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CHEM 112 - Ch 3 - 1st 2023

Mass Relationships in Chemical Reactions

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CHEM 112 - Ch 3 - 1st 2023

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  1. CHEM 112 Mass Relationships in Chemical Reactions Chapter 3

  2. Objectives By the end of this chapter, you should be able to • know and calculate Atomic Mass • calculate Molecular Mass • calculate Percent Composition of Compounds • Determine the Empirical Formula • Find Limiting Reagents • calculate Reaction Yield 2

  3. Outline • Atomic Mass • Avogadro’s Number and Molar Mass of an Element • Molecular Mass • Percent Composition of Compounds • Experimental Determination of Empirical Formula • Limiting Reagents • Reaction Yield Chemistry by Chang, 10th ed, Chapter 3

  4. Atomic Mass Mass of an atom = mass of p + mass of n + mass of e – binding energy mass loss mass of e = & binding energy mass loss is very small Mass of atom  mass of p + mass of n The atom is too small to be weighted. However, we can determine the mass of one atom relative to another. Atomic mass is the mass of an atom in atomic mass units (amu) which also known as “Dalton”. amu is the mass that exactly equal to one-twelfth the mass of one carbon-12 (12C) atom. By definition: 1 atom 12C “weighs” 12 amu On this scale; 1H = 1.008 amu 16O = 16.00 amu

  5. Average Atomic Mass • Atomic mass of carbon in the periodic table is 12.01; not 12.00 • The carbon exists in more than one form in nature (isotopes 12C & 13C) • Carbon is a mixture of Isotopes • Thus, average atomic mass for all these istopes is used in the periodic table. Natural Abundance

  6. How to find the Average Atomic Mass? • Average atomic mass of any element is: ∑ (The natural abundance x Atomic Mass) for each isotope • Example:calculate the average atomic mass of carbon? C-12 natural abundance = 98.90% , atomic mass = 12 amu C-13 natural abundance = 1.10% , atomic mass = 13 amu 98.90 x 12 + 1.10 x 13 = 12.01 amu 100 Average atomic mass = • Notes :carbon is mainly C-12 → the Average Atomic Mass is 12.01 amu → closer to 12 amu than 13 amu

  7. Average atomic mass for C = 12.011 amu Extra practice: Example 3.1,Chemistry by Chang, 10th ed

  8. 7.42% x 6.015 + 92.58% x 7.016 100 Every isotope has a different atomic mass. Natural lithium is: 7.42% (Abundance) 6Li (6.015amu) 92.58% (Abundance) 7Li (7.016amu) Average atomic massof lithium = = 6.941 amu

  9. Do you understand the concept of Average atomic mass? Chlorine has two isotopes: Cl-35 and Cl-37. The atomic masses of Cl-35 and Cl-37 are 34.969 amu and 36.966 amu. If the average atomic mass of the chlorine atom is 35.453 amu, which isotope is more abundant in nature? Cl-35 Cl-37 They present equally in nature. Cannot be judged by the available information.

  10. The mole Dozen = 12 Micro World atoms & molecules Macro World grams 1 dozen = 12 of anything 1 mole = 6.022 x1023 of particles The mole (mol) is the base unit of amount of substance ("number of substance") in the International System of Units (SI), defined as exactly 6.02214076×1023 particles, e.g., atoms, molecules, ions or electrons. 1 mol = NA = 6.02214067  1023 Avogadro’s number (NA) Thus: One mole of H atoms has 6.022 x 1023atoms One mole of H2 molecules has 6.022 x 1023molecules

  11. Molar Mass • The atomic mass of 12C is 12.00 amu • 1 mole of carbon-12 atoms has a mass of exactly 12 g • Thus: the Molar Mass (M) of 12C = 12.00 g/mol Molar mass (M)is the mass (in grams or kilograms) of 1 mole of a substance. For any element atomic mass (amu) and molar mass (g/mol) are numerically the same. • Example: • Atomic mass of Na is 22.99 amu. Then, Molar mass is 22.99 g/mol • Atomic mass of P is 30.97 amu. Then, Molar mass is 30.97 g/mol

  12. = molar mass in g/mol M Did You Understand Molar Mass? ⸪ One mole = NA (1 mol = NA = 6.02214067  1023) ⸫ How to Calculate the Number of Moles (n) NA= Avogadro’s number N NAn m Mn

