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Mass Relationships in Chemical Reactions

Mass Relationships in Chemical Reactions. 3.1 Atomic Mass 3.2 Avogadro’s Number and the Molar Mass of an Element 3.3 Molecular Mass 3.4 The Mass Spectrometer 3.5 Percent Composition of Compounds 3.6 Experimental Determination of Empirical Formulas 3.7 Chemical Reactions and

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Mass Relationships in Chemical Reactions

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  1. Mass Relationships inChemical Reactions 3.1 Atomic Mass 3.2 Avogadro’s Number and the Molar Mass of an Element 3.3 Molecular Mass 3.4 The Mass Spectrometer 3.5 Percent Composition of Compounds 3.6 Experimental Determination of Empirical Formulas 3.7 Chemical Reactions and Chemical Equations 3.8 Amounts of Reactants and Products 3.9 Limiting Reagents 3.10 Reaction Yield

  2. 3.1 Atomic MassThe mass of an atom depends on the number of electrons, protons, and neutrons it contains.By international agreement, atomic mass (sometimes called atomic weight ) is themass of the atom in atomic mass units (amu). One atomic mass unit is defined as amass exactly equal to one-twelfth the mass of one carbon-12 atom. Carbon-12 is thecarbon isotope that has six protons and six neutrons.3.2 Avogadro’s Number (mole) and the Molar Mass of an Element In the SI system the mole (mol) is the amount of a substance that contains a number of atoms, ( molecules, or other particles) as there are atoms in exactly12 g (or 0.012 kg) of the carbon-12 isotope. The actual number of atoms in 12 g ofcarbon-12 is determined experimentally. This number is called Avogadro’s number N A = 6.022 x 10 231 mole of hydrogen atoms contains 6.022 x 1023 H atoms.Figure 3.1 shows samples containing 1 mole each of several common elements.

  3. eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams)

  4. = molar mass in g/mol M Figure 3.2 The relationships between mass (m in grams) of an element and number of moles of an element (n) and between number of moles of an element and number of atoms (N) of an element. .m is the molar mass (g/mol) of the element and NA is Avogadro’s number NA= Avogadro’s number

  5. We have seen that 1 mole of carbon-12 atoms has a mass of exactly 12 g and contains 6.022 3 1023 atoms. This mass of carbon-12 is its molar mass ( M), define d as the mass (in grams or kilograms) of 1 mole of units (such as atoms or molecules) of a substance.Note that the molar mass of carbon-12 (in grams) is numerically equal to its mass in amu. Likewise, the atomic mass of sodium (Na) is 22.99 amu and its molar mass is 22.99 g; the atomic mass of phosphorus is 30.97 amu and its molar mass is 30.97 g; and so on. If we know the atomic mass of an element, we also know its molar mass.

  6. Do You Understand Molar Mass? How many atoms are in 0.32 g of Sodium (Na) ? N (atoms)=? m =0.32 g From periodic table N (atoms) = 0.32 (g)×6.022×1023/23 (g/mol) = 8.38 x 1021 atoms Na

  7. 1 mol K 6.022 x 1023 atoms K x x = 1 mol K 23.00 g K How many atoms are in 0.32 g of Sodium (Na) ? 1 mol K = 23.00 g K 1 mol K = 6.022 x 1023 atoms K 0.32 g K 8.38 x 1021 atoms Na 7

  8. n (mole)=? m =6.46 g From periodic table n (mol) = m (g) / M (g/mol) n = 6.46 / 4.003 = 1.61 mol He

  9. m (g)=? n=0.356 mole m (g) = n (mol) × M (g/mol) m = 0.356 × 65.39 = 23.28 g Zn طريقة أخرى Solution:1 mol Zn = 65.39 g Zn0.356 mole = ? G Znmass of Zn = 0.356x65.39= 23.28 g 1

  10. N (atom)=? m =16.3g N (atoms) = m (g) × NA (atoms) / M (g/mol) N = 16.3 × 6.022×1023 / 32.07 = 3.06×1023 S atoms

  11. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu Molecular mass(or molecular weight) is the sum of the atomic masses (in amu) in a molecule. element I element II Molecular mass= (No. of atoms× atomic mass) +( No. of atoms× atomic mass) For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

