
Chemistry 1020Lecture 8—Notes The MOLE One can draw some reasonable inferences from the table of relative atomic masses without needing to know the actual mass of an atom. For example, if an atom of C is 12 times as heavy as an atom of H, then 1000 atoms of C would be 12 times as heavy as 1000 atoms of H, and X atoms of C would be 12 time as heavy as X atoms of H. It follows that12 pounds of C should contain the same number of atoms as one pound of H, or that 12 grams of C should contain the same number of atoms as one gram of H. If we say that 12 grams of C contains N atoms of C, then one gram of H contains N atoms of H, and 16 grams of O contains N atoms of O, and 35.5 grams of Cl contains N atoms of Cl, and so on for all the elements. Furthermore, 18.0 grams of water (H2O) would contain N molecules of water. This realization leads to the definition of the term gram molecular weight (called mole for short, and abbreviated mol) which is the weight of a substance, in grams, equal to the atomic or molecular weight of the substance. (An earlier distinction between gram atomic weight when referring to elements, and gram molecular weight when referring to compounds or molecular forms of of a substance.) A mole is a quantity like a dozen is a quantity. It represents a count of something. While a dozen means twelve of anything, a mole means N of anything, where N is a definite number and is known as Avogadro’s Number. We do not need to know the value of N for this concept to be useful. For example, in determining the relative masses of substances reacting in a chemical equation, we can note that we can interpret a chemical equation in a second way. 2 H_{2} + O_{2} > 2 H_{2}O means both: "two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water" and: "two moles of hydrogen react with one mole of oxygen to produce two moles of water" In the combustion of methane: CH_{4} + 2 O_{2} > CO_{2} + 2 H_{2}O This equation tells you that one mole of methane (16.04 grams) reacts with two moles of oxygen (64.00 grams) to produce one mole of carbon dioxide (44.01 grams) and two moles of water (36.03 grams). It also demonstrates the Law of conservation of mass because: 16.04 grams + 64.00 grams = 44.01 grams + 36.03 grams To predict quantitative relationships like this, you must write the correct balanced equation, and calculate the molecular weights correctly. If you would like practice doing either of these things, then you can do some drill problems on balancing equations and calculating molecular weights. (Since ionic compounds do not contain molecules, we refer to the formula weight of an ionic compound. NaCl would have a formula weight of 22.99 + 35.45 = 58.44, and 58.44 grams of NaCl would contain N atoms of Na and N atoms of Cl). With this definition of the mole, we can give another unit besides atomic mass unit (amu) to atomic and molecular weights. This other unit is grams per mole (or gmol^{1}). Therefore, we can say the molecular weight of ammonia (NH_{3}) is 17.03 amu, or 17.03 grams/mole. (For most of the purposes of this class, unless you are specifically told to do otherwise, it is sufficient to carry atomic and molecular weight calculations to the nearest 0.1 amu). One can use simple ratio and proportion arguments to calculate quantities reacting in a chemical equation. Take the process of respiration, which occurs in the human body. Carbohydrates, such as glucose (C_{6}H_{12}O_{6}), are processed through a series of metabolic reactions which result in their overall combustion to CO_{2} and H_{2}O in the following balanced process. Note the numbers under the equations represent the relative amounts of substances reacting: C_{6}H_{12}O_{6} + 6 O_{2} > 6 CO_{2} + 6 H_{2}O one mole 6 moles 6 moles 6 moles 180.2 g 192.0 g 264.1 g 108.1 g How many grams of CO_{2} are produced in one day by an individual who consumes one pound (454 g) of glucose.? This can be solved in several ways: ratio and proportion ? grams CO_{2}/454 g glucose = 264.1 g CO_{2}/180.2 g glucose ? grams CO_{2} = 454 x 264.1/180.2 = 665.3795 g CO_{2} (round to 3 significant figures) dimensional analysis Avogadro’s NumberBut now we do know the value of N. It has the value of 6.02 x 10^{23} particles/mole and we can redefine the mole as 6.02 x 10^{23} particles of anything, just as a dozen means twelve of anything. How many CO_{2} molecules would be produced in the combustion on 1.00 mole of glucose? What is the mass of one molecule of CO_{2}? Again two ways: ratio and proportion ?g /molecule CO_{2} = 44.0 g/6.02 x 10^{23} molecules CO_{2} ? g = 7.31 x 10^{23} g. dimensional analysis So if we look at figure 3.8 and see that photosynthesis transfers 110 billion metric tons of carbon per year from CO_{2} to organic matter (equivalent to carbohydrate as glucose), you should easily be able to translate that into amount of CO_{2} utilized and amount of glucose produced: 110 bmt C x 44.0 bmt CO_{2}/12 bmt C = 403 bmt CO_{2} 110 bmt C x 180 bmt glucose/12 bmt C = 1.65 x 10^{3} bmt glucose One metric ton is 1000 kg or 10^{6} grams. How many molecules of CO_{2} is consumed annually by photosynthesis? 403 billion metric tons = 403 x 10^{9} x 10^{6} grams One can imagine many types of calculations like these—most involving calculating a molar mass at one stage or another. For some drill problems interconverting moles and masses of compounds, check the links on the interactive drill web page. Wrap up The last part of the chapter gives more information concerning the debate about CO_{2} production, other greenhouse gases, and global warming. The issue is by no means completely settled. To provide more information to your background reading, I have provided some links to several web pages which give more extensive data on the issue. You may well be called on in the future to vote on questions related to how much we should invest in trying to head off this issue. What additional information would you like to know beyond the data presented in this chapter an in these web pages? How might you go about getting this information in order to make an informed decision on your vote? Remember: Test 1 on Thursday. Help Session this evening at 5 pm. 
