Typo or am I missing something?

[Johnny Appleseed]

Which pair of formulas represents the empirical and molecular formulas of uracil, respectively?

A) CHNO and C4H4N2O2

B) C2H2N2O2 and C2H2N2O2

C) C4H4N2O2 and C4H4N2O2

D) C2H2N2O2 and C4H4N2O2

I thought the answer should be C2H2NO and C4H4N2O2..... what am I missing?! I feel like an idiot right now. @Altius Premier Tutor


SOLUTION:

D is the correct answer. Counting the number of hydrogens, nitrogens, and oxygens in the molecule of uracil the molecular formula is C4H4N2O2. Since each of these numbers can be divided by two the empirical formula must be C2H2N2O2 making D the correct answer. Answer A is not correct as the empirical formula is incorrect. Answer B is showing both options as empirical formulas. Answer C gives both options as the molecular formula.

---

Members don't see this ad.

 

[Johnny Appleseed]

BB Section, Practice test 9, Question 47

 

[freak7]

Looks like a typo to me.

 

[Johnny Appleseed]

While I'm at it here is another one I can't seem to figure out...


If 5.0 grams of citric acid (C6H8O7) are produced during the Citric Acid Cycle, how many moles of citric acid are formed?

A) 0.156

B) 0.182

C) 0.026

D) 0.208

I solved for moles and got answer C, but that was incorrect and D was the correct choice. I don't understand the portion I bolded in the solution... I feel like the stem is missing information like "how many moles of Hydrogen are produced" or something like that....

SOLUTION:

To perform this calculation, one must first calculate the molecular weight of citric acid based upon the formula given in the stem. Six carbons at 12.0 g/mol + 8 hydrogens at 1.0 g/mol + 7 oxygens at 16 g/mol = 192 g/mol. The number of moles is equal to: 5g/192.124(g/mol) = 0.026 moles. As is often true on the MCAT, the math may seem too difficult to perform without a calculator, but combining estimation with the use of scientific notation makes it reasonable. One way to do this is to consider 5.0g to be 5.0 x 100 g and divide this by 1.9 x 102 g/mol. 2.0 would divide into 5.0 exactly 2.5 times, so one can estimate 5.0/1.9 is slightly larger, say 2.6. The exponent is obtained by subtracting the exponents = 2.6 x 10-2 grams of citric acid. Per the formula given, there are 8 moles of hydrogen for every one mole of citric acid, so this must be multiplied by 8 to give 20.8 x 10-2 or 0.208, Answer D.

BB Section, Practice test 9, Question 44.

 

[Johnny Appleseed]

Looks like a typo to me.

I'm glad to hear that. Maybe I'm not as big of an idiot as I thought.

 

[aldol16]

To perform this calculation, one must first calculate the molecular weight of citric acid based upon the formula given in the stem. Six carbons at 12.0 g/mol + 8 hydrogens at 1.0 g/mol + 7 oxygens at 16 g/mol = 192 g/mol. The number of moles is equal to: 5g/192.124(g/mol) = 0.026 moles. As is often true on the MCAT, the math may seem too difficult to perform without a calculator, but combining estimation with the use of scientific notation makes it reasonable. One way to do this is to consider 5.0g to be 5.0 x 100 g and divide this by 1.9 x 102 g/mol. 2.0 would divide into 5.0 exactly 2.5 times, so one can estimate 5.0/1.9 is slightly larger, say 2.6. The exponent is obtained by subtracting the exponents = 2.6 x 10-2 grams of citric acid. Per the formula given, there are 8 moles of hydrogen for every one mole of citric acid, so this must be multiplied by 8 to give 20.8 x 10-2 or 0.208, Answer D.

Based on your bolded part alone, you should immediately be able to tell that they mean "hydrogen formed" and not "citric acid formed." It's a classical case of test prep companies trying to do too much with the question - they're trying to deceive you by making you convert grams to moles but in the process of getting that language correct, they accidentally changed the meaning of the entire question.

 

[freak7]

While I'm at it here is another one I can't seem to figure out...


If 5.0 grams of citric acid (C6H8O7) are produced during the Citric Acid Cycle, how many moles of citric acid are formed?

A) 0.156

B) 0.182

C) 0.026

D) 0.208

I solved for moles and got answer C, but that was incorrect and D was the correct choice. I don't understand the portion I bolded in the solution... I feel like the stem is missing information like "how many moles of Hydrogen are produced" or something like that....

SOLUTION:

To perform this calculation, one must first calculate the molecular weight of citric acid based upon the formula given in the stem. Six carbons at 12.0 g/mol + 8 hydrogens at 1.0 g/mol + 7 oxygens at 16 g/mol = 192 g/mol. The number of moles is equal to: 5g/192.124(g/mol) = 0.026 moles. As is often true on the MCAT, the math may seem too difficult to perform without a calculator, but combining estimation with the use of scientific notation makes it reasonable. One way to do this is to consider 5.0g to be 5.0 x 100 g and divide this by 1.9 x 102 g/mol. 2.0 would divide into 5.0 exactly 2.5 times, so one can estimate 5.0/1.9 is slightly larger, say 2.6. The exponent is obtained by subtracting the exponents = 2.6 x 10-2 grams of citric acid. Per the formula given, there are 8 moles of hydrogen for every one mole of citric acid, so this must be multiplied by 8 to give 20.8 x 10-2 or 0.208, Answer D.

BB Section, Practice test 9, Question 44.

Seems like a typo as well. The problem was almost too easy as stated.

 

[Johnny Appleseed]

Thanks guys, I've been stuck on these questions and getting frustrated. Should've just moved on and realized they were errors in the first place.

 

[Altius Premier Tutor]

I've gone over #44 with my students before and it used to ask for the moles of hydrogen. They are reviewing/updating all of the exams and have tried to make calculations a little easier, so I'm guessing they removed the extra step of going from citric acid moles to moles of hydrogen, but the solution didn't get updated properly. The correct answer is 0.026 (C) for moles of citric acid. #47 is definitely a typo, should be C2H2NO and C4H4N2O2 for Answer D.

I'll pass this along, they usually fix requests from tutors pretty fast. Thanks!