  13. Example 1. How many atoms are in 0.551 g of potassium (K) ? = = 0.0141 mol N = n x NA N = 0.0141  6.022  1023 = 8.49  1021 atoms K Extra practice: Example 3.2, 3.3 & 3.4, Chemistry by Chang, 10th ed

  14. Example 2 : Calculate the number of atoms in 6.46 grams of helium (He)? The molar mass of He is 4 g/mole First calculate number of moles of He: n = = 1.62 mole N = NA n = 6.022  10+23 1.62 = 9.73  1023 atoms m Mn N NAn

  15. 32.07 amu 1S 64.07 amu SO2 Molecular Mass The molecular mass(or molecular weight) is the sum of the atomic masses (in amu) in the molecule. 2O + 2 x 16.00 amu SO2 For any molecule molecular mass (amu) and molar mass (g/mol) are numerically the same. 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

  16. Example 3.6: How many moles of CH4 are present in 6.07 g of CH4? Molecular Mass CH4= (1  12 .01) + (4  1.008) = 16.04 g/mol =

  17. Example 5: How many hydrogen atoms are present in25.6 g of urea [(NH2)2CO]. The molar mass of urea is60.06 g/mol? We calculate the number of molecules of urea and use it to calculate the number of hydrogen atoms in the sample. nurea = Nurea= n  NA= 0.426x 6.022 x 1023 = 2.57 x1023 molecules of urea From the chemical formula of urea (NH2)2CO: 1 molecule of urea  4 atom of H 2.567 x 1023 molecules  ? Atom of H ? = 4  2.57  1023 = 1.03 x1024 atoms of H N NAn m Mn

  18. Different Strategies to solve Example 5 How many hydrogen atoms are present in25.6 g of urea [(NH2)2CO]. The molar mass of urea is60.06 g/mol? N ? NAn 0.426 m 25.6g M n 60.06 ? 1 molecule of urea  4 atoms of H 2.57 x1023 molecules  ? 1 mole of urea  4 moles of H 0.426 mole of urea  ? m 25.6g M n 60.06 ? N ? NAn 1.704

  19. What is the mass of one copper atom in grams? To solve the problem, consider: m = ? g - M = 63.55 g/mol – n = ? mol – N = 1 atom – NA = 6.022 x 1023 . Practice problem: How much grams are in 1 amu? m Mn N NAn m =1.66 x 10-22× 63.55 = 1.06 x 10-22 g n = 1 / 6.022 x 1023 = 1.66 x 10-22 Mass of one atom =

  20. Percent Composition of Compounds • The percent composition by mass isthe percent by mass of each element in a compound. Percent composition of an element in a compound = x 100% nis the number of moles of the element in 1 mole of the compound • The Percent Composition by Mass is important to determine: • → The Empirical Formula • → The Molecular Formula

  21. 2 x (12.01 g) 1 x (16.00 g) 6 x (1.008 g) %C = %O = %H = x 100% = 13.13% x 100% = 34.73% x 100% = 52.14% 46.07 g 46.07 g 46.07 g C2H6O Example : Calculate the percentage of each element in C2H6O? Molar mass of C = Atomic mass of C = 12.01 g /mol Molar mass of H = Atomic mass of H = 1.008 g/mol Molar mass of O = Atomic mass of O = 16.00 g/mol Molar mass of C2H6O = Molecular mass C2H6O = (2 x 12.01) + (6 x 1.008) + (1 x16.00) = 46.07 g/mol 52.14% + 13.13% + 34.73% = 100.001% Extra practice: Example 3.8, Chemistry by Chang, 10th ed

  22. Percent Composition & Empirical Formulas Precent composition can be obtained experimentally by elemental analysis using elemental analysers, mass spectrometry or other techniques. To determine the empirical formula from the percentage of elements in the compound: 1- change% to g 2- change g to mole. 3- divide by the smallest number of moles. 4-if there was fraction after division change to integer subscripts (multiply by 2, 3 or 4 etc until reaching integer).

