Mass and Moles of a Substance


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1 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
2 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 H 2 O contains 2 hydrogen atoms (at amu each) and 1 oxygen atom ( amu), giving a molecular weight of amu. For example, one formula unit of NaCl contains 1 sodium atom ( amu) and one chlorine atom ( amu), giving a formula weight of 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 carbon12. The number of atoms in a 12gram sample of carbon12 is called Avogadro s number (to which we give the symbol N a ). The value of Avogadro s number is 6.02 x Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 4 2
3 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, C 12 H 22 O carbon 12.01amu = hydrogen 1.01amu = oxygen 16.00amu = Formula mass of sucrose = amu Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 6 3
4 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 7 Mole calculations moles of " A" 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" Or n A = m A /mm A Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 8 4
5 Mass and Moles of a Substance Mole calculations Suppose we have grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? g Fe moles 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) 140 grams of (24.3 g/mol) Mg Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
6 Mass and Moles of a Substance Mole calculations Conversely, suppose we have 3.25 moles of glucose, C 6 H 12 O 6 (molecular wt. = g/mol). What is its mass? mass C6H12O6 (3.25 moles) (180.0 g/mol) 585 grams of C 6H12O6 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, mole of any substance contains particles of that substance. Avogadro s number (NA) = The molar mass of a compound is the formula mass expressed in grams Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
7 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: 1mole HCl 1mole Fe atoms Suppose we have a 3.46g sample of hydrogen chloride, HCl. How many molecules of HCl does this represent? 3.46g HCl Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, mole HCl 36.5g HCl Fe HCl molecules x 10 HCl molecules 1 mole HCl = 5.71x10 22 HCl molecules Which one of the following contains atoms? 1) 24.0 g O 2 2) 4.00 g He 3) 42.0 g N 2 4) 13.0 g C 2 H 2 5) 8.0 g CH 4 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
8 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 mass of the whole 100% 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, C 4 H 10. First, we need the molecular mass of C 4 H amu/atom amu/atom 1 molecule of C4H amu Now, we can calculate the percents amu C % C 58.0 amu total 10.0 amu H H % 58.0 amu total 100% 100% 48.0 amu 10.0 amu 82.8%C 17.2%H Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
9 Question The mineral leadhillite, which is essentially Pb 4 (SO 4 )(CO 3 ) 2 (OH) 2 (FW = 1079 amu), contains % oxygen by weight. 1) ) ) ) ) 17.8 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 17 Calculate the mass percent of nitrogen in urea, CH 4 N 2 O. Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
10 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,
11 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 = g S m O = 100 g x = g O Step 2 n s = g S x 1 mol S = g S Step 3 n O = g O x 1 mol O = g O Divide both mole values by smallest 1.247/1.247 = 1 S 3.750/1.247 = 3 O empirical formula is SO 3 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 21 The empirical formula of a carbohydrate is CH 2 O. If the molar mass is 180 g/mol, what is the molecular formula? The empirical formula mass is: g/mol Dividing the empirical formula mass by the molar mass: 180 g/mol n = g/mol CH O C H O 2 n Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
12 Table 3.1 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, ) Determining the formula of a compound by a Combustion method. Measuring the amount of hydrogen and carbon in a compound C x H y + excess O 2 y H 2 O + x CO 2 2 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
13 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 CO 2 and 5.58 mg H 2 O. a) Calculate the empirical formula of ethylene glycol. b) If its molecular mass is 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,
14 Molar Interpretation of a Chemical Equation The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as moletomole relationships. For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen. N2(g) 3 H2(g) 2 NH3(g) Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 27 Haber Process Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
15 Molar Interpretation of a Chemical Equation This balanced chemical equation shows that one mole of N 2 reacts with 3 moles of H 2 to produce 2 moles of NH 3. N 2(g) 3H 2(g) 2 NH 3(g) 1 molecule N molecules H 2 2 molecules NH 3 1 mol N mol H 2 2 mol NH 3 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 moles of " A" Converting the number of moles of a given substance into its mass, and vice versa. mass of "A" atomic (or molecular) mass of " A" Using shorthand symbols: n A = m A /Mm A or m A = n A x Mm A 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,
16 Molar Interpretation of a Chemical Equation Suppose we wished to determine the number of moles of NH 3 we could obtain from 4.8 mol H 2. (g) 3H (g) 2 NH (g) N2 2 3 Because the coefficients in the balanced equation represent moletomole ratios, the calculation is simple. 4.8 mol H 2 mol NH mol NH 3 3 mol H 2 Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 31 Ammonia, NH 3, and oxygen can be reacted together in the presence of a catalyst to form only nitrogen monoxide and water. The number Question of moles of oxygen consumed for each mole of NO that is produced is 1) ) ) ) ) Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
17 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) MnO2(s) 2 H2O(l) MnCl2(aq) Cl2(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,
18 Mass Relationships in Chemical Equations First, you write what is given (5.00 g MnO 2 ) and convert this to moles. Then convert to moles of what is desired (mol HCl). Finally, you convert this to mass (g HCl) g MnO 2 1 mol MnO 86.9g MnO 8.40 g HCl mol HCl 1 mol MnO g HCl 1 mol 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,
19 Limiting Reactant Zinc metal reacts with hydrochloric acid by the following reaction. Zn(s) 2 HCl(aq) ZnCl2(aq) H2(g) If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H 2 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 mol Zn 0.30 mol H 2 1 mol Zn 1 mol H mol HCl 0.26 mol H 2 2 mol HCl Because HCl is the limiting reactant, the amount of H 2 produced must be 0.26 mol. Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
20 If 50.0 g of O 2 are mixed with 50.0 g of H 2 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) 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 actual yield theoretical yield 100% Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
21 Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall that the theoretical yield of H 2 in the previous example was 0.26 mol (or 0.52 g) H 2. If the actual yield of the reaction had been 0.22 g H 2, then %Yield 0.22 g H 0.52 g H % 42% Copyright Houghton Mifflin Company. All rights reserved. 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,
22 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. 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. Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines, 3 43 Extra problems Copyright Houghton Mifflin Company. All rights reserved. Presentation of Lecture Outlines,
23 This document was created with Win2PDF available at The unregistered version of Win2PDF is for evaluation or noncommercial use only.
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