic favorability determines the direction of the reaction. There are two methods of determining thermodynamic favorability. The first, ∆Suniverse states that ∆Suniverse = ∆Ssystem + ∆Ssurroundings. If ∆Suniverse is positive, then the reaction is favorable. If a reaction increases the entropy of the system (∆Ssystem > 0) and is exothermic, then the process must be thermodynamically favored. The second method of determining favorability is through Gibbs free energy. Gibbs free energy equals the change in enthalpy minus the system temperature multiplied by the entropy change.

∆G = ∆H - T∆S

If ∆G is negative, then the reaction is thermodynamically favored. Gibbs free energy is a state function. ∆G˚f is the energy change that occurs when one mole of a compound is made from its elements in their standard states. ∆G˚f values can be used to determine the ∆G value for an equation using the equation above. 

Combing over these concepts and writing this blog post, none of these concepts seem incredibly difficult. However, when it comes to applying these concepts and choosing which equation or formula to use, I find myself at a complete loss. I struggled to complete the review task chains. In the Thermodynamics II worksheet as well as in the task chains, I found myself struggling the most on problems that involved stoichiometry. I feel I have adequate understanding of Hess' law and finding ∆H based on enthalpies of formation. I have had trouble determining whether or not reactions are endothermic or exothermic. The concept of specific heat and calorimetry also baffled me a bit. As I do not feel prepared in the slightest for the upcoming test this week, I will be going over the task chains multiple times and revisiting worksheets that were completed in class. I'm sure the test will be a sobering experience that will test the self inflated image of my self worth and knowledge.  

11/4/13-11/8/13

This week in AP Chemistry I continued to study intermolecular forces. The topics covered this week included viscosity, surface tension, cohesive and adhesive forces, different forms of solids, vapor pressure, and lattice energy.

Viscosity is a liquid's resistance to flow. It is the ease by which the molecules in a substance move past each other. The higher the viscosity, the less a liquid moves. Viscosity increases with stronger intermolecular and decreases with higher temperature. This topic was covered in the Liquids lecture.

Surface tension results from an imbalance of intermolecular forces at the surface of a liquid. Molecules on the surface are subject to only inward forces, whereas molecules in the center of a substance are surrounded on all sides by molecules and thus are subject to a full cadre of intermolecular forces. Of all the topics covered this week, I had the the weakest understanding of surface tension. I did not comprehend the portion in the Liquids lecture about surface tension well. I also feel as though I am unable to complete the questions in the worksheets I was given in class that regard surface tension. This website gave me a little clarity on the subject, although I was unable to find a good source that could give me a more in depth explanation that I could understand well.

Cohesive forces are intermolecular forces between molecules of the same kind. Adhesive forces are forces that are intermolecular forces between two molecules of a different kind. The strength of cohesive and adhesive forces can determine how a substance acts on a certain surface. For example, if you put a drop of water on a waxed surface and a drop of water on glass, the water on the waxed surface beads up (the cohesive forces between water molecules are greater than the adhesive forces between water and the waxed surface), while the water on the glass spreads out (the cohesive forces between water molecules are weaker than the adhesive forces between the water and the glass). Cohesive and adhesive forces were covered in the Liquids lecture as well as in the Intermolecular Forces II worksheet. I felt I had a strong understanding of this topic.

Another example of cohesive vs. adhesive forces. In the tube with water, the adhesive forces between water and glass are greater than the cohesive forces between the water molecules. Thus, water has a concave meniscus. In the tube with mercury, the cohesive forces between mercury molecules are greater than the adhesive forces between mercury and glass. Therefore, mercury has a convex meniscus.

