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Lecture 4: (Lec 3B) Stoichiometry Sections (Zumdahl 6th Edition) 3.6-3.7 Outline: •Mass Composition of Molecules •Conservation of Atoms • (Modern version of the law of definite proportions) •Balancing Chemical Reactions; Meaning Ch 3 Problems: 3.53-3.56. (at the end of Ch 3 in Text) (What is different about 3.56b and 3.56g?) Discussion Problem: 3.9 Why do we need to go between moles and grams? • Molecules go together as atoms (or moles) • Molecules can only be weighed to determine amount. • We need to convert back and forth because reactions preserve atoms (or moles of atoms) • Molecules represent the rearrangement and joining together of numbers of atoms of different types (or moles of atoms). Review: The RYP “molecule” Atomic Masses given on each atom What is the molecular Formula: R3 Y3 P2Red, Yellow, Purple From the molecular formula and Atomic Masses, we can compute the mass fraction of each element in the molecule. Our Goal: Find the Molecular Formula •Determine the Molecular Formula from the mass of the elements that make up the molecule. •Elemental Analysis: The method of finding the mass percent of each element in a molecule. •Determine the Empirical Formula from the mass percents. •Use the Mass Spectroscopy data, which gives the total molecular mass, to determine the molecular formula. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis (the process Used to get the mass percents of each element)! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula. Some Examples of Compounds with the same Elemental Ratio’s Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6 Empirical and Molecular Formulas Name water hydrogen Molecular H2O H2O2 peroxide ethane C2H6 sulfur S8 acetic acid CH3COOH Empirical Empirical and Molecular Formulas Name water Molecular H2O Empirical H2O hydrogen peroxide H2O2 HO ethane C2H6 CH3 sulfur acetic acid S8 CH3COOH S COH2 FIGURE 3.6: Molecules with Different Empirical and Molecular Formulas Ball-and-Stick Model: P4O10 Various Names Molecular Empirical Empirical Weight Molecular Weight Ken O'Donoghue Space Filling: Caffeine Molecule Name Formula Molar Mass C=black; N=blue; O=red; H=green Steps to Determine Empirical Formulas from masses Mass (g) of Element in sample Or a relative mass ÷ AW (g/mol ) for that element Moles of Element Use no. of moles as subscripts. Preliminary Formula Change to integer subscripts: ÷ smallest, conv. to whole #. Empirical Formula Determining Empirical Formulas from Measured Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the atomic masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: n(Na) = n(Cr) = n(O) = N.B.: The masses are equivalent to the mass percents or mass fractions. Determining Empirical Formulas from Measured Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the atomic masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: 5.678 g ( Na ) 5.678 g ( Na ) ⋅1mol n ( Na ) = = = 0.2469mol ( Na ) Aw ( Na ) 22.99 g 1 mol Cr = 0.12347 mol Cr 52.00 g Cr 1 mol O n(O) = 7.902 g O x = 0.4939 mol O 16.00 g O n(Cr) = 6.420 g Cr x Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Converting to integer subscripts (dividing all by smallest subscript): Rounding off to whole numbers: Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na0.2469 Cr0.1235 O0.4939 Converting to integer subscripts (dividing all by smallest subscript): Na1.99 Cr1.00 O4.02 Rounding off to whole numbers: Na2CrO4 Sodium Chromate Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the Molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = Mass Hydrogen = Mass Oxygen = Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the Molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O 99.989 g Cpd Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Moles of H = Moles of O = Constructing the preliminary formula: Converting to integer subscripts, ÷ all subscripts by the smallest: Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Mass of C x 1 mole C = 3.3306 moles C 12.01 g C 1 mol H Moles of H = Mass of H x = 6.6657 moles H 1.008 g H Moles of O = Mass of O x 1 mol O = 3.3294 moles O 16.00 g O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, ÷ all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = C1H2O1 = CH2O Alternative to finding Empirical Formula Begin with relative masses: 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. Preliminary-Empirical-formula is C 40.00 H 6.719 O53.27 = C3.33 H 6.67O3.33 12.01 1.008 Empirical 16 Formula C 3.33 H 6.67 O3.33 = C1 H 2O1 