CHEMISTRY Fall 2003 CHAPTER 3

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1 EMISTRY Fall 2003 APTER 3 Formulas, Equations and moles Working with the masses used in chemical reactions and relating them to the individual atom and molecules involved is not quite as easy as might be expected because atoms are very small and cannot be weighed using the usual laboratory balance. The atomic weights listed in the text book are not really weights but rather relative masses. At first they were measured relative to hydrogen = 1, then to oxygen = 16 and now to carbon-12 = 12 exactly. Since these are relative masses they have no units, however you will see atomic mass quoted as atomic mass units, a.m.u. or more properly simply u. 1 u = 1 12 the mass of a carbon-12 atom. Then why is carbon,, not listed as 12 but as ? Most elements occur naturally as a mixture of isotopes: so the observed relative mass is a weighted average of all three isotopes. For oxygen, O, the relative mass (atomic weight) is so the atomic mass is u (a.m.u.) atomic mass units. We could compare the mass of 12 6 and O in a sample of 12 6 O we get atomicmassof atomic mass of O = mass of 1 atom of =0.75 ratio 3:4 mass of 1 average atom of O At this stage we cannot know the mass of 1 atom but we can define the mass of a collective we call the mole (equivalent to saying 1 dozen = 12). 1 mole = number of atoms in exactly 12 g of carbon-12, aving said this, what is the relative mass of O? The molar mass of an element = the mass in grams of 1 mole of the element = the relative atomic mass in grams, so 1 mole has mass = g (natural) 1 mole O has mass = g (natural) Today the isotopic masses and abundance of the isotopes present in the natural elements can be measured very accurately using a mass spectrometer this device is used to compare the masses of isotopes (see page 103 fig. 3.10). The isotopic abundance of the isotopes of carbon in natural carbon are: Isotope Isotopic Abundance Isotopic mass % 12 exactly u % u 14 6 trace u 3.1

2 Natural fluorine, F, contains only one isotope and is found to have a mass that is times the mass of 12 6 so the relative mass of F = x 12 u = u exactly. If we look at the result for natural chlorine Isotopic Mass Abundance chlorine l u % 17 chlorine l u 24.47% 17 So, if we have 100 average atoms 35 l = u x = u 37 l = u x = u mass 100 atoms u so for one average atom the relative mass = u Now we can do the same for neon: Isotopic abundance 20 Ne Ne u 90.92% Ne Ne u 0.26% Ne Ne u 8.82% So 100 atoms 20.0 x = 1, x 0.26 = x 8.82 = for u â avg u 3.2

Mass Spectrometer (measures relative masses) The relative mass of a molecule or a formula unit is simply the sum of the relative masses of the constituent atoms.

3 Mass Spectrometer (measures relative masses) The relative mass of a molecule or a formula unit is simply the sum of the relative masses of the constituent atoms. So, the relative mass of the molecules of carbon dioxide, O 2 is x O 2 x O 2 total So how many is a mole? (like there 12 in a doz) The number can be determined approximately and is a very large number called the Avogadro onstant, N A, = x to 8 sig. figs. So the mass of 1 atom can be calculated as follows: The number of mole = mass g molar mass g mol 1 so if we have 3.01 g O 2 (g) we have = 3.01 g g mol 1 = 6.84 x 10-2 mol and is = 6.84 x 10-2 mol x x mol -1 = 4.12 x formula units... a very large number! Example alculate the mass of mole of magnesium sulfate trihydrate, MgSO O. First calculate the molar mass of magnesium sulfate: 1 Mg 1 x S 1 x x O 7 x Molar mass of MgSO O g mol -1 So the mass of 1 mole = g. Mass of mole = mol x g mol -1 = 52.3 g 3.3

