Chemistry Question Pack: Passage 1

1) This question asks about the possible bonds formed by each of the molecules in the answer choices. Additionally, going through our choices, we see four distinct molecules: Methane, glycine, carbon dioxide, methanol. We have to determine which molecules can form extensive networks of intermolecular hydrogen bonds with both participating. What do we know about hydrogen bonds? Hydrogen bonds are formed when a hydrogen atom bonds with a very electronegative atom. In a hydrogen bond, hydrogen atoms gain a partial positive charge, and the electronegative atom gains a partial negative charge. Let’s also pull up and reference Table 1.

We can also draw our four options here.

Carbon dioxide does not have any hydrogen atoms, while methane has 4 hydrogen atoms, all bonded to the same carbon. Alternatively, glycine and methanol have hydrogens bonded to different atoms, and are able to accept and donate hydrogens, so an extensive network of hydrogen bonds can be formed.

1. Methane and methanol. We already went through and mentioned that while methanol is a viable option, methane does not form a hydrogen bond between carbon and hydrogen. On the MCAT we always look for our FON (fluorine, oxygen, nitrogen) atoms when we’re looking at hydrogen bonding. Let’s try and find a better answer. 
2. Methane and glycine. Similar to answer choice A, this answer also contains methane. Glycine is a viable option, however, so this answer is only half correct, just like answer choice A.
3. Glycine and methanol. This answer choice supports what I said in the breakdown: glycine and methanol have hydrogens bonded to different atoms, and are able to accept and donate hydrogens, so an extensive network of hydrogen bonds can be formed. Answer choice C is now our best option.
4. Methanol and carbon dioxide. While we like methanol as a viable option, carbon dioxide doesn’t have any hydrogen atoms, and therefore we don’t expect hydrogen bonding. We can eliminate answer choice D as well. We’re left with our correct answer, answer choice C. 

2) HCl is one of the common strong acids, and this makes sense because archaebacteria thrive in extreme pH. We’re given the molarity of a pool and asked to solve for pH. After reading the question, you can tell right away that it is almost like a standalone question. It can be solved without using the passage. The only information needed to solve for pH is the molarity provided, and our general knowledge. 

pH can be solved as the negative log of hydrogen ion concentration. Because we know our HCl is going to dissociate completely, the hydrogen ion concentration ends up being the same as our molarity. Our molarity is 0.01 molar which can be rewritten using scientific notation. (If you don’t remember which way to move the decimal point, go back to our mnemonic LARS, which is Left add, right subtract). We start with 0.01 x 10^0. We move the decimal point two places to the right to get a whole number, meaning we subtract two from our exponent to give us 1.0 x 10^-2 molar.

Let’s plug in our numbers. We know ph is the negative log of our hydrogen ion concentration. Our hydrogen ion concentration, again, is the same as our molarity. So plug in 1x 10^-2 as our hydrogen ion concentration. So when we take our negative log, we find the pH is 2.

For the MCAT, pH is measured as the negative log of hydrogen ion concentration. We’re going to see values between 0 and 14. We know acids will have a pH below 7 on the scale, and bases will have a pH above 7 on the scale. So after solving for our pH of 2, we can go back to our answer choices to see which one of them make sense. Remember we said pH below 7 means an acid, a pH above 7 means a base.

1. 12. If you did the calculation and got a pH of 12, that means you incorrectly took the negative log of the hydroxide ion concentration, instead of the negative log of hydrogen ion concentration
2. 6. This represents a weak acid. Answer choice B is still better than answer choice A which is actually a base. 
3. 2. This matches our breakdown of the question and the calculated pH. 
4. 0.01. While this represents a very acidic pH, it does not match the calculated pH the way answer choice C does. We’re left with answer choice C as our best answer.

3) This is going to be similar to our last question. It’s tangentially related to what we read in the passage, but the passage isn’t completely necessary to answer the question. This is like a standalone question that most likely can be answered without the use of the passage. First and foremost, to be able to identify the compound with the same geometry as methane, we have to know the geometry of methane itself, then compare with the additional answer choices.

Methane has a central carbon atom bonded to four hydrogen atoms. The four hydrogens are organized as far apart as possible from one another at around 109-degree bond angles. 

We use VSEPR theory and know the atoms in the molecules achieve a geometry that minimizes repulsion between the electrons in the valence shell of the atom. We know methane has a tetrahedral molecular geometry, because it has four electron dense areas, and no lone pairs.