  23. Percent Composition and Empirical Formulas Example: Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75%, Mn 34.77% and O 40.51%. Extra practice: Example 3.9, Chemistry by Chang, 10th ed

  24. Empirical formula C3H4O3

  25. Percent Composition and Molecular Formulas To determine the molecular formula from the empirical formula of the compound: 2- calculate the molecular mass of the empirical formula. 3- calculate the ratio between molecular formula and empirical formula as following: 4- The molecular formula = (empirical formula)Ratio Present Composition by Mass ↓ Empirical Formula ↓ Molecular Formula

  26. Percent Composition and Molecular Formulas Example 3.11 A sample compound contains 1.52g of N and 3.47g of O. The molar mass of this compound is between 90g and 95g. Determine the molecular formula. Present Composition by Mass ↓ Empirical Formula ↓ Molecular Formula • The molar mass of the empirical formula (NO2) = 14.01 + (2x16.00) = 46.01g • The ratio between the empirical formula and the molecular formula: You can chose any number between 90 and 95 • The molecular formula = (empirical formula)Ratio (NO2)2 = N2O4

  27. Chemical Reactions and Chemical Equations • Chemical Reaction: is a process in which one or more substances is changed into one or more new substances. • Chemical Equation:It is symbolic representation, describing chemical reaction in the form of symbols and formulae. Reactants  Products

  28. 3 ways of representing the reaction of H2 with O2 to form H2O

  29. How to “Read” Chemical Equations? 2 Mg + O2 2 MgO Read it! a) Reaction of 2 atoms of Mg and 1 molecule of O2 produces 2 molecules of MgO b) Reaction of 2 mole of Mg and 1 mole of O2 produces 2 mole of MgO c) From the periodic table: Reaction of 48.6 grams Mg and 32.0 grams O2 produces 80.6 g MgO Calculating the mass from the number of moles Reaction of 2 grams of Mg and 1 gram of O2 produces 2 g of MgO

  30. C2H6 + O2 CO2 + H2O Balancing Chemical Equations • Identify all reactants and products and write their correct formula on the left side and right side of the equation. • Example. Ethane reacts with oxygen to form carbon dioxide and water • 2. Begin balancing by change the numbers in front of the formulas (coefficients) to make the number of atoms in each element the same on both sides of the equation. Do not change the subscripts. Example. NO2 when multiply by 2: 2 NO2 not N2O4 • Start by balancing those elements that appear in only one reactant and one product. • Balance those elements that appear in two or more reactants or products. • Check to make sure that you have the same number of each type of atom on both sides of the equation.

  31. Balancing Chemical Equations Example 1: Balance the following equation: C5H12 + O2 CO2 + H2O 1. Identify all reactants and products and write their correct formula on the left side and right side of the equation. 2. Start by balancing those elements that appear in only one reactant and one product. So, we should start with C or H but not with O C5H12 + O2 CO2 + H2O C5H12 +O25 CO2 + H2O C5H12 +O25 CO2 + 6 H2O 5 C on the left 1 C on the right  5 12 H on the left 2 H on the right  6

  32. 2 O on the left 52 O 6 O Balancing Chemical Equations 3. Balance those elements that appear in two or more reactants or products C5H12 +O2 5 CO2 + 6H2O C5H12 + 8 O25 CO2 + 6 H2O = 16 O on the right  8

  33. Reactants Products 5 C 5 C 12 H 12 H 16 O 16 O Balancing Chemical Equations 4. Check to make sure that you have the same number of each type of atom on both sides of the equation. C5H12 + 8 O2 5 CO2 + 6 H2

  34. Homework 1. What is the coefficient of H2O when the equation is balanced: _ Al4C3 + _ H2O  _ Al(OH)3 + 3CH4 • 13 • 4 • 6 • 12 2. What are the coefficients of Al4C3 ,H2O and Al(OH)3,respectively, when the equation is balanced: _ Al4C3 + _ H2O  _ Al(OH)3 + 3CH4 • 4,1,5 • 1,12,4 • 1,24, 4 • 4,12,1

  35. Amounts of Reactants and Products Two basic question in chemical laboratory: • How much product will be formed from specific amounts of starting materials (reactant)? 6.0 g reactant → ? product • Howmuch starting materials must be used to obtain a specific amount of product? ? reactant → 6.0 g product

  36. Amounts of Reactants and Products To relate the masses of two material in a reaction, you have to:. 1- write the balanced equation for the reaction 2- convert the given amount of compound A to moles. 3- use the mole ratio (coefficients) from the balanced equation to calculate the number of moles of compound B. 4- convert the number of moles of compound B to grams.