4 Percentage omposition By Mass eg. arbon dioxide O 2 In 1 mole g mol g mol -1 2 x O 2 x g mol g mol -1 O 2 Total g mol -1 â % g mol 1 = g mol 1 x 100 % = % â% O g mol 1 = g mol 1 x 100 % = % The total must always add to Total = % Finding the formula from the analysis A certain sample of sodium nitrate contains Na, g N, g O, g What is the formula for sodium nitrate? The formula represents the whole number ratio of the moles of each element present in the compound. Na N O Mass of element present g g g alculate the moles of each element present g g g g mol g mol g mol mol mol mol Divide by smallest to give the ratio Since the analysis is measured experimentally in the last line the nearest whole number (we can only have whole atoms). If the ratio gave numbers like 1.5 or 1.25 then the result would be multiplied by 2 or 4 respectively as the deviation is beyond experimental error. The simplest formula - called the EMPIRIAL FORMULA is Na 1 N 1 O 3. Leaving out the 1s as usual gives NaNO

5 Empirical Formula NaNO 3 The analysis only gives the ratio, so the Empirical Formula is not necessarily the molecular/formula unit formula which might be Na 2 N 2 O 6, Na 3 N 3 O 9, Na 4 N 4 O 12 etc. More information is needed to get molecular/formula unit formula. Any ideas? Percentage omposition When a sample of a compound has been submitted for analysis (only about 1 mg is used) the result is returned as the percentage composition by mass. The percentage composition by mass for nicotine is:, 74.1 %;, 8.6 %; N, 17.3 %. alculate the empirical formula for nicotine. The percentage composition can tell how much of each element present in a 100 g sample. Nicotine 74.1 % 8.6 % N 17.3 % per 100 g 74.1 g 8.6 g 17.3 g The calculation is then the same as the previous one: N 74.1 g 8.6 g 17.3 g 74.1 g g mol g g mol g g mol mol mol mol mol 1.00 mol The empirical formula is 5 7 N If the molar mass is known then the molecular formula can be found. For nicotine the molar mass is about 162 g mol -1. What is the molecular formula? molar mass empirical formula mass = =2 So the formula is (2 x 5 7 N) = N 2 The structure (cannot be obtained from the molecular formula) is shown on the right. N N 3 3.5

6 Example: A compound has analysis, 62.58%;,9.63%; O, 27.79% and a molar mass = 230 u. What is the molecular formula. O g 9.63 g g g 9.63 g g mol g mol g g mol mol mol mol Empirical formula is 6 11 O 2 Now, the empirical formula mass is (6 x )+ (11 x )+(2 x ) = molar mass empirical formula mass = =2 so the molecular formula is (2 x 6 11 O 2 ) = O 4 Ans: O 4 Balancing Equations hemical equations must always be balanced it they are to be useful. Each equation must: 1. be mass balanced. (All the atoms that go in must come out.) 2. be charge balanced. The total charge before and after the reaction must be the same. 3. be balanced with respect to electron transfer. (Explained later.) You must have same number of atoms of each element on each side of the arrow. When balancing you must NOT (no, never, ever..) change the formulas. 3.6

7 Example 1: Solid copper reacts with solid sulfur to give solid copper(i) sulfide as the product. Write a balanced equation for the reaction. u(s) + S(s) u 2 S(s) This is not balanced because there is 1 copper on the left and two on the right. The equation can be balanced by making 2 u and the left. 2 u(s) + S(s) u 2 S(s) The equation is now balanced. Example 2: Aluminum metal burns on air to give aluminum oxide. Write the equation for the reaction. Aluminum is Al 3+ in compounds and the oxide ion is O 2- so the formula for the oxide must be Al 2 O 3 to balance the charges. So the equation becomes Al(s) + O 2 (g) Al 2 O 3 (s) (not balanced) 2 Al(s) + O 2 (g) Al 2 O 3 (s) 2 Al(s) + 1½ O 2 (g) Al 2 O 3 (s) x 2 to give whole numbers 4 Al(s) + 3 O 2 (g) 2 Al 2 O 3 (s) now balanced 3.7

