Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Size-exclusion chromatography wikipedia , lookup
Computational chemistry wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Rate equation wikipedia , lookup
History of molecular theory wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Chapter 5 Chapter Three Calculations with Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows the chemist to carry out recipes for compounds based on the relative numbers of atoms involved. The calculation involving the quantities of reactants and products in a chemical equation is called stoichiometry. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 2 1 Chapter 5 The molecular weight (mass) of a substance is the sum of the atomic weights of all the atoms in a formula of the compound. For, example, a molecule of H2O contains 2 hydrogen atoms (at 1.008 amu each) and 1 oxygen atom (16.000 amu), giving a molecular weight of 18.016 amu. For example, one formula unit of NaCl contains 1 sodium atom (22.999 amu) and one chlorine atom (35.453 amu), giving a formula weight of 58.452 amu Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 3 Mass and Moles of a Substance The Mole Concept A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon-12. The number of atoms in a 12-gram sample of carbon-12 is called Avogadro s number (to which we give the symbol Na). The value of Avogadro s number is 6.02 x 1023. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 4 2 Chapter 5 Mass and Moles of a Substance The molar mass (Mm) of a substance is the mass of one mole of the substance. For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units. That is, one mole of any element weighs its atomic mass in grams. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 5 Calculate the molar mass of sucrose, C12H22O11 12 carbon atoms @ 12.01amu = 144.12 22 hydrogen atoms @ 1.01amu = 22.22 11 oxygen atoms @ 16.00amu = 176.00 Formula mass of sucrose Copyright © Houghton Mifflin Company. All rights reserved. = 342.34amu Presentation of Lecture Outlines, 3 6 3 Chapter 5 Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 7 Mole calculations Converting the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations. mass of " A" atomic (or molecular) mass of " A" moles of " A" Or nA = mA/mmA Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 8 4 Chapter 5 Mass and Moles of a Substance Mole calculations Suppose we have 100.0 grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? moles Fe 100.0 g Fe 55.8 g/mol 1.79 moles of Fe Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 9 Mass and Moles of a Substance Mole calculations Conversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass? mass Mg (5.75 moles) (24.3 g/mol) 140 grams of Mg Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 10 5 Chapter 5 Mass and Moles of a Substance Mole calculations Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = 180.0 g/mol). What is its mass? mass C6 H12O 6 ( 3.25 moles) (180.0 g/mol) 585 grams of C6 H12O 6 Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 11 1 mole of any substance contains 6.022×1023 particles of that substance. Avogadro s number (NA) = 6.022×1023 The molar mass of a compound is the formula mass expressed in grams Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 12 6 Chapter 5 Mass and Moles and Number of Molecules or Atoms The number of molecules or atoms in a sample is related to the moles of the substance: 1 mole HCl 6.02 10 23 HCl molecules 1 mole Fe 6.02 10 23 Fe atoms Suppose we have a 3.46-g sample of hydrogen chloride, HCl. How many molecules of HCl does this represent? 3.46g HCl 1 mole HCl 36.5g HCl 6.02 x 10 23 HCl molecules 1 mole HCl = 5.71x1022 HCl molecules Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 13 Which one of the following contains 1.20 1024 atoms? 1) 24.0 g O2 2) 4.00 g He 3) 42.0 g N2 4) 13.0 g C2H2 5) 8.0 g CH4 Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 14 7 Chapter 5 Determining Chemical Formulas The percent composition of a compound is the mass percentage of each element in the compound. We define the mass percentage of A as the parts of A per hundred parts of the total, by mass. That is, mass % " A" mass of " A" in whole 100% mass of the whole Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 15 Mass Percentages from Formulas Let s calculate the percent composition of butane, C4H10. First, we need the molecular mass of C4H10. 4 carbons @ 12.0 amu/atom 48.0 amu 10 hydrogens @ 1.00 amu/atom 10.0 amu 1 molecule of C4 H10 58.0 amu Now, we can calculate the percents. %C %H 48.0 amu C 58.0 amu total 10.0 amu H 58.0 amu total Copyright © Houghton Mifflin Company. All rights reserved. 100% 82.8%C 100% 17.2%H Presentation of Lecture Outlines, 3 16 8 Chapter 5 Question The mineral leadhillite, which is essentially Pb4(SO4)(CO3)2(OH)2 (FW = 1079 amu), contains ____ % oxygen by weight. 1) 10.4 2) 11.9 3) 13.3 4) 14.8 5) 17.8 Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 17 Calculate the mass percent of nitrogen in urea, CH4N2O. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 18 9 Chapter 5 Determining Chemical Formulas 1. Determining the formula of a compound from the percent composition. The percent composition of a compound leads directly to its empirical formula. An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 19 The empirical formula of a compound is the simplest whole number ratio of the atoms. The molecular formula is the actual number of each type of element in the compound. The molecular formula equals the empirical formula or is a whole number multiple of the empirical formula. