Topic 7: The Mole Concept Relating Mass to Numbers of Atoms

Topic 7: The Mole Concept Relating Mass to Numbers of Atoms (Chapter 3 in Modern Chemistry beginning on p.82) In order to understand the quantitative parts of chemistry, there are three very important concepts the mole, Avogadro s number, and molar mass. These provide the basis for relating masses in grams to number of atoms. The Mole The mole is the SI unit for amount of substance. A mole (abbreviated mol) is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. The mole is a counting unit, just like a dozen is. Avogadro s number The number of particles in a mole has been experimentally determined in a number of ways. The best modern value is 6.022 1415 x 10 23. This is the number of particles (atoms, ions, molecules, formula units) in a mole. It is called Avogadro s number. Avogadro s number 6.022 1415 x 10 23 is the number of particles in exactly one mole of a pure substance. For most purposes, Avogadro s number is rounded to 6.022 x 1023. Watch this video for an idea of how many particles Avogadro s number really is. http://www.youtube.com/watch?v=1r7niium2ti Molar Mass An alternative definition of mole is the amount of a substance that contains Avogadro s number of particles. The mass of one mole of a pure substance is called the molar mass of that substance. Molar mass is usually written in units of g/mol. The molar mass of an element is numerically equal to the atomic mass of the element in atomic mass units (which can be found in the periodic table). Using Chemical Formulas (Chapter 7 in Modern Chemistry beginning on p.237) As you have seen, a chemical formula indicates the elements as well as the relative number of atoms or ions of each element present in a compound. Chemical formulas also allow chemists to calculate a number of characteristic values for a given compound. In this section, you will learn how to use chemical formulas to calculate the molar mass, formula mass, and the percentage composition by mass of a compound. HN Chemistry Page 1

Molar Mass/Formula Mass Molecular and formula mass use the amu that we learned in Chapter 3. We use the atomic mass units from the periodic table. For example: Find the formula mass of potassium chlorate, KClO 3. 1 K atom x 39.10 amu = 39.10 amu 1 Cl atom x 35.45 amu = 35.45 amu 3 O atoms x 16.00 amu = 48.00 amu Formula mass of KClO 3 = 122.55 amu The term molecular mass is usually used for covalent compounds. The term formula mass is usually used for ionic compounds. Molar mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula. The units for these calculations are g/mole. This is calculated the same way as molecular or formula mass. Example: Find the molar mass of Ba(NO 3 ) 2. 1 mole Ba x 137.33 g/mole = 137.33 g Ba 2 moles N x 14.01 g/mole = 28.02 g N 6 moles O x 16.00 g/mole = 96.00 g O Molar mass of 1 mole of Ba(NO 3 ) 2 = 261.35 g/ mole Task 7a 1. Find the molar mass of the following. a. NaCl b. K 2 O c. NaOH d. Ca(OH) 2 e. (NH 4 ) 3 PO 4 HN Chemistry Page 2

Percent Composition Sometimes you need to know the percentage by mass of a particular element in a chemical compound. For example, suppose the compound potassium chlorate, KClO 3, were to be used as a source or oxygen. What is the percentage of oxygen in this compound? Use the following formula: mass of element in sample of compound Mass of sample of compound x 100 = % of element in compound The percentage by mass of each element in a compound is known as the percentage composition of the compound. Find the percentage composition of copper(i) sulfide, Cu 2 S. Find the molar mass of each element and of the compound. 2 Cu x 63.55 g/mol = 127.1 g/mole 1 S x 32.07 g/mol = 32.07 g/mole Molar mass of Cu 2 S = 159.2 g/mole Put these numbers into the formula for percent composition. Task 7b % Cu = % S = 127.1 g/mol 159.2 g/mole 32.07 g/mole 159.2 g/mol x 100 = 79.85% x 100 = 20.15% 1. Find the percent composition of each element in the following compounds. (Note: You already calculated the molar mass in Task 7a). a. NaCl b. K 2 O c. NaOH d. Ca(OH) 2 e. (NH 4 ) 3 PO 4 HN Chemistry Page 3