  12. a- SO2 = 32.07 amu + 2(16.00 amu) = 64.07 amu b- C8H10N4O2 = 8(12.01 amu) + 10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu) = 194.20 amu

  13. No. of moles ? CH4 m= 6.09 g n (mol) = m (g) / M (g/mol) = 6.07 / (12.01 + 4(1.008)) = 0.378 mol CH4

  14. No. of H atoms ? m= 25.6 g (NH2)2CO M = 60.06 g/mol N (atoms) = m (g) × NA (atoms) / M (g/mol) = 25.6 × 6.022×1023 × 4 / 60.06 = 1.03 ×1024 H atoms

  15. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? No. of H atoms ? m= 72.5 g C3H8O calculate M g/mol 60.09 N (atoms) = m (g) × NA (atoms) / M (g/mol) = 72.5 × 6.022×1023 × 8 / 60.09 = 5.81 × 1024 H atoms No. of H atoms in molecule

  16. 8 mol H atoms 6.022 x 1023 H atoms 1 mol C3H8O x x x = 1 mol C3H8O 1 mol H atoms 60 g C3H8O How many H atoms are in 56.8 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 56.8 g C3H8O 4.56x 1024 atoms H

  17. It must be noted that: Atomic mass Atoms Molar mass (g/mol) Al, As, Ni, H كتلة ذرية (من الجدول الدوري) molecular mass molecules HCl, H2, O2, NaCl كتلة جزيئية

  18. Percent compositionof an element in a compound = is the percent by mass of each element in a compound 1 x (16.00 g) 6 x (1.008 g) 2 x (12.01 g) n x molar mass of element %O = %H = %C = x 100% = 13.13% x 100% = 34.73% x 100% = 52.14% x 100% 46.07 g 46.07 g 46.07 g molar mass of compound C2H6O n is the number of moles of the element in 1 mole of the compound Molar mass= 2(12.01)+ 6(1.008)+ 16.0= 46.07 g 52.14% + 13.13% + 34.73% = 100.0% Check the answer!

  19. 3 x (1.008 g) 4 x (16.00 g) 30.97 g %H = %P = %O = x 100% = 3.086% x 100% = 31.61% x 100% = 65.31% 97.99 g 97.99 g 97.99 g Molar mass of H3PO4 = 3(1.008) + 1(30.97) + 4(16.00) = 97.99 g

  20. 2 Al 2x 26.98 3 S 3x 32.06 12 O + 12 x 16.00 3.3 Molecular Mass The molecular mass (sometimes called molecular weight ) is the sum of the atomic masses (in amu) in the molecule. For example, the molecular mass of H2O is 2(atomic mass of H) + 1 atomic mass of O or 2(1.008 amu) + 16.00 amu = 18.02 amu What is the formula mass of Al2(SO4)3? 342.14 amu

  21. Ex. 2 In 1 mole of hydrogen peroxide (H 2 O 2 ) there are 2 moles of H atoms and 2 moles of O atoms. The molar masses of H 2 O 2 , H, and O are 34.02 g, 1.008 g, and 16.00 g, respectively. Therefore, the percent composition of H 2 O 2 is calculated as follows: The sum of the percentages is 5.926% + 94.06% = 99.99%.

  22. 1Na 22.99 amu NaCl 1Cl + 35.45 amu NaCl 58.44 amu Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl

  23. 3 x 40.08 3 Ca 2 P 2 x 30.97 8 O + 8 x 16.00 What is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 310.18 amu

  24. 3.7 Chemical Reactions and Chemical Equations A process in which one or more substances is changed into one or more new substances is a chemical reaction reactants products A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H2 with O2 to form H2O

  25. How to “Read” Chemical Equations قراءة المعادلة الكيميائية 1- عدد الذرات أو الجزئيات 2- عدد المولات للذرات أو الجزئيات)) 3- الكتلة المولارية

  26. IS NOT How to “Read” Chemical Equations 2 Mg + O2 2 MgO √ 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO √ √ 2(24.31) 2(16.0) 2[24.31+16.0] reactant =80.6 Product = 80.6 X 2 grams Mg + 1 gram O2 makes 2 g MgO

  27. C2H6 + O2 CO2 + H2O NOT 2C2H6 C4H12 Balancing Chemical Equations نكتب الصيغه الصحيحه لكل متفاعل (على الطرف الايسر) ولكل ناتج (على الطرف الايمن) • 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 وزن المعادله الكيميائيه يكون بتغير الارقام التي بجانب الصيغه وليست التي تحتها بحيث يكون للعنصر نفس العدد على طرفي المعادله. • 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.