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 20 10 Chapter 5 A compound contains 40.00% of sulfur and 60.00% of oxygen. What is the empirical formula? Step1 In 100 g of compound: m S = 100 g x 0.4000 = 40.00 g S m O = 100 g x 0.6000 = 60.00 g O Step 2 ns = 40.00 g S x 1 mol S = 1.247 32.065 g S nO = 60.00 g O x 1 mol O = 3.750 16.00 g O Divide both mole values by smallest 1.247/1.247 = 1 S 3.750/1.247 = 3 O Step 3 empirical formula is SO3 Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 21 The empirical formula of a carbohydrate is CH2O. If the molar mass is 180 g/mol, what is the molecular formula? The empirical formula mass is: 12.0 2 1.0 16.0 30.0 g/mol Dividing the empirical formula mass by the molar mass: 180 g/mol n= 6 30.0 g/mol CH2 O n C 6H12O 6 Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 22 11 Chapter 5 Table 3.1 Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 23 2) Determining the formula of a compound by a Combustion method. Measuring the amount of hydrogen and carbon in a compound CxHy + excess O2 Copyright © Houghton Mifflin Company. All rights reserved. y H2O + x CO2 2 Presentation of Lecture Outlines, 3 24 12 Chapter 5 Q? Ethylene glycol is used as an automobile antifreeze and its formula contains only C,H and O. Combustion of 6.38 mg of ethylene glycol gives 9.06 mg CO2 and 5.58 mg H2O. a) Calculate the empirical formula of ethylene glycol. b) If its molecular mass is 62.068 g/mole, what is its molecular formula? Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 25 Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. Stoichiometry is based on the balanced chemical equation and on the relationship between mass and moles. Such calculations are fundamental to most quantitative work in chemistry. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 26 13 Chapter 5 Molar Interpretation of a Chemical Equation The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as mole-to-mole relationships. For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen. N 2 (g ) 3 H 2 (g ) Copyright © Houghton Mifflin Company. All rights reserved. 2 NH 3 ( g ) Presentation of Lecture Outlines, 3 27 Haber Process Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 28 14 Chapter 5 Molar Interpretation of a Chemical Equation This balanced chemical equation shows that one mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. N 2 (g) 3H 2 (g) 1 molecule N2 1 mol N2 + + 3 molecules H2 3 mol H2 2 NH 3 ( g ) 2 molecules NH3 2 mol NH3 Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 29 Mole calculations Converting the number of moles of a given substance into its mass, and vice versa. moles of " A" mass of " A" atomic (or molecular) mass of " A" Using shorthand symbols: nA = mA/MmA or mA = nA x MmA where n = # of moles M= mass of compound A Mm = molecular mass of A Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 30 15 Chapter 5 Molar Interpretation of a Chemical Equation Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2. N 2 (g) 3H 2 (g) 2 NH 3 ( g ) Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple. 4.8 mol H 2 Copyright © Houghton Mifflin Company. All rights reserved. 2 mol NH 3 3 mol H 2 3.2 mol NH 3 Presentation of Lecture Outlines, 3 31 Ammonia, NH3, and oxygen can be reacted together in the presence of a catalyst to form only nitrogen Question monoxide and water. The number of moles of oxygen consumed for each mole of NO that is produced is 1) 0.625. 2) 1.25. 3) 2.50. 4) 3.75. 5) 5.00. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 32 16 Chapter 5 Mass Relationships in Chemical Equations Amounts of substances in a chemical reaction by mass. How many grams of HCl are required to react with 5.00 grams manganese(IV) oxide according to this equation? 4 HCl(aq) MnO 2 (s ) 2 H 2O(l) MnCl 2 (aq) Cl 2 (g ) Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 33 Steps in a Stoichiometric Calculation Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 34 17 Chapter 5 Mass Relationships in Chemical Equations First, you write what is given (5.00 g MnO2) and convert this to moles. Then convert to moles of what is desired (mol HCl). Finally, you convert this to mass (g HCl). 5.00 g MnO 2 1 mol MnO 2 86.9g MnO 2 4 mol HCl 1 mol MnO 2 36.5 g HCl 1 mol HCl 8.40 g HCl Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 35 Limiting Reactant The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reactant ultimately determines how much product can be obtained. For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 36 18 Chapter 5 Limiting Reactant Zinc metal reacts with hydrochloric acid by the following reaction. Zn(s) 2 HCl(aq) ZnCl 2 (aq) H 2 (g ) If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced? Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 37 Limiting Reactant Take each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiting reactant. 1 mol H 2 0.30 mol Zn 0.30 mol H 2 1 mol Zn 1 mol H 2 0.52 mol HCl 0.26 mol H 2 2 mol HCl Because HCl is the limiting reactant, the amount of H2 produced must be 0.26 mol. Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 38 19 Chapter 5 If 50.0 g of O2 are mixed with 50.0 g of H2 and the mixture is ignited, what mass of water is produced? 1) 50.0 g 2) 56.3 g 3) 65.7 g 4) 71.4 g 5) 100.0 g Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 39 Theoretical and Percent Yield The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants. The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). %Yield Copyright © Houghton Mifflin Company. All rights reserved. actual yield 100% theoretical yield Presentation of Lecture Outlines, 3 40 20 Chapter 5 Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall that the theoretical yield of H2 in the previous example was 0.26 mol (or 0.52 g) H2. If the actual yield of the reaction had been 0.22 g H2, then %Yield Copyright © Houghton Mifflin Company. All rights reserved. 0.22 g H 2 100% 42% 0.52 g H 2 Presentation of Lecture Outlines, 3 41 Part B: If the reaction yielded 14,69 g of carbon disulfide what is the % yield of the reaction? Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 42 21 Chapter 5 Operational Skills Calculating the formula weight from a formula or model. Calculating the mass of an atom or molecule. Converting moles of substance to grams and vice versa. Calculating the number of molecules in a given mass. Calculating the percentage composition from the formula. Calculating the mass of an element in a given mass of compound. Copyright © Houghton Mifflin Company. All rights reserved. Calculating the percentages C and H by combustion. Determining the empirical formula from percentage composition. Determining the molecular formula from percentage composition and molecular weight. Relating quantities in a chemical equation. Calculating with a limiting reactant. Presentation of Lecture Outlines, 3 43 Extra problems Copyright © Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 44 22 This document was created with Win2PDF available at http://www.daneprairie.com. The unregistered version of Win2PDF is for evaluation or non-commercial use only.