The FRQs discussed in this episode are 2008 - Question 2 and 5 as well as 2019 question 5. These are released FRQs from previous exams and copyright of the College Board.
The FRQs discussed in this episode are 2008 - Question 2 and 5 as well as 2019 question 5. These are released FRQs from previous exams and copyright of the College Board (0:22).
Question 2 of the 2008 focuses on the experimental approaches, first the determination of the formula of a hydrate (2:06) and then gravimetric analysis (3:33). Question 5 of the 2008 exam starts with questions about ionization energies for fluorine in comparison with oxygen and xenon. The second part of the question asks for Lewis Diagrams of xenon compounds (7:59), the geometric shape (8:25) as well as hybridization (8:35) and polarity (8:55).
Question 5 of the 2019 exam provides a PES spectrum, which you will use to determine the electron configuration and identity of the element (9:08) and asks you to calculate the wavelength needed to remove an electron from the valence shell (10:01).
True or false: I have to answer the FRQ Questions from 1 to 7 as well as a through last letter in order.
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Unit 1 and 2 Free-response questions.
The FRQ questions on the AP exam often combine content from two or three different units. In today’s episode, we are focusing on questions that require knowledge and skills from Unit 1: Atomic Structure and Properties as well as Unit 2: Molecular and Ionic Compound Properties and Structure. The questions we are using are online accessible. Our suggestion: Answer the questions yourself and then listen to this episode to hear the explanations, as well as do’s and don’ts for answering questions of Unit 1 and 2. The questions used are: 2008 - Question 2 and 5 as well as 2019 question 5. These are released FRQs from previous exams and copyright of the College Board.
Lets Zoom in:
Let’s start with 2008, Question 2. This long-response question focuses on Unit 1, specifically the formula of a hydrate, gravimetric analysis, the experimental approach as well as relevant calculations. In part a you are prompted to explain why the student can correctly conclude that the hydrate was heated sufficiently. We can see in the table that there were three heatings. The difference in mass between the second and third heating is 0.0001g. This indicates that no additional water was driven off during the third heating and that all the water has been released.
Using the mass of the sample and container after the third heating and subtracting the initial mass of the sample and container, we can calculate that 1.848g of water has been lost. Using the molar mass of water, 18.02 g/mol, we calculate the number of moles of water lost to be 0.1026 moles, with four significant figures. In part ii) we are asked to determine the formula of the hydrate. The formula of the hydrate is the ratio of moles of anhydrous salt and water. We already know the moles of water. To determine the moles of anhydrous salt, we determine the grams of anhydrous salt by subtracting the mass of the container from the mass of the third heating, because after all the water has been driven off, what remains in the container is the anhydrous salt. This is 1.630g. We know the identity of the salt, magnesium chloride, and can therefore use the molar mass of 95.20g/mol to convert the grams to 0.01712 moles of magnesium chloride. To get the mole ratio, we divide both moles, the salt and the water, by the smaller number of moles, in our case 0.01712. This gives us a ratio of 1:6. The empirical formula of the hydrate is therefore magnesium chloride hexahydrate.
In part c) we are asked to describe and justify the answer if some solid would splatter out of the crucible. This would change the ratio of anhydrous salt to water. It specifically affects the loss of water: you would think you’ve lost more water than you actually did, because your crucible is lighter after the heating than it should be. The question continues with a second experiment, focusing on gravimetric analysis. It asks you to describe a procedure on how you can determine the mass of magnesium chloride in a mixture with potassium nitrate. It indicates that silver nitrate is used to precipitate silver chloride. Listen to our episode 48 on gravimetric analysis, if you need a refresher! For this question, you receive points if your answer includes that after adding silver nitrate and the formation of a precipitate you have to filter the mixture, dry the precipitate and determine the mass by difference.
In part e of the question, we are now looking at the calculations. First, we calculate the number of moles of magnesium chloride in the original mixture. We calculate that 5.48g silver chloride is 0.0382 moles of silver chloride. For each mole of silver chloride, we have 1 mole of Chloride and we need 2 moles of Chloride for each one mole of magnesium chloride. Therefore, we have 0.0191 moles of magnesium chloride. In the second part, we use the molar mass of magnesium chloride to convert the moles to grams. Since we have 1.82 grams of magnesium chloride in the 2.94 gram sample, we have 61.9% magnesium chloride by mass. And that’s a wrap on 2008 - Question 2. This was a long response question.