  37. 2 CH3OH + 3 O2 2 CO2 + 4 H2O M 32.0 g/mol Methanol burns in air according to the equation below. If 209 g of methanol are used up in the combustion, what mass of water is produced? Mass 209 g Mass ? g 235 g m 209 g n= = m = n x M = 13.0625 x 18 mole 13.0625 mole 6.53 From the equation 2 mol of CH3OH  4 mol of water 6.53 mol of CH3OH  ? Mol of water

  38. Limiting Reagent 2NO + O2 2NO2 • Limiting Reagentis the reactant used up completely in a reaction and thus determine the amount of product. • Excess Reagent is the reactant present in quantities greater than necessary to react with the quantity of the limiting reagent (the one that is left at the end of the reaction). • Limiting reagent is in a reaction of more than one reactant! NO is the limiting reagent O2 is the excess reagent

  39. Questions in Limiting Reagent: First:we have to determine which reactant is the limiting reagent and which is the excess reagent! Second:after we know which one is the limiting reagent, we could determine the amount of the product!! Third:after we know the excess reagent, we could determine how much excess of it is left at the end of the reaction!!!

  40. Do You Understand Limiting Reagents? 2Al + Fe2O3 2Fe + Al2O3 2Al + Fe2O3 Al2O3 + 2Fe M M 27 g/mol 160 g/mol 1 2 In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. Mass 601 g Mass 124 g Mass ? g 234.1 g m m n= n= m = n x M = 2.295 x 102 601 g 124 g 4.59 mol Al 3.76 mol Fe2O3 < mole 3.76 Limiting reagent < Excess Reagent mole 4.59 mole 2.295 From the equation 2 mol of Al  1 mol of Al2O3 4.59 mol of Al  ? Mol of Al2O3

  41. M M 17 g/mol 44 g/mol Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide. In one process, 637.2g of NH3 are treated with 1142 g of CO2 (A) which of the reactants is the limiting reagent (B) calculate the mass of urea formed (C) How much excess reagent (in gram) is left at the end of the reaction. 2 NH3 + CO2 (NH2)2CO + H2O Mass 1142 g Mass 637.2 g Mass ? g 1122 g m m n= n= m = n x M = 18.7 x 60 1142 g 637.2 g (37.41/2)<(25.95/1) mole 25.95 Limiting reagent < Excess Reagent mole 37.41 mole 18.7 From the equation 2 mol of NH3 1 mol of urea 37.41 mol of NH3 ? mol of urea

  42. M M 17 g/mol 44 g/mol 2 NH3 + CO2 (NH2)2CO + H2O Mass 1142 g Mass 637.2 g m excess CO2 = 25.95 mol– 18.7 = 7.25 mol m of CO2 = n x M = 7.25 x 44 = 319.1 g m n= n= 1142 g 637.2 g mole 25.95 Provided in the reaction mole 37.41 mole 18.7 2 mol of NH3 1 mol of CO2 Needed in the reaction 37.41 mol of NH3 ? mol of CO2

  43. Homework 1.When 22.0 g NaCl and 21.0 g H2SO4 are mixed and react according to the equation below, which is the limiting reagent?2NaCl + H2SO4 Na2SO4 + 2HCl (a) NaCl (b) H2SO4 (c) Na2SO4 (d) HCl (e) No reagent is limiting.

  44. % Yield = x 100 Actual Yield Theoretical Yield Reaction Yield • Theoretical Yieldis the amount of product that would result if all the limiting reagent reacted. • Actual Yieldis the amount of product actually obtained from a reaction. • Normally the actual yield is less than theoretical d. • To determine how efficient a given reaction is, we calculate the percent yield. • Normally actual yield is given in the question • We calculate the theoretical yield from the limiting reagent.

  45. Titanium is a strong, lightweight, corrosion-resistance metal that is used in rockets, aircraft, jet engines, and bicycle frames. Its prepared by the reaction of titanium (IV) chloride with molten magnesium between 950 °C and 1150 °C: TiCl4 + 2Mg Ti + 2MgCl2 In a certain industrial operation 3.54 x 107 g of TiCl4 are reacted with 1.13 x 107 g of Mg. Calculate the theoretical yield of Ti in grams. Calculate the percent yield if 7.91 x 106 g of Ti are actually obtained. Theoretical yield of Ti = 8.95106 g of Ti

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