Hydrates In some salts, water molecules from solution bind within their crystal structure. For example sodium carbonate forms a hydrate, in which 10 water molecules are present for every formula unit of sodium carbonate. The formula of the hydrate is Na 2 CO 3. 10 H 2 O. Note that the first part of the formula is the formula of the compound, then a dot. Following the dot, the number of water molecules that are attached to each formula unit is written. Hydrates are named by naming the compound, prefix hydrate. Again, using the example above; the hydrate is named sodium carbonate decahydrate. The molar mass of a hydrate is calculated by the following formula: Task 7c MM hydrate = MM salt + (# of water molecules x MM water ) Na 2 CO 3. 10 H 2 O 2(23) + 12 + 3(16) + 10[2(1) + 16] 1. What is the name of MgSO 4. 5 H 2 O? 2. Write the formula of copper(ii) chloride trihydrate. 3. Calculate the molar mass of CuSO 4. 5 H 2 O. 4. What is the percent of water in CuSO 4. 5 H 2 O? 106 + 10(18) 106 + 180 286 g/mol The Mole Why are M & Ms sold in a package rather than individually? List other items we package and then refer to as a package rather than the individual item: 100 pennies = 2 socks = 12 eggs = HN Chemistry Page 4

A chemist does the same thing with atoms! Why? A chemist s package is a mole. 1 mole = 6.02 x 10 23 particles. 6.02 x 10 23 is called Avogadro s number. It has been experimentally determined that when one measures out the mass of an element equal to its average atomic mass, the number of atoms contained in the sample is equal to 6.02 x 10 23 atoms. 6.02 x 10 23 atoms = 1 mole = Avogadro s number = molar mass Mole Conversions Particles to moles o Convert representative particles to moles and moles to representative particles. (Representative particles are atoms, molecules, formula units, and ions.) Mass to moles o Convert mass of atoms, molecules, and compounds to moles and moles of atoms, molecules, and compounds to mass. o Convert representative particles to mass and mass to representative particles. Volume of a gas to moles o Convert moles to volume and volume to moles at STP. Mass Moles Particles (grams) atoms, ions, molecules, formula units MM from periodic table Use 6.02 x 10 23 (g/mol) Volume of Gas (at STP) (particles/mole) (liters) 22.4 L/mole The mole is the central unit in all mole conversions. The other units are used as needed for conversion factors. HN Chemistry Page 5

The key to these problems is using dimensional analysis. Remember the key to dimensional analysis is to cancel out the units you do not need and keep the units you do need. Summary If grams are in the problem: molar mass from the periodic table g/mole If particles are in the problem: Avogadro s number 6.02 x 10 23 particles/mole If liters of gas are in the problem: molar volume 22.4 L/mole Examples: Ibuprofen, C 13 H 18 O 2, is the active ingredient in many nonprescription pain relievers. 1) If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle? 2) How many molecules of ibuprofen are in the bottle? 3) What is the total mass in grams of carbon in 33 g of ibuprofen? Answers 1) 33 g C 13 H 18 O 2 1 mole C 13 H 18 O 2 = 0.16 mole C 13 H 18 O 2 206.31 g C 13 H 18 O 2 2) 0.16 mole C 13 H 18 O 2 6.02 x 10 23 molecules = 9.6 x10 22 molecules C 13 H 18 O 2 mole 3) 0.16 mole C 13 H 18 O 2 13 moles of C 12.01 g C = 25 g C 1 mole C 13 H 18 O 2 mole Task 7d 1. How many atoms in 400.0g of sulfur? 2. What is the mass of 1.2 x 10 24 atoms of magnesium? 3. What is the mass of 2.5 moles of oxygen atoms? 4. Given 18 grams of water, how many molecules do you have? 5. What is the mass, in grams, of 1.2 x 10 24 formula units of sodium chloride? 6. What is the volume of carbon dioxide gas in 589.4 g of carbon dioxide? HN Chemistry Page 6

Empirical & Molecular Formulas Empirical Formula the formula for a compound expressed as the smallest possible wholenumber ratio of subscripts of the elements in the formula. Molecular Formula- the formula for a compound in which the subscripts give the actual number of each element in the formulas it truly exists. An empirical formula cannot be reduced any further. A molecular formula may or may not be reduced. Molecular Formula H 2 O Empirical Formula CH 3 COOH CH 2 O C 6 H 12 O 6 Notice two things: 1. The molecular formula and the empirical formula can be identical. 2. You scale up from the empirical formula to the molecular formula by a whole number factor. Calculating Empirical Formulas A Simple Rhyme for a Simple Formula by Joel S. Thompson Percent to mass Mass to mole Divide by small Multiply til whole In other words: 1. Since percent can be on any mass, assume a 100 g sample and change directly to mass. 43 % of 100 g is 43 g. 2. Use molar mass to perform a mole conversion to change gram to moles. 3. Divide by the smallest number of moles to try to get a whole number ratio. 4. If you do not get a whole number, multiply by a whole number until you get a whole number. Remember formulas cannot be in fractions. HN Chemistry Page 7