Additionally, I studied various properties of solids this week. Solids generally fall into two groups: crystalline and amorphous. Crystalline solids are solids in which the molecules have a highly ordered arrangement. Amorphous solids are solids in which there is no order in the arrangement of particles. Furthermore, there are other classifications of solids including molecular solids, ionic solids, and covalent network solids. Molecular solids are solids where molecules are held together by van der Waals forces. They tend to be soft and usually have low melting and boiling points. Ionic solids are the solid forms of ionic compounds. In ionic solids, ions pack themselves in a position that maximizes attraction between ions. Ionic solids typically have very high melting and boiling points. Covalent network solids are solids in which atoms are covalently bonded to each other in a large network. Substances like diamond, graphite, quartz, and glass are all network solids. These solids tend to be strong and have high melting and boiling points.

A vapor is the gas form of a substance that is normally liquid at room temperature. Pressure is the force by which gas particles hit a surface. Vapor pressure is simply the pressure of a vapor at a given temperature. The greater the number of gas particles, the greater the pressure of a gas. As temperature increases, the kinetic energy of molecules in a liquid increases and more particles "escape" the liquid and evaporate. Thus, the pressure of the surrounding atmosphere is increased. When the vapor pressure equals the atmospheric pressure, a substance boils (at sea level, this pressure is 1 atm). However, intermolecular forces influence boiling point. The greater the intermolecular forces, the greater the boiling point. Boiling point and vapor pressure are linked; the greater the boiling point, the lower the vapor pressure. The lower the boiling point, the higher the vapor pressure. All substances have the same vapor pressure at their boiling point. Initially, this concept threw me off. When we studied properties like viscosity and boiling point, as the intermolecular forces grew stronger, the viscosity and boiling point of that given substance increased as well. But with vapor pressure the relationship is different. Other than that small caveat, I understood this concept well. I completed several assignments that related to vapor pressure, including a lecture and the Intermolecular Forces II worksheet.

The final subject I studied this week was lattice energy. Lattice energy is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions. Lattice energy is associated with coulomb's law; it increases as charge magnitude increases and increases as the size of ions decreases (distance between oppositely charged ions decreases, so the charge between them is stronger). This process is very exothermic; it takes a considerable amount of energy to break ions from the solid.

I felt I had a strong grasp on most of the topics covered this week with the exception of surface tension. I will continue to look for a source that could give me some clarity in that subject. I also feel relatively prepared for the upcoming test on Tuesday, but none the less I will be furiously studying and stressing on Monday night.

10/28/13-11/1/13

This week in AP Chemistry I learned about intramolecular forces and intermolecular forces. Intramolecular forces are forces between atoms in a molecule such as covalent and ionic bonds. However, intramolecular forces were not the main focus this week. Intermolecular forces were.

Intermolecular forces are attractive and repulsive forces between molecules. Intermolecular forces are weaker than intramolecular forces but are strong enough to influence physical properties of substances. There are several different types of intermolecular forces. Most fall under the category of "van der Waals forces." In class, I learned of four important van der Waals forces: dipole-dipole interactions, dipole induced dipole interactions, induced dipole-induced dipole interactions (a.k.a. London dispersion forces), and Hydrogen bonding.

Dipole-dipole interactions are attractive forces between the partially negative end of a polar molecule and the partially positive end of another molecule. Although the strength of dipole-dipole interactions are dependent on the magnitude of a substance's dipole moment and the proximity of its molecules, in general dipole-dipole attractions are relatively weak. Dipole-dipole interactions are only present in polar substances.

Hydrogen bonds are closely related to dipole-dipole interactions. Hydrogen bonds occur when partially positive hydrogen atoms are attracted to very electronegative partially negative atoms like oxygen, fluorine, and nitrogen. Although in essence they are the same as dipole-dipole interactions, they have their own classification because they are much stronger than typical dipole-dipole forces. Hydrogen bonds are the strongest of all of the van der Waals forces.

When a polar molecule approaches a nonpolar molecule, the electrons in the nonpolar molecule, usually shared evenly, are attracted to the polar molecule. This results in an induced dipole in the nonpolar molecule. This interaction is aptly named a dipole-induced dipole interaction. Much like the dipole-dipole interaction, it too is relatively weak.