Keep in mind we used VSEPR theory to determine methane has a tetrahedral molecular geometry. It is imperative to know the concepts on the MCAT content outline, and this is no different. When given the number of electron dense areas and lone pairs, you should be able to determine the molecular geometry of a compound. 

1. H₂S. Answer choice A is hydrogen sulfide-Hydrogen sulfide has four electron dense areas, and has two lone pairs of electrons. That gives it a bent molecular geometry
2. CO₂ Carbon dioxide has two electron dense areas and no lone pairs. So that gives it a linear molecular geometry
3. XeF₄ Xenon tetrafluoride has six electron dense areas, with two lone pairs. Giving it a square planar molecular geometry
4. SiCl₄ Silicon tetrachloride has four electron-dense areas, with no lone pairs. It has a central silicon atom that is bonded to four chlorine atoms-it has the same molecular geometry as methane. It has a tetrahedral molecular geometry. Answer choice D is our correct answer. 

4) The question is ultimately asking us “which answer choice has similar chemical properties to oxygen?” Elements with similar chemical properties to oxygen will most likely accept electrons in a manner similar to oxygen. Elements that are found in the same group on the periodic table tend to have similar characteristics and chemical properties.

We can consult with the periodic table of elements, and it’s unlikely we have to revisit our passage to answer the question.

The best electron acceptors are those which have the highest electron affinity. Electron affinity increases to the right, and up the periodic table.

Looking at our periodic table, we see sulfur is found in the same group as oxygen, meaning it will have similar chemical properties, and also have a high electron affinity compared to our other answer choices. Elements that are found in the same group on the periodic table tend to have similar chemical properties. 

This periodic table is going to be readily available to you on test day, so make sure you get used to using it as necessary while you are practicing.

1. S We have answer A which is sulfur, which is found in the same group as oxygen and near the top right of the periodic table. That’s where electron affinity is highest. This is a good option right away.
2. He Next we have answer choice B, helium. Helium is a noble gas, and what did we say about noble gases? We said they’re unreactive and have very low electron affinity
3. H₂ Next we have hydrogen gas which is already a stable molecule, so it’s not going to want to accept any additional electrons
4. Fe Answer choice D, we have iron. Iron is more likely to donate electrons than be an electron acceptor. That should make sulfur the best answer choice. We can stick with answer choice A with our best option.

 

Chemistry Question Pack: Passage 2

5) This question is asking us to decide whether limonene or positive carvone will distill first. Why did I phrase it this way? Because even though the test-maker doesn’t mention the two components by name, we recall the components of caraway seed oil. 

Here we have our two structures from the passage. Like I mentione, we’ll look at the structure of both carvone and limonene to compare. Carvone has a hexane ring, just like limonene. The biggest difference is the additional carbonyl group, which I’ve circled in red. How does that affect the properties of the two compounds, and finally the boiling point? We’ll have to rely a bit on our content. Oxygen is more polar than carbon, which makes carvone more polar than limonene. More polar compounds have higher boiling points. Said differently, limonene has a lower boiling point than carvone. The additional double bond is also an indicator that carvone has a higher boiling point. Now what are we doing with this information? 

Let’s discuss the experiment as well. During distillation, temperature is slowly increased until one of the components begins to vaporize out of the solution. This vapor is then condensed into a different container. The component that vaporizes first will do so at a lower temperature than the second. Therefore, the component that has the lower boiling point (limonene) will vaporize first. So before even looking at our answer choices, we have a predicted answer, with some solid reasoning behind the prediction. We can go back to our answer choices and see if this matches any of our answers.  

1. Limonene, because it has a lower boiling point. Right away, this sounds like a good answer choice for two reasons. First of all, we said the component with the lower boiling point would distill first. Even if we didn’t know if this were limonene or positive carvone, we would still like this answer choice because the reasoning is sound. But the second reason we like this answer choice, is because it matches that prediction we came up with. We said limonene has a lower boiling point and will vaporize first. We like answer choice A. 
2. Limonene, because it has a higher boiling point. This answer choice also lists limonene, but does the reasoning work here? We said because of the carbonyl group on carvone, it has the higher boiling point. Even though the specific component matches our breakdown of the question, the reasoning does not. This is going to be an incorrect answer choice, and answer choice A is superior.
3. (+)-Carvone, because it has a lower boiling point. This would be an attractive answer choice, if it were factually correct. Positive carvone has that carbonyl group, and it has a higher boiling point. The component with the lower boiling point is going to distill first, but that’s limonene, not carvone. This is also an incorrect answer.
4. (+)-Carvone, because it has a higher boiling point. The reasoning here is factually correct. Carvone does have a higher boiling point. But we’re looking for the component that will most likely distill first. That’s limonene because of the lower boiling point. That makes this another incorrect answer. We’re left with our correct answer, answer choice A: Limonene, because it has a lower boiling point.