Example #1 Quantitative analysis shows that a compound contains 32.38 % sodium, 22.65 % sulfur, and 44.99 % oxygen. Find the empirical formula of this compound. Change % Use MM to change Divide by the Hopefully, get a to grams g to moles smallest moles whole number 32.38 g Na mole Na = 1.408 mole Na = 2 22.99 g 0.7063 mole Write the formula 22.65 g S mole S = 0.7063 mole S = 1 Na 2 SO 4 32.07 g 0.7063 mole 44.99 g O mole O = 2.812 mole O = 4 This is the 16.00 g 0.7063 mole empirical formula Example #2 Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula of this compound? Do not have to Change g Divide by the Didn t get a whole change to grams, to moles smallest number for both, already in grams so multiply by a number to get a whole number 4.433 g P mole P = 0.1431 mole P = 1 (2) = 2 Write the 30.97 g P 0.1431 mole formula 5.717 g O mole O = 0.3573 mole O = 2.5 (2) = 5 16.00 g O 0.1431 mole The grams of O came from subtracting 10.150 g 4.433 g P 2 O 5 HN Chemistry Page 8

Task 7e 1. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? 2. A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of the compound is known to be approximately 140 g/mol. What is the empirical formula? 3. A compound is found to contain 63.52 % iron and 36.48 % sulfur. Find its empirical formula. 4. Find the empirical formula of a compound found to contain 26.56 % potassium, 35.41 % chromium, and the remainder oxygen. Calculating Molecular Formulas To determine the molecular formula you must know the empirical formula and the molecular mass. Remember the molecular formula is a multiple of the empirical formula. To find the multiple: molar mass of the molecular formula = x molar mass of the empirical formula To write the molecular formula: multiply the subscripts in the empirical formula by x Example #3 In example problem #2, the empirical formula of a compound of phosphorus and oxygen was found to be P 2 O 5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound s molecular formula? Molecular formula s mass = 283.89 g/mol Empirical formula s mass = 141.94 g/mol 283.89 g/mol = 2 P 2 O 5 x 2 = P 4 O 10 141.94 g/mol The empirical formula was given in the problem. HN Chemistry Page 9

Example #4 A compound with a formula mass of 42.08 amu is found to be 85.64 % carbon and 14.36% hydrogen by mass. Find the molecular formula. Find the empirical formula first. 85.64 g C mole C = 7.131 mol C = 1 Empirical formula 12.01 g C 7.131 mol CH 2 14.36 g H mole H = 14.25 mol H = 2 1.008 g H 7.131 mol 42.08 amu = 3 CH 2 (3) = C 3 H 6 14.03 amu Empirical formula mass Molecular formula Task 7f 1. A 1.50 g sample of hydrocarbon undergoes complete combustion to produce CO 2 and H 2 O. The empirical formula of this compound is CH 3. Its molecular weight has been determined to be about 78. What is the molecular formula? 2. Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol. What is its molecular formula? Concentration of Solutions (Molarity) (Chapter 12 in Modern Chemistry beginning on p.418) The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution. In this section, we introduce two different ways of expressing the concentrations of solutions: molarity and molality. Sometimes, solutions are referred to as dilute or concentrated, but these are not very definite terms. Dilute just means that there is a relatively small amount of solute in a solvent. Concentrated, on the other hand, means that there is a relatively large amount of solute in a solvent. These terms are unrelated to the degree to which a solution is saturated. A saturated solution of a substance that is not very soluble might be very dilute. HN Chemistry Page 10

Molarity Molarity is the number of moles of solute in one liter of solution. Molarity is symbolized with a M or brackets, [ ]. The formula for finding molarity is: molarity (M) = moles of solute (mol) volume of solution (L) Note that the solute must be in moles and the volume of solution must be in liters. It is also important to realize that making a 1 molar (1 M) solution is not made by adding 1 mol of solute to 1 L of solvent. You need to dissolve the solute in a small amount of water then add enough water to make the total volume of solution 1L. Here are some examples of molarity problems. Ex. 1 You have 3.50 L of solution that contains 90.0 g of sodium chloride. What is the molarity of this solution? molarity (M) = moles of solute (mol) volume of solution (L) First you have to change the 90.0 g of sodium chloride to moles of sodium chloride. 90.0 g NaCl moles of NaCl 58.5 g of NaCl = 1.54 mol NaCl Then plug this information into the molarity formula. M = 1.54 mol NaCl 3.50 L solution OR = 0.440 M NaCl You can do the same problem using dimensional analysis. 90.0 g NaCl moles of NaCl 58.5 g of NaCl 3.50 L solution = 0.440 M NaCl HN Chemistry Page 11