Induced dipole-induced dipole interactions are the most common. Electrons are constantly in motion. Even in nonpolar molecules, there is a probability that at some point there will be an imbalance of electrons (this distortion in electron clouds is called polarizability). When such an imbalance occurs, a temporary dipole is produced. This temporary dipole induces a temporary dipole in other molecules. This intermolecular force, also called the London dispersion force, is present in all substances.

Additionally, I learned this week about ion-dipole interactions. Ion-dipole interactions are forces between ions and molecules. They are not van der Waals forces. Ion-dipole interactions make it possible for some ionic substances to dissolve in water. If the forces of attraction between atoms in an ionic compound are greater than the ion-dipole force between the ionic compound and water, the substance will not dissolve in water. If the forces of attraction between atoms in an ionic compound are lesser than the ion-dipole force between the ionic compound and water, the substance will dissolve in water.

Intermolecular forces have a great effect on the physical properties of substances. For example, the stronger the intermolecular forces, the higher the boiling and melting point of a substance. Intermolecular forces also affect the vapor pressures and viscosity. To help learn the concepts associated with intermolecular forces, I completed three lectures (one on dipoles and induced dipoles, one on hydrogen bonds, and one on ion-dipole interactions) as well as a the Intermolecular Forces POGIL and the Water POGIL. In addition, we went over the Intermolecular Forces I worksheet. Initially I found the concept of intermolecular forces to be confusing. I found the scanned textbook chapter you gave us in class on Tuesday to be challenging reading. I did not comprehend the text well (the fact that I was running on a minuscule amount of sleep certainly did not help in this regard). However, I found that the lectures brought me to understand these topics well.

A demonstration from class this past week. Butane is lit on fire. In the can, butane is in liquid form because it is under very high pressure. However, when it was sprayed into a test tube, it immediately started to boil, as butane normally has a very low boiling point. When lit on fire, the butane gas burned.  

10/21/13-10/25/13

This week in AP Chemistry we had the third test of the year. This exam covered the topics of covalent bonding, electron domain geometry, as well as and molecular geometry. To study for the test, I had to complete a task chain. We also went over a packet of review worksheets in class on these subjects. At home, I did the task chain multiple times and completed extra problems on the review worksheets. On Tuesday, I took the test and got an 80 percent on the multiple choice and a 98 percent on the free response. I was very satisfied with my performance on the free response but was a bit disappointed with my multiple choice test. I felt I had a fairly good grip on this subject matter (better than the material on the stoichiometry test in fact), and yet I got a score that I was not very satisfied with. Considering the fact that I have gotten relatively iffy scores on the last two multiple choice exams, I've surmised that I need to pay more attention to the multiple choice portions on tests and make a better effort to practice the types of the questions that may be found on them. All in all, I was a bit irked with my performance on the multiple choice. However, because of my results on the short answer and the fact that the multiple choice did not significantly affect my grade, I found some satisfaction in my performance on this test.

The day after the test was mole day. We were assigned an essay on the subject of polarity and hydrogen bonding as it related to an article we were given about paintball. I found the assignment to be difficult at times, as I found the prompt to be a bit vague. As only one section of the paintball article really related to polarity and hydrogen bonding, it was hard to find excerpts in the writing that supported the points in my essay. With that being said, I did learn about hydrogen bonding and how polarity relates to solubility. Hydrogen bonding is when a partially positive hydrogen atom bonded to an electronegative atom is attracted to a partially negative electronegative atom in a separate molecule. Unlike covalent bonding, there is no physical bond present, but there is attraction between the two atoms.

A hydrogen bond, represented by the dashed line.

After these two assignments, I was reintroduced to the topic of ionic bonding via a POGIL that I completed on class. Ionic bonds are bonds between a metal and a nonmetal. I learned that the size and individual charge of each ion has a profound effect on the strength of an ionic bond. The stronger the bonds in an ionic compound, the higher the melting point. For the most part, I felt I had a decent understanding of this topic. I was a bit confused about the sizes of cations and anions.