or CH 2O 3.33 3.33 3.33 Obtaining the empirical formula requires dividing the preliminary one by the smallest valued subscript; but this is not necessarily the complete answer. There is a bit of an art to it. Compounds with empirical formula CH2O Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula weight of the empirical formula is: Whole-number multiple = M compound empirical formula mass Therefore the Molecular Formula is: = Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula mass of the empirical formula is: 1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03 g/mol s= Whole-number multiple = s = MW of the compound = empirical formula mass 180.16 g (Mf) 30.03 g (Ef) Therefore the Molecular Formula is: C1x6H2x6O1x6 = C6H12O6 = 6.00 = 6 Mf Ef Obtain the Molecular Formula with no guessing • Eg. A hydrocarbon sample is analyzed, to contain 0.2778 g C, 0.0116g H, and 0.3699 g O. The Molecular Mass is 180.16 g (from Mass Spec). What is the molecular formula? • Strategy --- Scale the masses first to give the correct masses for the actual molecular mass. MW 180.16 s= = = 259.4589 M ( E .F . ) 0.2778 + 0.0467 + 0.3699 • Therefore the new masses are 72.0719 g C, 12.1063 g H, 95.9818 g O. = s ⋅ 0.2778 gC , s ⋅ 0.0467 gH , s ⋅ 0.3699 gO C 72.0719 H 12.1063 O95.9818 = C6.001 H12.01O5.999 12.01 1.008 16.00 How Chemical Analysis is done: eg : (n + 12 m)O2 + Cn H m → mH 2O + nCO2 Mass Carbon = fC*Mass CO2 Mass Hydrogen = fH*MassH2O 12 12 3 fC = = = = 0.273 12 + 2 ⋅16 44 11 1⋅ 2 2 1 fH = = = = 0.111 2 + 16 18 9 Chemical Analysis and empirical formula Know the mass of the sample From the chemical analysis: Obtain the mass of H2O and CO2 Mass Carbon = fC*Mass CO2 Mass Hydrogen = fH*MassH2O 12 12 3 = = = 0.273 12 + 2 ⋅16 44 11 1⋅ 2 2 1 fH = = = = 0.111 2 + 16 18 9 fC = Our Goal: Determine the fraction (by mass) of each element in the original sample e.g. Mass Carbon = mass CO2*(3/11) Mass Fraction Carbon = Mass Carbon / mass sample Mass Fraction Hydrogen = Mass Hydrogen / mass sample Mass Extra (e.g. N or O) = Mass sample – (mass C + mass H) Ascorbic acid ( Vitamin C ) - I contains only C , H , and O • Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O. • Calculate its Empirical formula! • Mass C: • Mass H: • Mass Oxygen = Ascorbic acid ( Vitamin C ) - I contains only C , H , and O • Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O. • Calculate its Empirical formula! • Mass C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C • Mass H: 2.64 x10-3g H2O x (2.016 g H/18.02 gH2O) = 2.95 x 10-4 g H • Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O = 3.54x10-3 g O Ascorbic acid combustion – cont’d • Moles C = • Moles H = • Moles O = • Divide each by smallest: • Moles C = • Moles H = • Moles O = Vitamin C combustion - II • C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C • H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H • O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O • • • • Divide each by smallest (2.21 x 10-4 ): C = 1.00 Multiply each by 3: C = 3.00 = 3.0 H = 1.32 (to get ~integers) H = 3.96 = 4.0 O = 1.00 O = 3.00 = 3.0 C3H4O3 If the empirical formula of ascorbic acid is C3H4O3, (empirical mass 88 g/mol) and the molecular mass of ascorbic acid is 176 g/mol, what is the molecular formula? 1. C3H4O3 2. C6H8O6 3. C9H12O9 4. C12H16O12 5. None of the above Change: The meaning of a chemical reaction • • • • • A rxn represents a change. Burn Sugar: Sugar becomes CO2 and Water. Sugar on the left (with O2) before reaction Products on the right after the change. Properties of New state must equal properties of old state plus the change between them. • For example; There are 300 people in this room: We count noses. Two people leave. How many in the room? – We do not need to recount the number of people in the room to know there are 298 people. We kept track of the change, and do the math. New = Initial + Change (NIC Table) Change ≡ X ≡ ΔX = New − Initial How to Balance Equations • Mass Balance (or Atom Balance)- same number of each element on each side of the equation: (1) start with simplest element (or largest molecule) (2) progress to other elements (3) make all whole numbers (4) re-check atom balance 1 CH4 (g) + __O2 (g) 1 CO2 (g) + __H2O (g) 1 CH4 (g) +__ O2 (g) 1 CO2 (g) + 2 H2O (g) 1 CH4 (g) + 2 O2 (g) 1 CO2 (g) + 2 H2O (g) • Make charges balance. (Remove “spectator” ions.) Ca2+ (aq) + 2 OH- (aq) + Na+ Ca(OH)2 (s) + Na+ DEMO: Methane Bubbles Extra Information • More Information and extra problems follow. • When working extra problems (or