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Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Presentation on theme: "Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

1 Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = amu 16 O = amu Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams 3.1

3 Natural lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) 7.42 x x = amu 3.1 Average atomic mass of lithium:

4 Average atomic mass (6.941)

5 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly grams of 12 C mol = N A = x Avogadro’s number (N A )

6 1 O 2 molecule 1 mol O 2 molecules How many atoms of He are in 5.36 moles of He? 5.36 mol He x x atoms He 1 mol He = 3.23 x atoms He How many atoms of O are in 5.36 moles of O 2 ? 5.36 mol O 2 molecules x x molecules O 2 x 6.46 x atoms O 2 atoms O =

7 Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = x atoms = g 1 12 C atom = amu 1 mole 12 C atoms = g 12 C 1 mole lithium atoms = g of Li For any element atomic mass (amu) = molar mass (grams) 3.2

8 One Mole of: C S Cu Fe Hg 3.2

9 1 amu = 1.66 x g or 1 g = x amu 1 12 C atom amu x g x C atoms = 1.66 x g 1 amu 3.2 M = molar mass in g/mol N A = Avogadro’s number

10 Do You Understand Molar Mass? How many atoms are in g of potassium (K) ? 1 mol K = g K 1 mol K = x atoms K g K 1 mol K g K x x x atoms K 1 mol K = 8.49 x atoms K 3.2

11 Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S32.07 amu 2O+ 2 x amu SO amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = amu 1 mole SO 2 = g SO 2 3.3

12 Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1N amu 3O+ 3 x amu HNO amu 1 molecule HNO 3 = amu 1 mole HNO 3 = g HNO HNO 3 1H1.008 amu

13 2 N2 x amu 6 O+ 6 x amu Ba(NO 3 ) amu 1 Ba(NO 3 ) 2 formula unit = amu 1 mole Ba(NO 3 ) 2 = g Ba(NO 3 ) Ba(NO 3 ) 2 1Ba amu - OR - Ba amu 2 NO x amu Ba(NO 3 ) amu How many atoms are in 1.00 mol of Ba(NO 3 ) 2 ? 1 Ba(NO 3 ) 2 1 mol Ba(NO 3 ) 2 x x Ba(NO 3 ) 2 x 5.42 x atoms 9 atoms= 1.00 mol Ba(NO 3 ) 2

14 Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol H = x atoms H 5.82 x atoms H mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x x H atoms 1 mol H atoms x =

15 KE = 1/2 x m x v 2 v = (2 x KE/m) 1/2 F = q x v x B 3.4 Light Heavy Similar to JJ Thomson’s Experiment + Very small amount of sample Different masses deflect differently Can look at atoms, molecules or fragments of molecules. Can see isotopes of atoms.

16 Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C2H6OC2H6O %C = 2 x (12.01 g) g x 100% = 52.14%H = 6 x (1.008 g) g x 100% = 13.13%O = 1 x (16.00 g) g x 100% = 34.73% 52.14% % % = 100.0% 3.5

17 Empirical Formulas Weight % of ElementsRatio of Elements 1 - Start with weight percent of elements (make sure they total to 100%). 2 - Assume 100 gram sample and convert weight percent of each element to grams of each element. 3 - Convert to moles of each element using atomic weights. 4 - Calculate relative ratio of elements by dividing by the smallest number of moles. 5 - Multiply to get rid of any fractions, if necessary. When in doubt, convert to moles !!!

18 A compound is found to contain: 20.2 wt. % Al and 79.8 wt. % Cl. What is its empirical formula? 1 - Make sure total is 100% % % = 100.0% 2 - Convert weight percent of each element to grams of each element g Al and 79.8 g Cl 3 - Convert to moles of each element using atomic weights g Al = mol Al 79.8 g Cl = 2.25 mol Cl g/mol Al g/mol Cl 4 - Calculate relative ratio of elements by dividing by the smallest number of moles. => mol Al is smaller mol Al = mol Cl = 3.01mol Cl / mol Al mol Al => AlCl 3 Round to 3

19 3.6 g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O

20 3.7 3 ways of representing the reaction of H 2 with O 2 to form H 2 O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction reactantsproducts

21 3.7 reactants products Starting materials Substances formed a A + b B c C + d D plus to yield Conservation of Mass Must have the same number of atoms on each side of the reaction !!