The final topic we covered this week was metals. Metals are elements that conduct heat an electricity well. They are typically malleable and ductile. They exhibit several other unique physical properties, such as luster. Metals bond together through metallic bonds. In metals, electrons called conduction bond electrons form a "sea" of electrons around atoms. These conduction bond electrons are d-orbital electrons that are promoted into the outer p-orbitals of atoms. Conduction electrons are the reason that metals conduct heat and electricity well. The more conduction electrons, the higher the melting point of the metal and the harder the metal is. Most ordinary uses of metals involve alloys. Alloys are mixtures of elements that have the properties of metals. There are two main types of alloys: substitutional alloys (solute particles take the place of solvent metal atoms) and interstitial alloys (solute particles fit in holes in between solvent metal atoms). To help learn about metals, I completed a POGIL on metals on Friday and watched a lecture on metals this weekend. I found the subject straightforward and easy to understand.

    

10/14/13-10/18/13

This past week in AP Chemistry was the most confusing so far. I was introduced to the concept of hybridization and sigma and pi bonds.

In covalent bonds, electrons can only be shared if the electron orbitals of two atoms overlap. Hybridization occurs when these orbitals combine to form new, hybrid orbitals. For example, the s and p orbitals combine to form an sp orbital. There are three widely accepted forms of hybridization: sp, sp², and sp³. Hybridization is a theory; whether or not it works beyond the second period is questioned. Additionally, hybridization is central to valence bond theory, a.k.a. sigma and pi bonding, which is explained later. 

sp hybridization.

I did not understand the concept hybridization well, which would explain my rather lackluster explanation. I do know that you can determine the hybridization of a molecule by simply counting the number of electron domains around the central atom. If there are two electron domains, the hybridization is sp. If there are three electron domains around the central atom, the hybridization is sp². If there are four electron domains, the hybridization is sp³. Whether hybridization occurs beyond that is hotly contested within the scientific world and does not concern me. However, conceptually I do not understand hybridization beyond the notion that orbitals can combine to make "hybrid" orbitals. To cover these concepts, I watched a lecture on hybridization and completed the Hybrid Orbitals POGIL for homework. While what I know may be enough to get by on the test, I still would like to understand the concept of hybridization more.


I found the concept of sigma and pi bonds to be more straightforward. Sigma bonds are bonds with head to head overlap and cylindrical symmetry of electron density about the internuclear axis. Pi bonds are bonds with side to side overlap. Electron density is above or below the internuclear axis. In a single bond there is one sigma bond. In a double bond there is one sigma bond and one pi bond. In a triple bond there is one sigma bond and two pi bonds. I was confused about this concept until I completed the task chain quiz that involved sigma and pi bonds. When I got an answer wrong regarding a sigma and pi bond, the information window that popped up gave a good explanation of sigma and pi bonds. To cover sigma and pi bonds, I watched a lecture on the subject. This website also gave me some clarity on the subject.

There is a pi bond between the blue p orbitals, as there is side to side overlap between them. There is a sigma bond between the green orbitals, as there is head to head overlap between them. 

Throughout the week in class I worked on a lab/activity involving the program Web MO in which I modeled molecules from the VSEPR Theory lab with my table group. Using the program, I determined the bond angles, polarity, dipole moment, and more for several molecules. The activity really helped me better understand the concepts of bond and molecular polarity, as well as how unbonded electron pairs affect the shape of molecules.

10/7/13-10/11/13

This week in AP Chemistry we continued work on Lewis structures and VSEPR theory. Additionally, we learned the concepts such as formal charge and polarity.

