Lecture 2: Counting Atoms and Organizing the Elements

Lecture 2: Counting Atoms and Organizing the Elements Summary This chapter, we introduced the concept of the mole. This is a constant number, also referred to as Avogadro’s number, equal to 6.0221409 × 1023 . The atomic mass units (AMUs) assigned to an element are equal to the mass in grams of a mole of atoms of that element. Just as Avogadro’s number is just a number, the AMU is just a mass, and 1 AMU is equal to 1.67377 × 10−24 grams. Just like the mole, an AMU is another special unit that chemists use to connect the very small world of atoms with the comparatively large world of things we work with. This link between the macro and atomic worlds means that by simply measuring the mass of a substance, we can know how many atoms of that material we have. In class we used the example of the combustion of a candle, otherwise known as the nicely balanced reaction C25 H52 + 38O2 → 25CO2 + 26H2 O. We used this example to emphasize how the concept of the mole gives us a link between the atomic and macroscopic worlds, between grams and atoms. We also used this example to re-emphasize the concepts of limiting reagent and yield. Next, we focused on organizing the periodic table. Unlike in the last chapter, here organization means more than coming up with a naming scheme or a simple categorization of the elements, as Dalton, Lavoisier, and many others had already done. In this lecture what we mean by organization is really pattern recognition. Scientists were starting to see patterns emerging among the elements, and the more elements were discovered, the more effort was put to understanding these patterns. Dmitri Mendeleev developed the precursor to today’s periodic table. His big breakthrough was that he arranged the then-known (63) elements not just by their mass, but also by the repeating (periodic!) patterns he observed in their properties. This led him to leave gaps in the periodic table that would later be filled with newly discovered elements. Because of the periodicity in properties, he could even predict the properties of many as-of-yet undiscovered elements. The rows of your periodic table are called periods, and the columns are called groups. The Main Group Elements are those in periods two and three. There is a general trend from metallic to non-metallic elements along the periods. The Transition Elements are in groups 3-12 and periods 4 and 5. Why this matters The population of humans on this planet has seen a dramatic, exponential growth. Let’s zoom in on where the growth really started to kick into high gear. See that uptick, around the early 1900’s? Notice the massive change in slope. That’s when we figured out how to make nitrogen in a way that plants can use. In other words, with the discovery that is now known as the Haber-Bosch process (named after the German 9 People vs. time plot © Population Reference Bureau. All rights reserved. chemists Fritz Haber and Carl Bosch), the world was able to make fertilizer in the abundance needed to feed billions. The key comes down to something called “fixing” nitrogen. Plants need nitrogen to grow, but most plants cannot use any of that plentiful 78% N2 that’s in the air. Instead, they need it in a form that serves up single N atoms, like ammonia (NH3 ). SO how can we go from N2 to NH3 ? Easily, with the following chemical reaction: N2 + H2 → N H3 Ah, sorry, let’s balance that: N2 + 3H2 → 2N H3 That’s better. What Haber and Bosch actually did was develop a catalyst and process that allowed this reaction to be carried out much more efficiently than previously. But catalysis is getting a little ahead of ourselves. For now, let’s focus on the balanced reaction and count some atoms. In 2019, 500 million tons of nitrogen is produced for fertilizer each year using the Haber-Bosch process. Half of the protein in all human beings on this planet comes from nitrogen that was fixed by this process! So here’s my question: how long can we continue fixing nitrogen this way? And since the reaction tells us we need both N2 and H2 , which one would be the limiting reagent? From all of our discussion so far in this lecture we know that we can use the periodic table to determine that 1 mole of NH3 is 17 grams. Just for simplicity, suppose in the future we continue needing to make the same amount we’re making today, namely 500 million tons of NH3 per year. There are roughly a million grams per ton, so 500 million tons per year 500 × 106 × 106 grams per year. Dividing by 17 grams per mole, this means we need to make roughly 30 × 1012 moles of NH3 /year. Given the balanced reaction above, this means we’ll need as input to the reaction 15 × 1012 moles of N2 per year. Now, the mass of the atmosphere is about 5.15 ∗ 1021 grams, of which 78% is N2 , so that makes about 4 × 1021 grams of N2 in the air. And since the atomic mass for a nitrogen atom is 7, then it’s 14 for N2 which means there are 14 grams per mole of N2 . So the number of moles available in the atmosphere is 4 × 1021 grams of N2 / 14 grams/mole = 3.5 × 1020 moles of N2 in the atmosphere. Well that’s good news! Taking 3.5 × 1020 moles of N2 total and dividing that by 15 × 1012 moles of N2 needed per year, we find that we could keep at it with the Haber-Bosch