6) Here we have an excerpt from our passage, as well as the figure. When I pull up these excerpts from the passage, it’s not so you go back and re-read every single detail, but rather so I can demonstrate my thought process. I show you exactly what I’m referencing from the passage so you can follow along at the same time. When you’re actually practicing, and when you’re taking the exam, I don’t recommend going back to the passage unless you’re looking for details in the form of data or figures. 

The passage says an ebulliator was lowered into the distillation flask to introduce small air bubbles into the system. The author gives us the explicit function of the ebulliator. That ebulliator creates an area of imperfections for gases to collect and start boiling. So essentially it’s a boiling chip. In perfectly clean, smooth glassware there’s nowhere for gas bubbles to start forming at boiling temperatures. 

Without the ebulliator, you might super-heat your liquid, or there may be bumping or splashing. The ebulliator is also helping the solution begin boiling at the correct temperature.

1. keep the condensed vapors cool in the receiving flask. Does this answer choice sound like what we broke down about the ebulliator? We didn’t mention keeping the vapors cool in the receiving flask. But that’s not always a reason to get rid of an answer. There’s a very small chance you’ll be able to predict what every correct answer will be on an exam, so let’s dive a bit deeper. The author says the ebulliator was lowered into the distillation flask. That’s near the bottom left of our figure. Is this ebulliator going to be affecting the vapors in the receiving flask? That receiving flask is on the other side of the apparatus, so not likely. Let’s keep going through our answer choices and we’ll see if we can find a superior answer choice.
2. promote the establishment of a high vacuum in a system. If the ebulliator did this, that could certainly be helpful. Remember we mentioned using a vacuum allows the experiment to work at a lower pressure and for the compounds to boil at a lower temperature. But the ebulliator isn’t the one establishing this vacuum-instead it’s the vacuum source on the right side of the apparatus. Just like answer choice A, not a great option, so let’s keep comparing.
3. prevent superheating of the liquid to be distilled. This answer choice matches something we said in our readthrough. For those of you that may have taken organic chemistry lab, or any general science labs, think of a boiling chip – the entire purpose of the chip is to stabilize the boiling and prevent the solution from bumping / splashing everywhere. When it comes down to it, the ebulliator is just a chip that you throw into your solution to help it boil. We explicitly mentioned it prevents superheating during our breakdown, but what if you didn’t make that connection? This answer choice is still superior to answer choices A and B, because it directly relates to the ebulliators function and location in the experiment. We finally have a superior answer, so we can cross off answer choices A and B.
4. provide an outlet when the pressure inside the system becomes too high. This is essentially saying that chip allows for ventilation, or even the same function as the vacuum. This is going to be similar reasoning to answer choices A and B. The author tells us the function of the chip and we tied that function to the figure in the passage. The ebulliator isn’t functioning to provide ventilation when pressure is too high. This is another incorrect answer. We’re left with our correct answer, answer choice c: prevent superheating of the liquid to be distilled.

7) We know limonene and positive carvone are separated based on their boiling points, but we can revisit the passage to see what the author says specifically about the purpose of the experiment. 

Here we have an excerpt from the passage. It says Because the two compounds have very different boiling points, the chemist decided to separate them by vacuum fractional distillation. The entire experiment revolves around this boiling point difference, and separating the two components of caraway seed oil by vacuum fractional distillation.

We’re looking for the answer choice that would most likely improve the separation between the limonene and positive carvone. Let’s go through our answer choices and see which would most improve the experiment.

1. Heating the distillation flask at a slower rate. What would be the effect of heating the distillation flask at a slower rate? The experiment would run the exact same way, only the temperature of the caraway seed oil would raise more slowly. If that happens, could that improve the separation between the limonene and positive carvone? Of course! The entire purpose of the experiment is to separate the components of caraway seed oil by taking advantage of boiling point differences. 
2. Using a vacuum source that can achieve a lower pressure inside the distillation apparatus. Using a vacuum source to lower pressure will affect the boiling point of both limonene and carvone. We said during our readthrough, that lower pressure as a result of the vacuum source will also cause lower boiling points. Although both boiling points are lower, they will be relatively closer together than when they were at higher pressures. This will make less separation, not more. That means A remains superior.
3. Cooling the condenser with ice water. Cooling the condenser with ice happens after separation, and not during the separation portion of the experiment. This modification would not improve the separation more than answer choice A. We can eliminate this answer choice.
4. Using a shorter fractionating column. Shortening the column will actually make the separation less. We can visualize this. There’s less room for the vaporization to occur, so we don’t improve the separation between limonene and carvone. That means we’re left with answer choice A, which takes advantage of our boiling point differences. And we can improve the experiment by Heating the distillation flask at a slower rate. 