Near the beginning of the week, we finished up the balloon/gumdrop activity, which pertained to VSEPR theory. While Lewis structures help to understand the composition and covalent bonds of molecules, VSEPR theory (short for "valence shell electron pair repulsion") allows us to predict the shape of molecules. There are two main categories in VSEPR theory: electron domain geometry and molecular domain geometry. In predicting a molecules electron domain geometry, we assume that electron pairs are placed as far apart as possible. Electron pairs are referred to as electron domains; one electron pair is equal to one electron domain. Double and triple bonds also only count as one electron domain. To determine electron domain geometry, one first counts the number of electron domains and then chooses the corresponding shape. Pictured below are several electron domain geometries.  

Different electron domain geometries. For two electron domains, the shape is linear. For three domains, the shape is trigonal planar. For four domains,  the shape is tetrahedral. For five domains, the shape is trigonal bipyramidal. For six electron domains, the shape is octahedral.  

Molecular domain geometry is slightly different. It is defined by the positions of only the atoms and not the nonbonding electron pairs. Here are some examples of molecular geometries:

When there are two electron domains, the following molecular geometries are possible:

• AB2 - linear

When there are three electron domains, the following molecular geometries are possible:

• AB3 - trigonal planar
• AB2E - bent
• ABE3 - linear (this is a rare occurrence)

When there are four electron domains, the following molecular geometries are possible:

• AB4 - tetrahedral
• AB3E - trigonal pyramidal
• AB2E3 - bent
• ABE3 - linear

When there are five electron domains, the following molecular geometries are possible:

• AB5 - trigonal bipyramidal
• AB4E - seesaw
• AB3E2 - t-shaped
• AB2E3 - linear

When there are six electron domains, the following molecular geometries are possible:

• AB6 - octahedral
• AB5E - square pyramidal
• AB4E2 - square planar

To help learn the concept of molecular and electron domain geometry, I completed a POGIL in class and made models of molecules using gumdrops and toothpicks. I also watched a lecture on VSEPR theory. While I feel I have a decent understanding of this subject, I feel I could use more practice determining the molecular geometry and electron domain geometry. I sometimes get confused with differentiating between the two kinds of geometries, as the electron domain geometry involves lone pairs of electrons and the molecular geometry does not.

We also covered the concept of formal charge in class.  Formal charge is equal to the number of valence electrons in a free atom minus the number of bonds plus the number of nonbonding electrons. The sum of each individual formal charge of each atom in a molecule must be equivalent to the total charge of the molecule. If there are multiple structures for a molecule, the molecule with the lowest formal charge is preferred. If there is a negative formal charge, it should be on the least electronegative atom if there is a choice. If the central atom is in period three or higher, multiple bonds are possible if it will reduce the formal charge of the molecule. To help me learn formal charge, I had to view a lecture on formal charge. In class, I completed the Lewis Structures III POGIL and the Lewis Structures IV POGIL, both of which dealt with formal charge. I felt I had a relatively strong grasp of this concept, although I sometimes have trouble finding which Lewis structure has the lowest formal charge when there is more than one possible structure.

Finally, we touched on the concept of polarity in a lecture that was assigned last week. An molecule is polar when its electrons are not shared equally. Shared electron pairs in polar covalent bonds are not shared equally, while electron pairs in nonpolar covalent bonds are shared equally. Between the atoms of molecule, the greater the difference in electronegativity, the more polar the bond. Molecular polarity is possible and is calculated by adding up the individual bond dipoles. One thing I did not understand about polarity is the "dipole moment." From the lecture and from various sources online I was unable to gather a clear definition. I know it pertains to electrical charge and electrons, but I feel I need a concrete definition in my head to best understand what a dipole moment is.

9/30/13-10/4/13

This week in AP Chemistry, I continued to learn about Lewis structures. Over the course of the week I was introduced to the topics of bond order, resonance, hypervalency, and covalency. Bond order is the number of chemical bonds between a pair of atom. A single bond has a bond order equal to one, a double bond has a bond order equal to two, and so on and so forth. To help learn the concept of bond order, I completed the Lewis Structures Part 2 lecture at home and completed a POGIL on Lewis structures in class.