  28. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O Balancing Chemical Equations توزن أولا العناصر الاقل ظهورا • Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 multiply H2O by 3

  29. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O 2CO2 + 3H2O C2H6+ O2 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 Balancing Chemical Equations ثم توزن العناصر الاكثر ظهورا • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2 2

  30. 4 C (2 x 2) 4 C Reactants Products 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) 4 C 4 C 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O Balancing Chemical Equations الخطوه الاخيره هي التأكد من ان لديك نفس العدد من الذرات لكل عنصر على طرفي المعادله 5.Check to make sure that you have the same number of each type of atom on both sides of the equation.

  31. Amounts of Reactants and Products Reactant Product Use gram ratio of A and B From balanced equation 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

  32. 2CH3OH + 3O2 2CO2 + 4H2O Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? 4×Molar mass = 4[(2 × 1.008)+ (1 ×16.00)] 2×Molar mass = 2[(1 × 12.01)+(4 × 1.008)+ (1 × 16.00)] 64 g of CH3OH 72 g of H2O 209 g x g x = 72×209/64 = 235 g H2O

  33. 2CH3OH + 3O2 2CO2+ 4H2O grams CH3OH moles CH3OH moles H2O grams H2O 4 mol H2O 18.0 g H2O 1 mol CH3OH x = x x 2 mol CH3OH 32.0 g CH3OH 1 mol H2O Methanol burns in air according to the equation If 123 g of methanol are used up in the combustion, what mass of water is produced? molar mass CH3OH molar mass H2O coefficients chemical equation 123 g CH3OH 138.38 g H2O

  34. limiting reagents Excess Reagents Limiting Reagent

  35. 2NO + 2O2 2NO2 Limiting Reagents is the reactant used up first In a reaction Excess reagents are the Reactants present in quantities Greater than necessary to react With the present quantity of The limiting reagent. NO is the limiting reagent O2 is the excess reagent 3.9

  36. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3needed g Fe2O3 needed g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160. g Fe2O3 1 mol Al = x x x 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. OR 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

  37. Identify the Limiting Reactant تحديد الكاشف المحدد .1-حساب عدد المولات للمواد المتفاعلة وذلك بقسمة الوزن بالجرام على الكتلة المولارية 2- تقسم عدد مولات كل مادة متفاعلة على عدد مولاتها الموجودة في المعادلة. 3- المادة اللي تكون ناتجها أقل هي الكاشف المحدد. Calculate the amount of product obtained from the Limiting Reactant . حساب كمية المادة الناتجة بناء على الكاشف المحدد 4- يستخدم الكاشف المحدد لحساب كمية المادة الناتجة المطلوبة تبعا للمعادلة الموزونة

  38. Do You Understand Limiting Reagents? 2Al + Fe2O3 Al2O3 + 2Fe In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. n (mol) = m (g) / M (g/mol) nAl = 124 / 26.98 = 4.596 mol nFe2O3 = 601 / 160 3.756 mol نسبة كل مركب Al = 4.596/2 = 2.298 Fe2O3 = 3.756/1 = 3.756 الأصغر عدديا هو الكاشف المحدد limiting reagent وبناء على النتائج فإن Al هو الكاشف المحدد

  39. لحساب كمية Fe2O3 نستخدم الكاشف المحدد 53.96 g of Al 101.96 g of Al2O3 124 g x g x = 124×101.96/53.96 = 234.3 g Al2O3

  40. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 1 mol Al x 27.0 g Al 1 mol Al2O3 102. g Al2O3 = x x 2 mol Al 1 mol Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. 234 g Al2O3 124 g Al At this point, all the Al is consumed and Fe2O3remains in excess.

  41. Reaction Yield • Quantities of product calculated represent the maximum amount obtainable (100 % yield) • Most chemical reactions do not give 100 % yield of product because of: • 1-Side reactions (unwanted reactions) • 2-Reversible reactions ( reactants products) • 3-Losses in handling and transferring

  42. % 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. The percent Yield is the proportion of the actually yield to the theoretical yield which can be obtained from the following relation: 3.10

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