8) To answer this question, we’re going to look at our answer choices and see which carbons we want to look at specifically. We know right away by glancing at the answer choices that our efforts should be focused on carbons 2, 5 and 7 because those are the only carbons listed in the answers. We have to remember the distinction between our positive and negative carvone enantiomers. The difference is the positive carvone is the enantiomer that rotates plane-polarized light in the positive direction, and the negative carvone is the enantiomer that rotates plane-polarized light in the counterclockwise, or negative direction. 

Here we have carvone with carbons 2, 5, and 7 labeled. We’re told Carvone exists as two different enantiomers in the passage. Enantiomers are non-superimposable mirror images of each other that differ in configuration. Enantiomers have opposite absolute configurations at every chiral carbon. Chiral carbons have 4 distinct substituents. So, with that in mind, let’s dig into the structure of our carvone, so we can find the ways positive and negative carvone differ. The passage mentions the two are enantiomers, meaning they will have opposite configurations at chiral carbons.

Let’s first look at carbons 2, 5 and 7 to see which have a stereocenter and four distinct substituents.

    –  Carbon 2 has a double bond, so not a stereocenter
    –  Carbon 5 has 4 different substituents
    –  Carbon 7 has a double bond, so not a stereocenter

Ultimately, we’re going to look for an answer choice that has carbon 5. 

1. Carbon 2 only We know this is incorrect for two reasons: the orientations differ around carbon 5, and they do not differ around carbon 2. Let’s try and find a better answer choice.
2. Carbon 5 only Answer choice B is consistent with what we’re looking for in our correct answer. That means answer choice A can be eliminated.
3. Carbons 2 and 5 only. Again, we know carbon 2 has a double bond and is not a stereocenter. We already said we are going to eliminate any answer choices that include carbon 2.
4. Carbons 2, 5, and 7 only. We eliminated answer choice C because of carbon 2, so we know we can also eliminate option D here. Even if we didn’t know about carbon 7, we can eliminate this answer choice knowing carbon 2 is not one of our correct options. Also, we actually said carbon 7 has a double bond and is not a stereocenter. We’re left with our correct answer, answer choice B: Carbon 5 only

9) The test-maker is asking the effect of a malfunction in the vacuum distillation in which the vacuum leaks. If the vacuum is not working correctly, it’s not properly lowering the pressure inside our distillation apparatus. We want to know what happens to the boiling points of carvone and limonene at higher pressures. The question is simply asking the correlation between pressure and boiling point. 

Vacuum distillation is distillation performed under reduced pressure, which ultimately decreases the boiling point of compounds. Why would we do this? Because sometimes compounds can have very high boiling points that are hard to reach. These high temperatures could adversely affect the compound itself-we can see denaturing, or the experiment just won’t work. A leak in the apparatus will undo this vacuum effect. That means a higher pressure, and we should see an increase in the boiling point of the compounds.

1. both increase. This answer choice is consistent with what we came up with as we broke down this question. We said a leak in the apparatus will undo the vacuum effect. That means a higher pressure, and an increase in the boiling point of the compounds. We like this answer choice for now, let’s quickly go through the remaining answers to confirm whether this is our best answer.
2. both decrease. We know that’s not true because a properly working vacuum is what decreases boiling points. We broke down and visualized the experiment, and that allowed us to determine this is an incorrect answer choice. Answer choice A remains the superior answer choice.
3. both remain the same. Once again, a leak in the vacuum will alter the boiling points by increasing them. We’re not expecting boiling points to remain the same. That means we’re still sticking with our best answer choice, answer choice A.
4. become more similar. In reality, the boiling points will likely become further apart, rather than become more similar. Why is that? Because there’s a wider range of boiling points available to these compounds. With the vacuum, both boiling points decreased and were closer to one another. With a malfunctioning vacuum, the boiling points would be increase, and would now be further apart. This is also an incorrect answer. The only answer choice that matches what we said during our breakdown of the question is answer choice A: both boiling points would increase.

 

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