Resonance is a term used to describe a situation in which there are more than one valid Lewis structures for a given molecule. Initially I was unsure of the exact definition of resonance, but this website gave a nice concise definition. The Lewis Structures II POGIL and the Lewis Structures Part 2 lecture dealt with this topic.

Possible resonance structures of NO3. The double headed arrows indicate interchangeability (all three structures are valid). 

If there are multiple valid Lewis structures, the bond order is affected. This was a point of confusion for me. Initially, after watching the Lewis Structures Part 2 lecture, I came under the impression that the bond order was equivalent to the number of bonds divided by the number of possible structures for a molecule. However, after starting the POGIL on Monday I discovered that this was certainly not the case. I learned that bond order between two atoms was simply equivalent to the number of bonds present. I am still confused as to how resonance affects bond order. I have searched online to find an answer, but I have not found one that makes it clear to me.

Hypervalency is a situation in which there is more than eight electrons around an atom in a molecule. When making a Lewis structure, if you satisfy the octet rule for all of the outer and central atoms and still have extra valence electrons, distribute them about the central atom. This is only possible if the central atom is in period 3 or greater.

In the PCl5 molecule, the Phosphorus atom is hypervalent. It exceeds the octet rule and has 10 valence electrons.  

When considering hypervalency, size is important. The larger the central atom, the larger the number of electrons that can surround it. Expanded octet most often occurs when the central atom is bound to atoms with high electronegativity such as Fluorine, Chlorine, or Oxygen. To learn the concept of hypervalency, I watched the Lewis Structures Part 3 lecture. The concept was also part of the VSPER Theory POGIL we started in class on Friday. I found most of this topic easy to understand. With that being said, I thought the portion in the lecture that explained how the expanded octet is possible was very confusing.

Covalency was the final new topic we covered this week. In the lecture on Covalency, I learned that molecules with covalent bonds stay together because the force of repulsion between protons is weaker than the force of attraction between protons and electrons. Electrons are said to be "paired" when they have opposite spins and enter the domain of the other. Additionally I learned that as bond order increases, strength increases.

Overall I though I understood most of the concepts covered this week well, with the exception of the relationship between resonance and bond order as well as the part of hypervalency that I explained earlier.

In addition to learning these concepts, we did a stoichiometry-related lab in class on Wednesday and Thursday. The purpose of the lab was to find the percentage mass of copper in a brass screw. In the lab, we took a brass screw and dissolved it in nitric acid under the fume hood, which produced both toxic NO2 gas and a liquid, Cu(NO3)2.

The brass screws react with nitric acid to produce a blue liquid, Cu(NO3)2, and a brown gas, NO2.

Then, half of the groups made dilutions with the Cu(NO3)2 liquid, measured their absorbance, calculated concentration, and made a calibration curve. The other groups (mine included) tested the visual comparison method of calculating concentration. In this process, you filled two beakers with solution. One beaker had stock solution and the other had the product of the brass screw/nitric acid reaction, known as the "mystery solution". The beaker with mystery solution was filled until its color matched that of the stock solution. The depth of each solution was then measured and then the concentration of the mystery solution was calculated with the equation:

(Molarity1)(Depth1) = (Molarity2)(Depth2)    

The lab was fairly straight forward. I think my work on it is an improvement over my work on the previous lab. Now that I have a little bit of experience doing labs, I am more comfortable and the work in my lab notebook is much tidier. I am unsure when the lab is due and would like to know so I know when to start working on the post-lab questions.