Ex. 2 You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Remember that the capital M means moles of solute per liters of solution so 0.5 M means 0.5 moles of HCl in 1 liter of solution. Writing the molarity this way allows you to see that you can use dimensional analysis to cancel liters and solve for moles. 0.5 moles HCl 0.8 L solution = L solution 0.4 mol HCl You could also find the grams of HCl by using molar mass if needed. Ex. 3 To produce 40.0 g of silver chromate, you will need at least 23.4 g of potassium chromate in solution as a reactant. All you have on hand is 5 L of 6.0 M K 2 CrO 4 solution. What volume of the solution is needed to give you the 23.4 g K 2 CrO 4 needed for the reaction? First, it is important to realize that the 40.0 g of silver chromate is not important information. We need to change 23.4 grams to moles. 23.4 g K 2 CrO 4 1 mol K 2 CrO 4 194.2 g K 2 CrO 4 = 0.120 mol K 2 CrO 4 Now plug the information into the molarity formula. molarity (M) = moles of solute (mol) volume of solution (L) 6.0 moles K 2 CrO 4 L solution = 0.120 mol K 2 CrO 4 x x = 0.020 L K 2 CrO 4 soln OR You can do the same problem using dimensional analysis. 23.4 g K 2 CrO 4 moles of K 2 CrO 4 L of soln 194.2 g of K 2 CrO 4 6.0 mol K 2 CrO 4 = 0.020 L K 2 CrO 4 soln HN Chemistry Page 12

Task 7g 1. What is the molarity of a solution composed of 5.85 g potassium iodide, KI, dissolved in enough water to make 0.125 L of solution? 2. How many moles of H 2 SO 4 are present in 0.500 L of a 0.150 M H 2 SO 4 solution? 3. What volume of 3.00 M NaCl is needed for a reaction that requires 146.3 g of NaCl? Molality Molality is the number of moles of solute per kilogram of solvent. Molality is symbolized with a m. The formula for finding molality is: molality (m) = moles of solute (mol) Kilograms of solvent (kg) Note that the solute must be in moles and the solvent must be in kilograms. It is also important to realize that making a 1 molal (1 m) solution is made by adding 1 mol of solute to 1 kg of solvent. Here are some examples of molality problems. Ex. 1 A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution. molality (m) = moles of solute (mol) Kilograms of solvent (kg) First you have to change the 17.1 g of sucrose to moles of sucrose. 17.1 g C 12 H 22 O 11 moles of C 12 H 22 O 11 342.34 g of C 12 H 22 O 11 = 0.0500 mol C 12 H 22 O 11 Then plug this information into the molality formula, remembering that you have to use kg of solvent. 125 g of water is 0.125 kg of water. m = 0.0500 mol C 12 H 22 O 11 0.125 kg solvent OR = 0.400 m C 12 H 22 O 11 HN Chemistry Page 13

You can do the same problem using dimensional analysis. 17.1 g C 12 H 22 O 11 moles of C 12 H 22 O 11 342.34 g of C 12 H 22 O 11 0.125 kg solvent = 0.400 m C 12 H 22 O 11 Ex. 2 A solution of iodine, I 2, in carbon tetrachloride, CCl 4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.480 m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used? Remember that the m means moles of solute per kilograms of solvent so 0.480 m means 0.480 moles of I 2 in 1 kg of CCl 4. Writing the molality this way allows you to see that you can use dimensional analysis to cancel kilograms and solve for moles. 0.480 moles I 2 0.100 kg solvent = kg solvent 0.480 mol I 2 It is then possible to change moles of iodine to grams of iodine by using the molar mass of iodine. 0.480 mole I 2 253.8 g I 2 mol I 2 = 12.2 g I 2 OR You can do the same problem using dimensional analysis. 0.480 moles I 2 0.100 kg solvent 253.8 g I 2 kg solvent mol I 2 = 12.2 g I 2 Task 7h 1. What is the molality of acetone in a solution composed of 255 g of acetone, (CH 3 ) 2 CO, dissolved in 200. g of water? 2. What quantity, in grams, of methanol, CH 3 OH, is required to prepare a 0.244 m solution in 400. g of water? HN Chemistry Page 14

Dilution Sometimes you will need to dilute a more concentrated solution. The formula for dilution is: M 1 V 1 = M 2 V 2 For example: What volume of 6.0 M KCl will be needed to make 300.0 ml of 2.4 M KCl? (6.0 M)V 1 = (2.4 M) (300.0 ml) V 1 = 120 ml Task 7i 1. How would you make 300.0 ml of 2.4 M KCl from 6.0 M KCl? HN Chemistry Page 15