The second question from 2008 we are looking at is Question number 5. This question combined Unit 1 and Unit 2. Since ionization energy is the removal of an electron, the equation shows fluorine yields fluorine cation - yes, I know, weird - plus 1 electron. In part (b) you are asked to discuss why the first ionization energy of atomic fluorine is greater than the energy for oxygen. In your answer, you have to discuss the greater effective nuclear charge of fluorine compared to oxygen. Both remove the electron from the same sublevel, namely 2p, but fluorine has 9 protons and oxygen has 8 protons. Therefore the force of attraction between the nucleus and the electrons is stronger for fluorine and it takes more energy to remove an electron. Sticking with first ionization energy we are asked to predict the first ionization energy for xenon compared to fluorine. If you predicted that xenon has a lower first ionization energy, you are right. Now let’s look at the justification: When removing an electron from fluorine, we are removing it from the 2p orbital. For xenon it would be the 5p orbital. The 5p orbital is a higher energy level further from the nucleus. According to Coulomb’s Law, the attraction decreases with increasing distance. Therefore it will take less energy to remove it from xenon than from fluorine.
Part (d) now brings us into the realm of Unit 2. The Lewis diagram of Xenon trioxide shows a central xenon with a lone pair and three single bonds to the three oxygen atoms. Each oxygen has three lone pairs. Xenon tetrafluoride shows a central xenon atom with two lone pairs and 4 single bonds to each fluorine. Each fluorine is surrounded by three lone pairs. The geometric shape of Xenon trioxide is trigonal pyramidal, since it has three bonding and one non-bonding domain. The second part of e asks us about hybridization. Since the overhaul of the AP content, you only need to know hybridization for up to 4 domains, sp3. Therefore, you don’t have to know that this is sp3d2, unless you are a time traveller. In the last section of this question, you are asked about the polarity of xenon trioxide, which is a polar molecule, because the bond dipoles do not cancel out. Xenon will have a partially positive charge.
Question number 5 of the 2019 exam gives you a PES spectrum. For these it is always important to take a closer look at the x-axis labels as well as the scaling of the peaks. In this spectrum, we can see that the energy needed to remove the electron decreases from left to right. This indicates that the electrons on the left of the spectrum are the closest to the nucleus. Looking at the scaling, we can see that the first peak on the left reaches the first line on the y-axis. Since this is the 1s2 orbital, it indicates that one horizontal marking is 2 electrons. Now we can write the electron configuration, going from left to right and using the Aufbau Principle: 1s2 2s2 2p6 3s2 3p6 4s2. This element is Calcium.
In the second part of the question we are asked to calculate the wavelength needed to remove an electron from the valence shell. We can see that the energy needed for this is 0.980 x 10-18 J. We can use the equation E = planck's constant x frequency to calculate the frequency. Then we use speed of light = frequency times wavelength to calculate the wavelength. We can also combine these two equations and use wavelength = planck's constant x frequency over Energy. Plugging in your values you get a wavelength of 2.03 x 10-7m. In this case it wasn’t necessary to change any units. However, generally be mindful of the units. Plank’s constant is in J, so the energy has to be in joules. The speed of light is in meters per second, so the calculated value is in meters.
To recap…
For calculations, show all your work and steps. Use Units! As you know from earlier episodes, units are friends! Report your answers in the correct number of significant figures.
For periodic trends, keep in mind that the trend itself is not an explanation or justification. Your explanation of the trend needs to incorporate principles of Atomic Structure, Coulomb’s Law, effective nuclear charge and electron shielding.
For Unit 2 topics, be specific with your vocabulary, e.g. when you write ion, make sure you mean a charged atom or molecule and not an atom and vice-versa. Don’t forget to add lone pairs to your Lewis diagrams and be aware of the difference between Lewis Diagram, which is a 2D representation and molecular geometry, which is a 3D representation. Just because it is drawn as a diagram in 2D, doesn’t make it that shape in 3D!
Coming up next on the Apsolute RecAP Chemistry Edition: Types of solids and their properties.
To day’s Question of the day is about FRQ Questions.
True or false: I have to answer the FRQ Questions from 1 to 7 as well as through letter in order.