9/23/13-9/27/13

This week in AP Chemistry I took the second exam of the school year. The test covered stoichiometry, which I have been learning since the beginning of the year, and was held on Wednesday. To prepare for the test, we reviewed many of the concepts of stoichiometry in class using socrative.com. Additionally, to prepare for the test I watched a lecture on no-calculator math and completed three no-calculator math hotpots. I used the hot potato quizzes to study, doing each quiz two times. After starting to do the hotpots I became worried about the exam. I was able to complete the hotpots with most of the correct answers, but it was taking me too much time. Upon looking at a problem, it would take me a while to delineate what calculations to use to find the solution. However, I resolved this problem by simply practicing the quizzes again and by figuring out the shortest way to achieve the correct solution. I felt comfortable with the majority of the concepts of stoichiometry, so I primarily studied using the no-calculator math quizzes. On the test I got a B on the multiple choice portion. I have not seen the corrected copy of my test, however, I assume that I lost points on the questions where I had to chose an appropriate diagram of a given reaction. These were the problems on the hot potato quizzes that I had the most trouble with. On the free response I got a perfect score. Despite the fact that I got an 86 on the multiple choice, overall, I was satisfied with my performance on the stoichiometry exam. I was anticipating getting a far worse score prior to taking the exam. Nevertheless, I still would like to come in after school to see what problems I got incorrect.

After the test on Wednesday, we moved on to learning about Lewis dot diagrams. Lewis dot diagrams are renderings of substances that also show the number of valence electrons per atom and covalent bonds in the substance. To make one of the diagrams, you first write the element symbols. Then, you draw dots around the elements where there are valence electrons and dashes where there are covalent bonds. The total number of valence electrons in the substance should be accurately represented in the diagram. The diagrams should also follow the octet rule (atoms tend to have eight electrons in their outer shell). Sometimes, Lewis dot diagrams do not include the dots that represent valence electrons. However, for AP Chemistry it is always necessary to include dots in the diagrams.

Examples of Lewis Dot Diagrams. Valence electrons are represented by dots, while covalent bonds are represented by dashes.

To help learn about the Lewis dot diagrams, I completed a POGIL in class with my group and watched two lectures on the diagrams at home this weekend. I also found this website which offers a good summary on how to properly draw the diagrams. At first, I was unclear as to how to determine how many bonds to place in the Lewis dot structure, but the lecture helped to clear this up for me. Other than that, I believe I understood the Lewis dot diagrams well.

9/16/13-9/20/13

This week in AP Chemistry, I continued to learn about stoichiometry. However, this week it became more complicated as limiting reactants, excess reactants, and yield were added into the mix. To help learn the concept of limiting reactants, I completed several worksheets, including Stoichiometry 6 & 7. I also viewed a lecture on limiting reactants. Limiting reactants are reactants that limit the amount of substance that is produced in a chemical reaction. To determine the limiting reactant, you calculate the amount of product produced by each reactant. Whichever reactant produces the least amount of product is the the limiting reactant.

An example of how to find the limiting reactant from the worksheet Stoichiometry 6. 153.5 g of carbon monoxide produces in 175.6 g of methanol. 24.50 g of hydrogen produces 194.0 g of methanol. Since carbon monoxide produces a lesser amount of methanol, it is the limiting reactant.  

I found this concept straightforward and easy to understand. The concept behind excess reactants is similar. When the limiting reactant is used up there is excess of the other reactant. This is the excess reactant. Initially, I did not fully understand how to determine the amount excess reactant. However, after searching online, I found this website which helped me to understand the how to calculate excess reactant. To help learn the concept of excess reactants, I completed the Reaction Particle Diagrams worksheet in class.

I found the concept of yield straightforward as well. There are two kinds of yield: theoretical yield and actual yield. Theoretical yield is the calculated maximum amount of product possible from a given amount of reactant. Actual yield is the measured amount of product experimentally produced from a given amount of reactant. Percentage yield can be determined by the following equation:

In order to understand the concept of yield, I had to complete the Stoichiometry 8 worksheet, which is comprised of problems in which one must calculate the percentage yield of reactions.