process of fixing nitrogen from the atmosphere for more than 20 million years or so before using it all up. That’s just cool. Think about this: we just went from atom to macroscopic world and it’s all because we have this concept of the mole. But we never really answered the question, namely which one is the limiting reagent in the reaction, the N2 or the H2 ? We now know how long N2 will last us if we were to use it all up in this reaction, so what’s left is to do the same thing for H2 . Speaking of which, where will we even get H2 ? Certainly not from the air. I’ll let you all work through this but here’s a hint: the oceans contain a whole lot of water, like more than 1018 tons. Maybe you don’t even need to do any calculations to figure this out! Why this employs How does counting or organizing atoms lead to a job? Well, for one thing we’re talking about order, precision, and measurement. After all, Mendeleev’s success relied on combining careful measurement of properties with precise ordering. This skill is needed for many different jobs. One place with measurement as its mission is the National Institute of Standards and Technology (NIST). Check out the tag line in their Materials Measurement Laboratory: “MML serves as the national reference laboratory for measurements in the chemical, biological and material sciences. Activities range from fundamental and applied research to the development and dissemination of certified reference materials and data to assure the quality of measurement results.” And that’s just one of their many divisions. Another job that comes to mind is one related to the fertilizer discussed in our Why This Matters. The three fundamental constituents of commercial fertilizers used throughout the world are nitrogen, phosphorus, and potassium. As we know, nitrogen requires fixing. Potassium must be obtained in other ways. For example, it comes nicely with chlorine as potassium chloride (KCl), but we’d prefer not to have chlorine in the soil since it can lead to toxicity in the crops and high acidity and salinity in the soil. However, getting potassium in other forms is expensive and requires, as you might guess, chemistry. To make matters more complicated, both the natural sources for potassium and the specific soil needs are highly dependent on geography. For farmers in tropical regions in Brazil and some countries in Africa, the soil and rock compositions require fertilizers that better match the chemical needs of plants than are currently available. As a solution, more fertilizer is added to achieve as much crop yield as counterparts in the north, but many of the basic ingredients, especially the potassium, are produced in more northern regions and therefore are much more expensive in the Southern Hemisphere. The need for local, sustainable fertilizing solutions to increase yields is urgent! But this is starting to sound more like a second edition of Why This Matters and less like a Why This Employs. My point is that the development of fertilizers is a problem of chemistry, and the world is in desperate need of new solutions. In terms of jobs, this could mean becoming something called an “Analytical Chemist” which means you apply the principles of chemistry to figure out what stuff is made of, what the precise compositions are, how to monitor it, modify it, etc. This could also mean working at a company or research lab that develops cutting edge ideas (like recent work on using electrochemistry to make potassium). Extra practice 1. Solar cells sometimes contain a titanium (Ti) coating that stops the cell from short-circuiting. The Kroll Process for making titanium metal out of titanium oxide is: T iCl4 + M g → M gCl2 + T i a) Balance the reaction. b) If you have 50g of both TiCl4 and Mg, what is the limiting reagent in the reaction? c) What is the yield of the reaction in grams? d) Plot the yield of Ti (g) as a function of increasing Mg available for reaction. Solutions 1. a) T iCl4 + 2M g− > 2M gCl2 + T i b) Molar mass of Mg: 24.3 g/mol Molar mass of TiCl4 : 189.9 g/mol Find the moles of each reactant by dividing 50g by the individual molar masses: 50g/24.3 = 2 mol M g 50g/189.9 = 0.3 mol T iCl4 Use the mole ratio from the balanced reaction (1 mol TiCl4 /2 mol Mg) to see how much TiCl4 could be used up with the amount of Mg you have available: (2mol M g available)/(2mol M g/1 mol T iCl4 ) = 1 mol T iCl4 1 mol TiCl4 would be used up if we had an endless supply of it with the amount of Mg we have. This amount of TiCl4 is more than the amount we actually have available, meaning that the TiCl4 will run out before the Mg. TiCl4 is therefore the limiting reagent. c) To calculate the Ti yield of the reaction, we use the ratio of Ti to TiCl4 (because these two have a 1:1 molar ratio) and then convert from moles to grams of Ti: 0.3 moles of TiCl4 are being used (it’s the limiting reagent, so it will be used in its entirety) and because of the 1:1 molar ratio we know that 0.3 moles of Ti will be produced. Multiply this number by the molar mass of Ti: 0.3 ∗ 47.9g/mol = 14.37g T i d) The plot of Ti grams (on the y axis) vs. Mg grams (on the x axis) starts at 0 and increases linearly until the Ti grams reach 14.37 g, at which point we run out of TiCl4 and the reaction will stop.