22 How to “Read” Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg grams O 2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 3.7

23 Balancing Chemical Equations 1.Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2.Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts C 2 H 6 NOT C 4 H 12

24 Balancing Chemical Equations 3.Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O 3.7 start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O

25 Balancing Chemical Equations 4.Balance those elements that appear in two or more reactants or products oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 = 7 oxygen on right C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 4.Remove all fractions (generally by multiplying everything by 2) and reduce all stoichiometric coefficients to try to get one equal to 1 (generally by dividing everything by 2).

26 Balancing Chemical Equations 5.Check to make sure that you have the same number of each type of atom on both sides of the equation C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 x 2)4 C 12 H (2 x 6)12 H (6 x 2) 14 O (7 x 2)14 O (4 x 2 + 6) Always show a check of your solution on a quiz or exam !!

27 Balancing Chemical Equations 1.Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. 2.Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 3.Start by balancing those elements that appear in only one reactant and one product. 4.Balance those elements that appear in two or more reactants or products. 4.Remove all fractions (generally by multiplying everything by 2) and reduce all stoichiometric coefficients to try to get one equal to 1 (generally by dividing everything by 2). 5.Check to make sure that you have the same number of each type of atom on both sides of the equation.

28 Stoichiometry Quantitative study of chemical reactions. How much in ? => How much out ? You may have used these principles baking cookies, doing carpentry, ……….. Chemicals react in ratios of moles, NOT in weight ratios !! When in doubt, convert to moles !!! CH O 2 CO H 2 O Get a set of relationships between all reactants and products: 1 mol CH 4 = 2 mol O 2 = 1 mol CO 2 = 2 mol H 2 O These are exact conversions !!!

29 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Mass Changes in Chemical Reactions 3.8 When in doubt, convert to moles !!!

30 CH O 2 CO H 2 O 1 mol CH 4 = 2 mol O 2 = 1 mol CO 2 = 2 mol H 2 O How many mole of CO 2 can be generated from moles of O 2 ? mol O 2 x 1 mol CO 2 = mol CO 2 2 mol O 2 How many grams of CH 4 are needed to produce 3.85 mol of CO 2 ? 3.85 mol CO 2 x 1 mol CH 4 x g CH 4 = 1 mol CO g CH 4 1 mol CH 4 MW CH 4 = (1.008) = g / mol CH 4

31 Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 235 g H 2 O 3.8

32 6 green used up 6 red left over Limiting Reagents 3.9 Who gets used up first? When one reactant runs out the reaction stops. Excess reagent is the reagent left over when the reaction is complete.

33 N H 2 2 NH 3 1 mol N 2 = 3 mol H 2 = 2 mol NH 3 How many mole of NH 3 can be generated by mixing 2.35 mol N 2 and 6.50 mol of H 2 ? 2.35 mol N 2 x 2 mol NH 3 = 4.70 mol NH 3 1 mol N mol H 2 x 2 mol NH 3 = 4.33 mol NH 3 3 mol H 2 Lower number H 2 is limiting reagent and the amount of NH 3 that can be generated is 4.33 mol

34 Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent 3.9

35 Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 3.9

36 Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 % 3.10 Did you get as much as you thought you were going to get ??? NOTE: Law of conservation of mass dictates that % Yield < 100 % Can calculate using weight or moles?

37 N H 2 2 NH 3 If the theoreticall yield is 4.33 mol NH 3 and 3.62 mol NH 3 was recovered, what is the % yield ? % Yield = Actual Yield Theoretical Yield x 100 % 3.62 mol NH mol NH 3 X 100 % = 83.6 %

38 2 Mg + O 2 2 MgO If 26.5 g Mg was reacted with an excess of O 2 and a 75.2 % yield was achieved, how much MgO was recovered ? 26.5 g Mg x 1 mol Mg = 1.09 mol Mg g Mg 2 mol Mg = 2 mol MgO 1.09 mol Mg x 2 mol MgO = 1.09 mol MgO theoretical yield 2 mol Mg 1.09 mol MgO x = mol MgO MW MgO = ( ) g/mol = g/mol MgO mol MgO x g MgO = 33.0 g MgO 1 mol MgO


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