8 Example 3: page 66 text. Balance the equation for the reaction between hydrazine and dinitrogen tetraoxide N 2 4 (l) + N 2 O 4 (g) N 2 (g) + 2 O(l). First examine the equation. Notice that all the O on the left becomes water so we can add 4 2 O. N 2 4 (l) + N 2 O 4 (g) N 2 (g) O(l). all the comes from the N 2 4 (l) so at least 2 are needed 2 N 2 4 (l) + N 2 O 4 (g) N 2 (g) O(l). and now there are 6 N on the left so we should make it the same on the right and the equation is balanced! 2 N 2 4 (l) + N 2 O 4 (g) 3 N 2 (g) O(l). Example: The mineral chalcopyrite is found in Buchans and several other places in Newfoundland and Labrador. halcopyrite has the formula ufes 2. ow much copper could be obtained from kg of the mineral? Molar mass of ufes 2 = (2 x ) g mol -1 = g mol kg = kg x 1000 g kg -1 = 1000 g 1000 g Moles of ufes 2 in the sample = = g mol 1 9 mol Now each mole of the mineral contains 1 mole of u Moles of copper in sample = mol x 1 of u = mol of u Mass of copper = mol x g mol -1 u = g u Ans: g Using the reaction equation to predict the amount of product Magnesium metal burns in oxygen to give magnesium oxide as a white solid. ow much magnesium oxide will be formed when 3.21 g of magnesium reacts? Words 2 Mg (s) 3.21 g excess + O 2 (g) 2 MgO(s) how? much in g N.B. It is useful to focus on the species of interest - the Mg so divide by 2. Mg(s) + ½ O 2 (g) MgO at once it is clear what products one mole of Mg will give. 3.8

9 Moles of Mg used = MgO formed 3.21 g = g mol 1 1 mol and 1 mol Mg gives 1 mol of MgO so moles of = mol Mass of MgO formed = mol x ( ) g mol -1 = mol x g mol -1 = 5.32 g ow much O 2 was used? Each mole of magnesium requires ½ moles of oxygen so moles of mol O 2 required = ½ x mol = mol â mass O 2 = mol x (2 x ) g mol -1 = mol x g mol -1 = 2.11 g Another way? Subtraction should give the same answer 5.32 g g = 2.11 g Analysis of organic compounds ombustion Analysis x y O z XQT + excess O 2 x O 2 + ½ y 2 O + other products from XQT The amount of O 2 and 2 O produced in the reaction can be determined but not the oxygen because there are two sources one of which is an excess. If the substance contains only, and O then the amount of oxygen can be found by difference and the analysis is complete. Example: Advil (Ibuprofen) contains, and O. On complete combustion mg of Advil gives mg of O 2 and mg 2 O. What is the empirical formula for Advil? Based on the amount of O 2 formed the amount of present can be found: Mass in compound. Molar mass of O 2 = g mol -1 and contains g mol -1 of carbon x10 3 g Moles of O 2 formed = = x 10-4 mol g mol 1 so mass of is = x 10-4 mol x g mol -1 carbon = x 10-3 g = mg mg or as % of original sample x 100 % = % 5.00 mg 3.9

Mass in the compound. Use the method above or since the molar mass of 2 O is 18.0152 g and each mol contains 2.0158 g mol -1 2.0158 g so mass = x 3.926 mg 18.0152 g = 0.4393 0 mg 0.

10 Mass in the compound. Use the method above or since the molar mass of 2 O is g and each mol contains g mol g so mass = x mg g = mg mg or as % = x 100 % mg = % Mass O in compound = mg ( mg mg) = mg = 100 % (75.65 % %) = % Then the problem is the same as we have done earlier. O x 10-3 g x 10-4 g x 10-4 g x 10-4 mol x 10-4 mol x 10-5 mol O 2 The empirical formula is O 2. Redo the calculation using the % calculated above to show that you get the same answer. Limiting Reactant The amount of product that can form in a reaction depends on the amount of starting materials that are available to react. onsider the reaction between N 2 (g) and 2 (g) to give N 3 (g) as shown in the diagram... There a 4 N 2 (g) and 6 2 (g) (blue). If all the 2 (g) (at total of 12 ) then only 2 N 2 (g) (4 N) can react so there are 2 N 2 (g) in excess. The amount of 2 (g) limits the amount of product that can form. 3.10

