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The Mole Concept A. Atomic Masses and Avogadro’s Hypothesis 1. We have learned that compounds are _________________________________________ ________________________________________________________________________ Therefore, compounds must be composed of ___________________________________ ________________________________________________________________________ During a chemical reaction, the atoms that make up the starting materials ____________ ________________________________________________________________________ The question that arises however, is how many atoms and molecules are involved in the reaction or how much of one element will combine with another element? In addition, since atoms are so small, how can we count them? Early experimental work by English chemist _________ ______________ (1766–1844) was concerned with how much of one element could combine with a given amount of another element. He proposed the following hypotheses: _____________________________________________________________________ If compound “B” contains twice the mass of element “X” as does compound “A”, then compound “B must contain twice as many atoms of “X”. Simple compounds are made up of only ____________________________________ _____________________________________________________________________ _____________________________________________________________________ VERSION: October 18, 2000 2 2. CHEMISTRY 11 Dalton did not attempt to figure out the mass of an individual atom of any element. Instead, he assigned an ____________________ __________ to each element. He made the assumption that _______________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Carbon was found to be 6 times heavier than hydrogen so it was assigned a mass of 6. Oxygen was found to be 16 times heavier than hydrogen so it was assigned a mass of 16. e.g. The reaction between 2.74 g of hydrogen gas and 97.26 g of chlorine gas makes 100 g of hydrogen chloride gas. If we assume that hydrogen chloride contains one atom each of hydrogen and chlorine, the relative mass of chlorine is 97.26 g 2.74 g = __________ times heavier than ____________________ Since hydrogen is assigned a mass of “1”, chlorine has a mass of “__________”. If 46.0 g of sodium react with 71.0 g of chlorine, the relative mass of sodium is 46.0 g 71.0 g = __________ times the mass of ____________________ Since chlorine is assigned the mass of “__________”, the mass of sodium is __________ x __________ = __________ In this way, Dalton was able to calculate the “____________________ _____________” for several elements. UNIT V THE MOLE CONCEPT 3. 3 Dalton’s atomic mass scale was partly in error because ___________________________ ________________________________________________________________________ During the time that Dalton’s mass scale was just being introduced, the French chemist ______________________________ began to study how gases reacted. When Gay– Lussac reacted pairs of gases at the same temperature and pressure, he found that ________________________________________________________________________ 1 L of hydrogen gas reacts with 1 L of chlorine gas to make 2 L of HCl(g) 1 L of nitrogen reacts with 3 L of hydrogen gas to make 2 L of NH3(g) 2 L of carbon monoxide gas react with 1 L of oxygen gas to make 2 L of CO2(g) By itself, Gay-Lussac’s findings did not seem to be related to atomic mass but then the Italian chemist____________________ ____________________ proposed the following explanation for Gay-Lussac’s data. AVOGADRO’S HYPOTHESIS ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ In other words, ___________________________________________________________ ________________________________________________________________________ Therefore, the molecule formed by reacting A with B is__________ Similarly, if 2 L of gas A reacts with 1 L of gas B, the molecules formed have the formula __________ 4 CHEMISTRY 11 B. The Mole 1. Avogadro’s hypothesis _____________________________________________________ ________________________________________________________________________ ________________________________________________________________________ e.g. If 1 L of nitrogen reacts with 3 L of hydrogen to form ammonia, then its formula is __________ If 2 L of hydrogen reacts with 1 L of oxygen to form water, then its formula is __________ So if we want to make a particular compound, all we need to do is react volumes of gases in the ratio given by their formulas BUT how do we determine how much of one element reacts with another element when they are not gases? e.g. How much iron is required to react with sulphur to produce iron (II) sulphide, FeS, so that neither element is left over? Since the easiest way to measure solids is to measure their mass, we need to relate mass to the number of atoms. 2. The periodic table shows us the ______________________________________________ Its units are “_____” which stands for “___________________________________”. Unlike Dalton’s mass scale, the present day scale is not based on hydrogen. Instead, _________________________________________________________________ ________________________________________________________________________ UNIT V THE MOLE CONCEPT 5 A _______________ is the number of carbon atoms in exactly _________________________ Molar Mass is the _____________________________________________ 6 Atomic Number C Symbol Atomic Mass 12.0 one “C” atom has a mass of __________ one MOLE of “C” atoms has a mass of __________ The periodic table gives us _________________________________________________ ________________________________________________________________________ 8 O 16.0 20 26 Ca Fe 40.1 55.8 16 17 S Cl 32.1 35.5 Finding the molar mass of a compound simply involves __________________________ ________________________________________________________________________ ________________________________________________________________________ (remember the units of molar mass are grams). 6 CHEMISTRY 11 EXAMPLE V.1 CALCULATING MOLAR MASSES Problem: What is the molar mass of iron (III) sulphate? Solution: Iron (III) sulphate is Fe2(SO4)3 To calculate the molar mass we need to add the masses of 2 Fe + 3 S + 12 O 2 (__________) + 3 (__________) + 12 (__________) = __________ Alternatively, 2 Fe + 3 (SO4) 2 (__________) + 3 (__________ + 3 (__________)) 2 (__________) + 3 (__________) = __________ SAMPLE PROBLEM V.1 Problem: CALCULATING MOLAR MASSES Calculate the molar masses of (a) Na2Cr2O7 Solution: (b) Ag2SO4 (c) Pb3(PO4)4 (a) __________________________________________________ (b) __________________________________________________ (c) __________________________________________________ UNIT V THE MOLE CONCEPT 7 C. Relating Moles to Mass, Volume of Gas, and Number of Particles 1. The molar mass of a compound ______________________________________________ ________________________________________________________________________ ________________________________________________________________________ Since we know that __________ of “X” has a mass of ______________________________ we have two conversion factors: 1 mol (molar mass of "X") g (molar mass of "X") g or 1 mol EXAMPLE V.2 Problem: Solution: RELATING MASS AND MOLES (a) What is the mass of 3.25 mol of CO2? (b) What is the mass of 1.36 x 10-3 mol of SO3? (c) How many moles of N2 are there in 50.0 g of N2? (d) How many moles of CH3OH are there in 0.250 g of CH3OH? (a) 1 mol CO2 = __________ mass CO2 = __________ x __________ = __________ (b) 1 mol SO3 = __________ 8 CHEMISTRY 11 mass SO3 = ____________________ x __________ = __________ (c) 1 mol N2 = __________ mol N2 = __________ x __________ = __________ (d) 1 mol CH3OH = __________ mol N2 = __________ x __________ = _______________ SAMPLE PROBLEMS V.2 Problem: Solution: RELATING MASS AND MOLES (a) What is the mass of 0.834 mol of FeSO4? (b) What is the mass of 2.84 x 10-2 mol of Na3N? (c) How many moles of CH4 are there in 27.5 g of CH4? (d) How many moles of Ca(NO3)2 are there in 35.0 g of Ca(NO3)2? (a) 1 mol FeSO4 = __________ mass FeSO4 = __________ x __________ = __________ (b) 1 mol Na3N = __________ mass Na3N = ____________________ x __________ = __________ UNIT