GCC CHM 130LL: Mole Relationships page 1 of 9 CHM 130LL: Mole Relationships Introduction Moles A mole, known as the “chemist’s dozen,” relates the submicroscopic world of atoms and molecules to the much larger world of weighable amounts of chemicals. Just like a dozen is always 12 items, a mole is always 6.02 x 1023 items. Atoms and molecules are so small that it takes a very large number of them to have enough chemical to weigh out or use in a reaction. To obtain the weight of a mole of atoms, you simply need to look up the molar mass of the appropriate element on the periodic table. For example, a mole of C atoms has a mass of 12.01 grams while one mole of Cu atoms has a mass of 63.55 grams. One “mole” always contains the same number of atoms, 6.021023 (known as Avogadro’s number). Mole Calculations Most chemical calculations do not require the use of Avogadro’s number, only an understanding of its concept. We can see this by again using the “dozen” analogy. A person does not need to use the number “12” if he is buying something that comes in dozens. For example, if he wanted to buy 3 dozen eggs that cost $3 per dozen, he will need to pay 3$3.00 or $9.00. 3 dozen eggs $3.00dozen eggs $9.00 Likewise, a chemist does not need to use Avogadro’s number to determine the mass of carbon to weigh out a given number of moles in the lab. One simply needs to know the mass of one mole of carbon (12.01 grams), or the molar mass of carbon, 12.01 g Cmole C, to determine the mass of carbon required to get 3.00 moles of C atoms, as shown below: Example 1: Calculate the mass of carbon needed to get 3.00 moles of carbon. C g 36.0C moleC g 12.01C moles 3.00  One can also calculate how many moles of carbon are contained in a certain mass of carbon. This is analogous to a shopper determining how many dozen eggs she can get given $15. eggs dozen 5$3eggs dozen$15  Likewise, if a chemist has 15.0 grams of carbon, the chemist can determine the number of moles of carbon present in the following way: Example 2: How many moles of carbon are present in 15.0 grams of carbon? 15.0 g C mole C12.01 g C 1.25 moles C The examples above use the element carbon, which exists as individual atoms, but compounds generally exist as molecules or ionic compounds. For example, water exists as H2O molecules, while table salt, NaCl, exists as a three-dimensional network of Na+ and Cl– ions. To determine the mass of one mole of a compound, one must multiply theGCC CHM 130LL: Mole Relationships page 2 of 9 molar mass of each element by the number of atoms present for the element, then add all the masses of the elements to get the total molar mass for the compound. For example, one can calculate the molar mass (MM) for Ca(NO3)2 as follows: MM of Ca(NO3)2 = MM of Ca + 2(MM of N) + 6(MM of O) = (40.08 gmol) + 2(14.01 gmol) + 6(16.00 gmol) = 164.10 gmol Ca(NO3)2 Balanced Chemical Equations and Moles Balanced chemical equations give the relationship between the number of molecules in the reactants and products. For example, the balanced equation 3 H2 (g) + N2 (g)  2 NH3 (g) tells us that 3 molecules of H2 react with 1 molecule of N2 to produce 2 molecules of NH3. We can also “read” the balanced equation in terms of moles. That is, 3 moles of H2 react with 1 mole of N2 to produce 2 moles of NH3. From this statement, we can always relate the moles of any reactant or product to the moles of any other reactant or product using a mole-to-mole ratio obtained from the equation. For example, if we want to know how many moles of H2 it takes to react with 2.00 moles of N2, we would simply use the mole ratio between H2 and N2 from the equation above, being sure to set up the unit factor to cancel moles of N2 and leave moles of H2. The problem is set up below: Example: How many moles of H2 are required to react with 2.00 moles of N2? 200 moles N23 moles H21 mole N2 600 moles H2 What if we have more than 6.00 moles of H2 available to react with the 2.00 moles of N2? The extra H2 simply does not react. So if we mix our 2.00 moles of N2 with 10.00 moles of H2, we would have an extra 4.00 moles of H2 left over at the end of the reaction. In this case, chemists refer to H2 as the reactant in excess (leftover), and N2 is the limiting reactant (because it limits the amount of product formed). We could then calculate the number of moles of NH3 formed, as shown below: 200 moles N22 moles NH31 moles N2 400 moles NH3 Now we can combine molar mass and mole calculations to determine the mass of product that can be produced given the mass of reactants used. First we use molar mass to convert the mass for each reactant to moles, then use the appropriate mole-to-mole ratio, then finally we use the molar mass of the product to convert from moles to the mass of the product. Since we solve for the same product, we simply compare the two numbers. Whichever number is smaller is the amount of product made because as soon as that amount of product is made, the reactant is completely used up, and no more product can be made. Example: What mass of ammonia can be produced when 10.0 g of H2 reacts with 10.0 g of N2?GCC CHM 130LL: Mole Relationships page 3 of 9 Procedure 10.0 g N2mole N228.02 g N22 moles NH31 mole N217.04 g NH3 mole NH3 122 g NH3 10.0 g H2mole H22.02 g H22 moles NH33 mole H217.04 g NH3 mole NH3 562 g NH3 In this example, the N2 produced the smaller amount of NH3, 12.2 g, so that is the amount of product formed. Once 12.2 g of NH3 are produced, the N2 is completely used up, so N2 is the limiting reactant. Since there is enough H2 to make 56.2 g of NH3, once 12.2 g of NH3 are produced, there is still some H2 remaining, so H2 is the reactant in excess. In this experiment, you will prepare and mix solutions of lead(II) nitrate and potassium iodide to form a precipitate, PbI2(s). The equation for the reaction is: Pb(NO3)2 (aq) + 2 KI (aq)  PbI2 (s) + 2 KNO3 (aq) You and your partner will obtain a card indicating the mass of each reactant, Pb(NO3)2 and KI, to weigh out. You will calculate the mass of precipitate that should be produced, and thus,