130+ C/P Scorer AAMC MCAT Passage Dissection: Chemistry Question Pack Passage 1 Breakdown

MCAT passages are easily one of the most challenging parts to MCAT prep.

In fact, the AAMC makes sure that MCAT passages are designed to be confusing and challenging.

But, what if we told you that we have a solution that will have you approaching MCAT passages like a top scorer?

One of the quickest and most efficient ways to get good at MCAT passages is learning from and adopting the skills and strategies of a top scorer.

How To Make The Most Out Of This MCAT Passage Dissection

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This paragraph opens by talking about how archaebacteria live in extreme conditions by getting energy from a redox reaction. We see the definition of chemoautotrophs as bacteria that get energy from a redox reaction, so we know that archaebacteria are chemoautotrophs. They also tell us about the bond energy between carbon double bonded to oxygen in CO2 and the carbon-hydrogen bond in CH4, which we know is the amount of energy that would be required to break the bond.

This is a short sentence telling us that Table 1 has some chemicals that are in the same extreme environments as archaebacteria. We will keep in mind that Table 1 shows us the name and formula of some compounds we may not be familiar with.

This tells us that methanogens make methane and that methanogens can be found in the organism that digests cellulose. They also tell us that these organisms can be found in symbiotic relationships, which means a biological relationship where at least one of the organisms benefits. Then it says that using an isotope carbon-14, they could find methane in the earth’s crust. Archaebacteria may have been in the earth’s crust to make this methane. Overall, this passage seems to only give background information surrounding Table 1. Onto the questions.

The first thing we should do is summarize the question into just the most important parts. This deletes the superfluous information that can become confusing the more we reread the question.

The next thing we should do is predict the answer. Predicting the answer means having at least a vague idea of what should or should not be in the question. This helps us target our focus when reading the answers, so the right answer will stick out more. 

The last thing that we do is reword the answers in the same format as how we worded our question or prediction. For instance, if our prediction is something that increases current and the answers are something like decreasing the resistance, using a thinner wire, increasing the temperature of the circuit, or adding another resistor, we would have to translate answers into what effect they have on the current. It would be decreasing current for all of the above except for decreasing resistance. 

When considering answers for any of the science sections, we have 3 things to check off our list of requirements: is it physically possible, is it true, and does it answer the question. Oftentimes, there are 2 answers that are both physically possible and true, but only one answers the question. When we get down to 2, we have to consider the differences between them and refer back to the question to see which one fits best.

The more we practice this, the better we will get at developing that fast “instinct” or “gut feeling.” Soon, the answer will stick out sooner and you’ll be able to answer quickly and accurately. 

OK so let’s move on to the questions!

Translating this question, it is asking which pair in the answers can form the most hydrogen bonds with each other. We can use Table 1 to look at the molecular formulas. We must remember that hydrogen bonding comes from hydrogen bonded to a partially negative electronegative atom (oxygen and nitrogen are common ones) interacting with a lone pair from another partially negative electronegative atom. The difference in electronegativity between the hydrogen and the electronegative atom causes a partial positive charge on hydrogen and a partial negative on the electronegative atom. We will be looking for hydrogen and electronegative atoms in each of the pairs.

A: Methane is CH4, so this would not fit our requirements of an electronegative atom. Even though methanol fits our requirements, we must eliminate this answer.

B: We established CH4 is not able to make hydrogen bonds, so eliminate it.

C: Glycine has an amine and carboxylic group, and those two fit our requirements for hydrogen bonding. Methanol also has a hydroxyl group that fits our requirements as well. This answer would work.

D: Though methanol is able to make hydrogen bonds, carbon dioxide does not have the net dipole required for hydrogen bonding. Because CO2 is linear, the equal dipoles from the oxygen pull the electrons in opposite directions, canceling each other out. This makes carbon dioxide nonpolar and unable to have a hydrogen bond. This would be eliminated.

This question asks about the pH of 0.01 M HCl(aq) water. HCl is a strong acid, so all of it is expected to dissociate into H+ ions. This means there are 0.01 M H+ ions in the water. We should remember the equation for pH from the concentration of hydrogen ions. The equation is pH = -log[H+]. This would give us pH = -log[0.01]. Since we do not have a calculator, we have to use a shortcut to solve a logarithmic equation. We can do –log(m x 10-n) ≈ n – 0.m. Since 0.01 is 1 x 10-2, this would mean pH ≈ 2 - 0.1 ≈1.9. Our closest answer is pH 2, answer C. We could also double-check this using content knowledge that HCl solutions stabilize at around pH 2.

A: For this question, in particular, we do not even have to do the math. If we remember that HCl is a strong acid, we can eliminate pH 12 because this is extremely basic.

B: pH 6 is only slightly acidic compared to neutral pH 7. We can eliminate this because HCl is a strong acid and will be strongly acidic in water.

C: This is our answer from the elimination of all the others and our equation.

D: This can be eliminated because we know that pH does not equal the concentration of H+ ions. We had to take the negative log of the concentration.

This question asks for the geometry of methane and another molecule with the same geometry. Geometry refers to the positions of the atoms, not the lone pairs. For this, we can look at Table 1 to see the molecular formula if we don’t remember. Methane is CH4, which means it is tetrahedral with carbon having 4 covalent bonds to 4 hydrogens and 0 lone pairs. We will look for another molecule with 4 molecules bonded to 1 central atom and 0 lone pairs.

A: This only has 2 molecules on 1 central atom, making it bent from the 2 lone pairs.

B: This also has 2 molecules on 1 central atom, making this linear.

C: This has 4 molecules on 1 central atom, but Xe has 2 lone pairs. We can draw the lewis structure if we need to. We would need to remember that Xe is a noble gas and has a full set of 8 valence electrons. 4 of them are bonded to fluorine, one electron per fluorine covalent bond. Xe can also break the octet rule because it is past the 2nd row of the periodic table. The 2 lone pairs make this geometry a square planar.

D: This has 4 molecules on 1 central atom and 0 lone pairs. This geometry would also be tetrahedral, so this is our answer.

This question asks which element can act as an electron acceptor. We would want to go by electronegativity. The larger electronegativity, the more the atom would want to accept the electrons. The electronegativity trend increases left to right and down to up except for the noble gases because they already have a full shell of valence electrons. We will look for an element closest to the top right corner.

A: Sulfur is rather close to the top right corner and is often seen as S2-, similar to oxygen in the row above with the same number of valence electrons. This is our answer.

B: Though helium is in the top right corner, it is a noble gas. Helium has a full outer shell of 2 electrons, filling the 1s orbital. It would not want to accept more electrons.

C: Hydrogen gas with 2 atoms already has a full shell of 2 electrons as well. This would also be eliminated.

D: Iron and the rest of the transition metals would much rather lose electrons to become a positively charged metal cation than take electrons. Iron would commonly become Fe2+ or Fe3+. This would also be eliminated.

The only thing we needed to read from this whole passage was Table 1. If we knew the structures of the compounds in question 1, we would not even have to look at the passage at all. This is why I would go straight to the questions. It saves time and stamina by minimizing how much I have to read.