11 When sodium chloride solution is mixed with silver nitrate solution a white precipitate od silver chloride is formed. ow much silver chloride can form if 10.0 g of Nal and 10.0 g of AgNO 3 are present in the solutions. Nal(aq) + AgNO 3 (aq) Agl(s) + NaNO 3 (aq) 10.0 g 10.0 g From the equation: 1 mole of Nal reacts with 1 mole of AgNO 3. Moles of Nal available = 10 g g mol 1 = mol so if all the Nal could react the maximum yield of Agl = mol 10 g Moles of AgNO 3 available = = g mol mol so If all the AgNO 3 could react the maximum yield of Agl = mol. learly there is only enough AgNO 3 to form mol of Agl although there is enough Nal to form more. AgNO 3 is the LIMITING REAGENT and there is an excess of Nal present. Excess Nal = mol mol = mol Thus the maximum amount of Agl that can form = mol. This is the TEORETIAL YIELD. Theoretical yield of Agl in grams = mol x g mol -1 = 8.44 g Ans: 8.44 g Example: (A bit more complicated.) alculate the theoretical yield of phosphorus tribromide that can form when 10.0 g of phosphorus reacts with 50.0 g of bromine. 2 P(s) + 3 Br 2(l) 2 PBr 3(l) (1) P(s) + 3/2 Br 2(l) 1 PBr 3(l) (2) 2/3 P(s) + 1 Br 2(l) 2/3 PBr 3(l) Mass of available P: 10.0 g Br: 50.0 g 10.0 g 50.0 g Moles available g mol 1 2 x79.904g mol 1 Moles available mol mol Max product (1) 1 x mol = mol (2) 2/3 x mol So the bromine is the limiting reagent and the theoretical yield of PBr 3 (l) is.. Theoretical yield = mol x g mol -1 PBr 3 = mol Theoretical yield = 56.5 g PBr 3 (l) Some of the phosphorus is not used (there was an excess): Excess phosphorus = amount available - amount used Excess phosphorus = (2/3 x ) mol = mol In grams Excess phosphorus = mol x g mol -1 = 3.54 g 3.11

12 Percentage yield In the real world experiments rarely give the theoretical yield. Often there are competing reactions or product is lost in the workup. To give an indication of how successful a reaction is the percentage yield is calculated: Percentage Yield = Actual Yield Theoretical Yield x100% Say only 45.3 g of PBr 3 obtained from the above experiment the percentage yield is = 45.3 g 56.5 g x 100% = 80.2% The Avogadro onstant You might have noticed that nothing that we have done so far requires a knowledge of the number of single atoms in a mole and for most of chemistry it is now essential information. owever, the Avogadro Number, N, the number of atoms in exactly 12 g of cabon-12 isotope has been determined in a number ways. The most direct method (Rutherford and Geiger) involves counting the number of α-particles produced in the radioactive decay of radium. The α-particle, 4 2 e 4+, becomes helium gas, 4 2 e, which is collected and measured. The result is an approximate value for N of x mol -1 a very large number. Once we have this value we can find the mass of 1 u (a.m.u. in our book). 12 g mol 1 1 u = = x g u u x x mol 1 this is (6 protons + 6 neutrons + 6 electrons)/6 The mass of an average atom is = x g and this is 1 proton + 1 neutron + 1 electron... hmmn

Chapter 3. Stoichiometry

Chapter 3. Stoichiometry 1 hapter 3 Stoichiometry 2 Atomic Mass Avogadro s Number and Molar Mass 3 Atomic Mass: Mass of 1 atom in atomic mass units 1 amu = 1/12 of the mass of 1-12 atom = 1.661X10-24 g Naturally occurring carbon