V THE MOLE CONCEPT 9 (c) 1 mol CH4 = __________ mol CH4 = __________ x __________ = __________ (d) 1 mol Ca(NO3)2 = __________ mol Ca(NO3)2 = __________ x __________ = __________ 2. Calculations involving gas volumes are simplified by Avogadro’s hypothesis. Recall that AVOGADRO’S HYPOTHESIS ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ The MOLAR VOLUME of a gas is the _____________________________________ ______________________________________________________________________ Since the volume of a gas is _________________________________________________ ________________________________________________________________________ ________________________________________________________________________ STANDARD TEMPERATURE AND PRESSURE (_________) = _________ and _________ 10 CHEMISTRY 11 Avogadro’s hypothesis can be interpreted to mean that all gas samples with the same temperature, pressure, and numbers of particles occupy the same volume. This can be re–stated as ______________________________________________________________ ________________________________________________________________________ Experimentally, it is determined that __________ of any gas _______________ has a volume of __________ In other words, the ________________________________________________________ We can obtain two conversion factors: 1 mol 22.4 L or 22.4 L 1 mol EXAMPLE V.3 Problem: Solution: 3. RELATING VOLUME OF A GAS AND MOLES (a) How many moles of gas are contained in a balloon with a volume of 10.0 L at STP? (b) What volume will 0.250 mol of CO2 occupy at STP? (a) mol of gas = __________ x __________ = __________ (b) volume of CO2 = __________ x __________ = __________ The mole is the fundamental unit in chemistry for measuring the ___________________ ________________________________________________________________________ UNIT V THE MOLE CONCEPT 11 In a sense, the mole is simply a ____________________ ____________________. Just as a dozen = 12 experimentally, __________ = ____________________ This value, 6.02 x 1023, is called ____________________ number. Notice that there are no units in the same way a “dozen” stands for “12”. Conversion factors: 1 mol particles 6.02 x 1023 particles or 1 mol particles 6.02 x 1023 particles EXAMPLE V.4 Problem: RELATING NUMBER OF PARTICLES AND MOLES (a) How many molecules are there in 0.125 mol of molecules? (b) How many moles of N are there in 5.00 x 1017 N atoms? (c) How many atoms are in 5 molecules of CuSO4•5H2O? (a) molecules = __________ x Solution: 23 6.02 x 10 molecules 1 mol molecules = ____________________ molecules (b) moles N2 = ____________________ x 1 mol atoms 23 6.02 x 10 atoms = ____________________ 12 CHEMISTRY 11 (c) atoms atoms = ____________________ x 21 molecule = __________ SAMPLE PROBLEMS V.3 Problem: Solution: RELATING VOLUME OF GAS / NUMBER OF PARTICLES AND MOLES (a) How many moles of gas are contained in a balloon with a volume of 17.5 L at STP? (b) What volume of gas will 0.074 mol of gas occupy at STP? (c) How many atoms are there in 0.0185 mol of atoms? (d) How many moles of Fe2O3 are there in 8.75 x 1020 Fe2O3 molecules? (e) How many atoms of H are in 30 molecules of Ca(H2PO4)2? (a) mol of gas = __________ x __________ = __________ (b) volume of gas = __________ x __________ = __________ (c) atoms = _______________ x ________________________ = ____________________ (d) moles Fe2O3 = _______________ x ____________________ = ____________________ (e) H atoms = ___________________ x ___________________ = _______________ UNIT V THE MOLE CONCEPT 4. 