Additionally, this week I was introduced to the concept of empirical formulas. An empirical formula is a formula that represents the simplest ratio among the elements of a compound. (for example, the empirical formula of glucose, C6H12O6 is CH2O). Ionic compounds are always empirical formulas. Molecular formulas are sometimes empirical formulas. I was required to complete the Empirical Formulas 1 worksheet, in which I had to determine the empirical formula of compounds when given the mass composition of each compound.

I found in class this week that we completed many of the stoichiometry worksheets by having different groups writing the answers on whiteboards. While this is one way to learn the concepts, I found myself mindlessly copying down the answers. I think that I learn more effectively when there is a lecture quiz that offers guidance on the worksheet and I must complete most of the worksheet on my own.

With that being said, I feel as though I have a relatively strong understanding of all of these concepts. However, at the same time I think that some more practice would certainly not hurt. I will review the worksheets I did this week before the test this upcoming week.

9/9/13-9/13/13

This week in AP chemistry, I learned more about the mole and was introduced to the concept of molarity. A mole of anything equals 6.022 x 10^23 of that thing (For example, a mole of atoms is 6.022 x 10^23 atoms, a mole of molecules is 6.022 x 10^23 molecules, etc. There could even be a mole of moles, 6.022 x 10^23 moles). The mole is not a measure of mass, but is rather a quantity. The mole's mass depends on the mass of the substance. The mass of a mole of a substance equals the atomic mass of the substance, but in grams instead of atomic mass units. For example, an atom of Hydrogen has a mass of 1.01 amu while a mole of Hydrogen has a mass of 1.01 g. To help learn the concept I had to complete a lecture quiz on the mole as well as the Stoichiometry 1 & 2 worksheet. Additionally, we worked on an in class worksheet that included calculations involving the mole.

The Stoichiometry 1 & 2 tied into another concept that I learned this week, dimensional analysis. Dimensional analysis is a method of calculating that involves units. By using dimensional analysis, you can make sure that you are doing the correct kind of calculations and ensure that you are ending with the correct unit. In addition to the Stoichiometry 1 & 2 worksheet, I also had to complete a lecture quiz on dimensional analysis.

In addition to the mole, I learned about molarity. As one can tell from the name, molarity is related to the mole. Molarity (M) equals the moles of the solute divided by the volume of the solution in liters. It is one way to measure the concentration of a solution. Molarity was a central part of a lab that I completed over the course of the latter half of the week. In the lab, I measured absorbance of solutions composed of stock solution (water with blue #1 dye) and water, as well as the absorbance of blue Gatorade. I accomplished this with a partner using a device called a Spectrophotometer. As a class, we calculated the concentration of the solutions using the equation M1 x V1 = M2 x V2. I was initially unclear as to how the concentration was calculated. However, as I finished up the lab this weekend I figured out how this was accomplished. Additionally, I was not entirely sure I did some of the post-lab questions correctly. I struggled with the question that asked you to determine the molar concentration of the blue #1 in the sports drink. I know that we covered how to do that using Beer's law in class, but I still do not fully understand Beer's law. From the notes I took in class I was able to calculate the concentration of these substances, but I am unsure if my calculations are correct. This also made me unsure of the subsequent question (which read: "Determine the mass of the blue dye #1 found in 500 mL of the drink.") I understood how to calculate this because we went over it in class of Friday, however because I was unsure of my calculations in the previous question I was unsure my solution for this question was correct. The numbers I got seemed to be larger than the ones that we calculated in class. I also feel I did a insufficient job on the lab. I think my procedure was not in depth enough and in general I did a sloppy job in formatting the entire lab. That is certainly something I need to improve on next time.

Of the concepts that we covered this week I feel as though I have a sufficient understanding of the mole and dimensional analysis. I don't have an as strong understanding of Molarity and I think that I need to go over that more. With that being said, this week I got a better understanding of the mole and I have learned to think of it more as a quantity rather than a unit with mass.