13 All of the previous problems have involved single–step conversions between moles, mass, volume, or number of particles. The following is a summary of the conversion factors needed: CONVERSION CONVERSION FACTOR MOLES NUMBER OF PARTICLES 23 6.02 x 10 particles 1 mol (molar mass) g 1 mol MOLES MASS 22.4 L 1 mol MOLES VOLUME (gases @ STP) MOLECULES ATOMS (atom count) atoms 1 mol or 1 mol 23 6.02 x 10 particles 1 mol or (molar mass) 1 mol or 22.4 L 1 mol or (atom count) atoms The following flow chart will help to simply calculations that involve multiple conversions. PARTICLES (atoms/molecules) 6.02 x 10 23 MOLES molar mass MASS (in grams) 22.4 L/mol VOLUME (at STP) 14 CHEMISTRY 11 EXAMPLE V.5 Problem: MOLE CALCULATIONS INVOLVING MULTIPLE CONVERSIONS (a) What is the volume occupied by 50.0 g of NH3(g) at STP? (b) What is the mass of 1.00 x 10 (c) 12 atoms of Cl? How many oxygen atoms are contained in 75.0 L of SO3(g) at STP? Solution: (a) MASS MOLES VOLUME ? L = __________ x __________ x __________ = __________ (b) ATOMS MOLES MASS ? g = _____________ atoms g x = (c) 1 mol 23 6.02 x 10 atoms x ________ _______________ VOLUME MOLES MOLECULES ATOMS of O 23 ? O’s = _________ x _________x 6.02 x 10 1 mol = ____________________ SO3 x _________ UNIT V THE MOLE CONCEPT 4. 15 So far, all the volumes have been of gases at STP. If ____________________ is mentioned at any point in a problem, recall that d = m/V • __________________________________________________, calculate the volume ( from V = m/d ) . (Note you cannot use the molar volume of a gas, 22.4 L, when calculating the volume of a solid or liquid.) • ______________________________, you will need both mass and volume to ( calculate density from d = m/V • ). ________________________________________, use the density and volume to calculate mass from (m = d • V) and then convert mass to moles. • ________________________________________, it can be calculated by the (molar mass) equation: d = (molar volume) EXAMPLE V.6 Problem: DENSITY AND MOLE CALCULATIONS (a) What is the volume occupied by 3.00 mol of ethanol, C2H5OH? (d = 0.790 g/mL) (b) How many moles of Hg(l) are contained in 100 mL of Hg(l)? (d = 13.6 g/mL) Solution: (c) What is the density of O2(g) at STP? (a) We want volume of liquid so V = m /d , and mass can be related to moles, so ? mL = __________ x __________x __________= __________ (b) To get moles, we need to calculate mass first and then convert to moles. ? mol = __________ x __________x __________= __________ 16 CHEMISTRY 11 (c) There are no numbers given but the term STP suggests a volume of 22.4 L for one mole of gas and the mass of one mole of O2(g) is 32.0 g. m density = V mass of 1 mol (molar mass) = volume of 1 mol (molar volume) = __________= __________ EXAMPLE V.7 Problem: MORE DENSITY CALCULATIONS (a) A 2.50 L bulb contains 4.91 g of a gas at STP. What is the molar mass of the gas? (b) Al2O3(s) has a density of 3.97 g/mL. How many atoms of Al are in 100 mL of Al2O3? Solution: (a) Density can be found from d = m /V , so d = __________= __________ (b) Change volume to mass, then to moles, then to molecules, and finally atoms. 6.02 x 10 ? Al =_______ x _______x _______x 23 Al2O3 1 mol = ____________________ atoms Al x 2 atoms Al Al2O3 UNIT V THE MOLE CONCEPT 17 C. Percentage Composition 1. The PERCENTAGE COMPOSITION is _____________________________________ ________________________________________________________________________ EXAMPLE V.8 Problem: Solution: PERCENTAGE COMPOSITION (a) What is the percentage composition of H2SO4? (b) What is the percentage composition of water in CuSO 4 • 5H2O? (a) Assume that there is 1 mole of the compound. Molar mass = __________ Total mass of H = _______________ = __________ Total mass of S = _______________ = __________ Total mass of O = _______________ = __________ % H = __________ x 100% = __________ % S = __________ x 100% = __________ % O = __________ x 100% = __________ (b) Assume that there is 1 mole of the compound. Molar mass = __________ Total mass of H2O = _______________ = __________ % H2O = __________ x 100% = __________ 18 CHEMISTRY 11 D. Empirical and Molecular Formulas 1. The empirical formula is called the ___________________________________________ ________________________________________________________________________ ________________________________________________________________________ CH2, C2H4, C3H6, C4H8, and C5H10 all contain twice as many H’s as there are C’s. The empirical formula (simplest formula) for all of these molecules is __________ Finding empirical formula is essentially the opposite of determining percentage composition. EXAMPLE V.9 Problem: Solution: EMPIRICAL FORMULAS (a) What is the empirical formula of a compound consisting of 80.0% C and 20.0% H? (b) A compound contains 58.5 % C, 7.3% H, and 34.1 % N. What is the empirical formula of the compound? (a) Assume __________ of the compound mass of C = ____________________= __________ mass of H = ____________________ = __________ Use the mass to determine the moles of each element mole C = __________ x __________ = __________ mole H = __________ x __________ = __________ UNIT V THE MOLE CONCEPT 19 Determine smallest ratio by dividing by smallest number of moles (÷ 6.67) C = _______________ = __________ H = _______________ = __________ Empirical Formula is __________ (b) Assume 100.0 g of the compound mass of C = __________, mass of H = __________, mass of N = __________ Use the mass to determine the moles of each element 1 mol mole C = 58.5 g x 12.0 g = __________ 1 mol mole H = 7.3 g x 1.0 g = __________ 1 mol mole N = 34.1 g x 14.0 g = __________ Determine smallest ratio by dividing by smallest number of moles (÷ 2.44) 4.88 mol C = 2.44 mol = __________ 7.3 mol H = 2.44 mol = 2.99 __________ 2.44 mol N = 2.44 mol = __________ Empirical Formula is __________ 20 CHEMISTRY 11 EXAMPLE V.10 Problem: MORE EMPIRICAL FORMULAS What is the empirical formula of a compound consisting of 81.1% C and 18.2% H? Solution: Assume 100.0 g of the compound mass of C = __________, mass of H = __________ Use the mass to determine the moles of each element 1 mol mole C = 81.1 g x 12.0 g = __________ 1 mol mole H = 18.2 g x 1.0 g = __________ Determine smallest ratio by dividing by smallest number of moles (÷ 6.82) 6.82 mol C = 6.82 mol = __________ 18.2 mol H = 6.82 mol = __________ DO NOT just round off ratio, you must multiply both number by 2, 3, 4, or 5 until both are whole numbers Multiplying both numbers by _____ gives whole numbers _____ : _____ Empirical Formula is __________ ALWAYS ______________________________________________________________ ________________________________________________________________________ Improper round-off calculations will cause you to multiply by the wrong number when trying to obtain whole numbers. UNIT V THE MOLE CONCEPT 2. 21 The ____________________ ____________________ can be found be using the molar mass of the empirical formula; that is, the EMPIRICAL MASS. The molecular formula is ___________________________________________________ ________________________________________________________________________ CH2, C2H4, C3H6, C4H8, and C5H10 all have the same empirical formula CH2. The whole number multiple (N) is given by the formula molar mass Multiple = N = empirical mass _________________________ = _________________________ 3. It may be necessary to calculate the molar mass from information that is given in the question. (a) Finding molar mass from density of a gas at STP If: density of gas “X” = 1.43 g/L (at STP) Then: molar mass of “X” = 1.43 g/L x 22.4 L/mol = __________ (b) Finding molar mass from mass and volume of a gas at STP If you are told: “0.0425 L of gas ‘X’ at STP has a mass of 0.135 g” 0.135 g Then: density of gas “X” = 0.0425 L = 3.176 g/L And: molar mass of “X” = 3.176 g/L x 22.4 L/mol = __________ 22 CHEMISTRY 11 (c) Finding molar mass from mass and a given number of moles If you are told: “0.0250 mol of ‘X’ has a mass of 1.775 g” 1.775 g Then: molar mass = 0.0250 mol = __________ (d) Finding molar mass from of the molar mass if given as a multiple of a known molar mass If you are told: “X” has a molar mass which is 1.64 times that of CO2 Then: molar mass of CO2 = __________ And: molar mass of “X” = 1.64 x 44.0 g/mol = __________ EXAMPLE V.11 MOLECULAR FORMULA Problem: The empirical formula of a compound is SiH3. If 0.0275 mol of the compound has a mass of 1.71 g, what is the compound’s molecular formula? Solution: Empirical mass Molar mass of SiH3 = __________ + __________ = __________ Molar mass of compound Molar mass = _______________ = __________ Molecular formula N = __________ = _____ _____ x __________ = __________ UNIT V THE MOLE CONCEPT 23 E. Molar Concentrations 1. SOLUTIONS are ________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Most solutions contain a solid (_______________) dissolved in a liquid (_______________); however, there are solutions of gases as well. The CONCENTRATION of a substance in solution _____________________________ ________________________________________________________________________ ________________________________________________________________________ Chemists use the “mole” to describe the amount of substance in a solution. _______________ ____________________ or ____________________ of a substance is the _____________________________________________________________________ e.g. If 2.0 L of solution contain 5.0 mol of NaCl, what is the molarity of the NaCl? molar concentration = _______________ = __________ mol The unit symbol for L is “M” When expressed in words, the unit symbol “M” is written as “_______________”. The short-hand symbol for “______________________________” is a set of brackets: […], so [NaCl] means the “________________________________________” 24 2. CHEMISTRY 11 The definition of molar concentration leads to the following equation: moles n molar concentration = volume or M = V mol M mol V M mol V M mol V M V c = molar concentration, in mol/L, n = number of moles, V = volume, in litres EXAMPLE V.12 Problem: Solution: CALCULATING CONCENTRATION What is the [NaCl] in a solution containing 5.12 g of NaCl in 250.0 mL of solution? In order to find molarity (c), the moles (n), and volume are needed. The volume is given and the mass must be converted to moles. moles of NaCl = __________ x __________ = __________ n [NaCl] = M = V = _______________ = __________ EXAMPLE V.13 Problem: Solution: CALCULATING MASS CONTAINED IN SOLUTIONS What mass of NaOH is contained in 3.50 L of 0.200 M NaOH? The molarity (c) and volume (V) are given so moles can be found. Moles can then be converted to mass. n solving M = V for n gives M = c • V moles of NaOH = _______________ x __________ = __________ mass of NaOH = _______________ x __________ = __________ UNIT V THE MOLE CONCEPT EXAMPLE V.14 25 CALCULATING CONCENTRATION FROM DENSITY Problem: What is the molarity of pure sulphuric acid, H2SO4, having a density of 1.839 g/mL? Solution: Since both density and molarity both have units of amount/volume, density = amount (as mass) volume and molarity = amount (as moles) volume therefore all we need to do is convert from mass to moles using molar mass. [H2SO4] = _______________ x __________ = __________ F. Dilution Calculations 1. When two solutions are mixed, ______________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (total moles of chemical) molarity of mixture = (total volume of mixture) Consider the following dilution, 26 CHEMISTRY 11 Mole (in itia l) = Mi x V i H 2O Mole (fi nal) = Mf x V f Diluted Solution Stock Solution since the ________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ M1V1 = M2V2 this equation can be rearranged to give the following dilution equation M2 = EXAMPLE V.15 Problem: M1 x V1 V2 SIMPLE DILUTION CALCULATIONS If 200.0 mL of 0.500 M NaCl is added to 300.0 mL of water, what is the resulting [NaCl] in the mixture. Solution: M2 = M1 x V1 V2 [NaCl] = __________ x _______________________ = __________ EXAMPLE V.16 MAKING DILUTE SOLUTIONS FROM CONCENTRATED SOLUTIONS Problem: What volume of 6.00 M HCl is needed to make 2.00 L of 0.125 M HCl? Solution: Since UNIT V THE MOLE CONCEPT 27 M1V1 = M2V2 then, VHCl = _________________________ = __________ 2. When two solutions having different concentrations of the same chemical are mixed, ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ EXAMPLE V.17 Problem: Solution: MIXING SOLUTIONS OF DIFFERENT CONCENTRATION If 300.0 mL of 0.250 M NaCl is added to 500.0 mL of 0.100 M NaCl, what is the resulting [NaCl] in the mixture? For each solution, the diluted concentration is given by M2 = M1 x V1 V2 300.0 mL [NaCl]first solution = __________ x (300.0 + 500.0) mL = __________ 500.0 mL [NaCl]second solution = __________ x (300.0 + 500.0) mL = __________ [NaCl]total = [NaCl]first solution + [NaCl]second solution [NaCl]total = _______________ + _______________ = __________