problems in the text): Be sure to cover up the solution before you start the problem. Calculate MW and % composition of NH4NO3. • • • 2 mol N x 4 mol H x 3 mol O x Molar mass = M = %N = %H = %O = g N x 100% g cpd gH x 100% g cpd = = g O x 100% = g cpd Check: 100% total ? Calculate M and % composition of NH4NO3. • • • 2 mol N x 14.01 g/mol = 28.02 g N 4 mol H x 1.008 g/mol = 4.032 g H 3 mol O x 15.999 g/mol = 48.00 g O 80.05 g/mol %N = 28.02g N x 100% 80.05g = 35.00% %H = 4.032g H x 100% = 80.05g %O = 48.00g O x 100% = 59.96% 80.05g 99.997% 5.037% Calculate the Percent Composition of Sulfuric Acid H2SO4 Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol 2(1.008g H) x 100% = 98.09g 2.06% H %S = 1(32.07g S) x 100% = 98.09g 32.69% S %O = 4(16.00g O) x 100% = 98.09 g 65.25% O %H = Check = 100.00% Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C per mole sucrose = mass of H / mol = mass of O / mol = total mass per mole = Finding the mass fraction of C in Sucrose & % C : mass of C per mole Mass Fraction of C = = mass of 1 mole sucrose = To find mass % of C = Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C per mole sucrose = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H / mol = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O / mol = 11 x 16.00 g O/mol = 176.00 g O/mol total mass per mole = 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : mass of C per mole 144.12 g C/mol Mass Fraction of C = = mass of 1 mole sucrose 342.30 g Cpd/mol = 0.4210 To find mass % of C = 0.4210 x 100% = 42.10% Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = mol H x M of H x 100% = mass of 1 mol sucrose Mass % of O = mol O x M of O x 100% = mass of 1 mol sucrose (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = mol H x M of H x 100% = 22 x 1.008 g Hx 100% mass of 1 mol sucrose 342.30 g = 6.479% H Mass % of O = mol O x M of O x 100% = 11 x 16.00 g Ox 100% mass of 1 mol sucrose 342.30 g = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose x 0.421046 g C 1 g sucrose = 10.25 g C Calculating the Mass of an Element in a Compound: Ammonium Nitrate How much Nitrogen is in 455 kg of Ammonium Nitrate? Ammonium Nitrate = NH4NO3 The Formula Mass of Cpd is: 4 x H = 4 x 1.008 = 4.032 g 2 x N = 2 X 14.01 = 28.02 g Therefore mass fraction N: 3 x O = 3 x 16.00 = 48.00 g 28.02 g Nitrogen 80.052 g = 80.052 g Cpd Mass N in sample = Calculating the Mass of an Element in a Compound: Ammonium Nitrate How much Nitrogen is in 455 kg of Ammonium Nitrate? Ammonium Nitrate = NH4NO3 The Formula Mass of Cpd is: 4 x H = 4 x 1.008 = 4.032 g 2 x N = 2 X 14.01 = 28.02 g Therefore mass fraction N: 3 x O = 3 x 16.00 = 48.00 g 28.02 g Nitrogen 80.052 g = 0.35002249 g N / g Cpd 80.052 g Cpd 455 kg x 1000g / kg = 455,000 g NH4NO3 455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen or: 455 kg NH4NO3 X 28.02 kg Nitrogen = 159 kg Nitrogen 80.052 kg NH4NO4 Determining a Chemical Formula from Combustion Analysis - I Problem: Erthrose (M = 120 g/mol) is an important chemical compound used often as a starting material in chemical synthesis, and contains Carbon, Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded: 1.027 g CO2 and 0.4194 g H2O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO2 = Mass fraction of H in H2O = Calculating masses of C and H: Mass of Element = mass of compound x mass fraction of element Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: mol C x M of C Mass fraction of C in CO2 = = mass of 1 mol CO2 = 1 mol C x 12.01 g C/ 1 mol C = 0.2729 g C / 1 g CO2 44.01 g CO2 mol H x M of H = mass of 1 mol H2O 2 mol H x 1.008 g H / 1 mol H = = 0.1119 g H / 1 g H2O 18.02 g H2O Mass fraction of H in H2O = Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element Determining a Chemical Formula from Combustion Analysis - III Mass (g) of C = Mass (g) of H = Calculating the mass of O: Calculating moles of each element: C= H= O= Determining a Chemical Formula from Combustion Analysis - III Mass (g) of C = 1.027 g CO2 x 0.2729 g C = 0.2803 g C 1 g CO2 0.1119 g H Mass (g) of H = 0.4194 g H2O x = 0.04693 g H 1 g H2 O Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O Calculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd => C4H8O4 Chemical Equations Qualitative Information: Reactants Products States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H2 (g) + O2 (g) But also Quantitative Information! 2 H2O (g) Using A Table to Balance Molecular Reaction: aCH 4 + bO2 → cCO2 + dH 2O Molecular Species CH4 O2 CO2 H2O Coefficients a=1 b=2 c=1 d=2 C 1 0 1 0 H 4 0 0 2 O 0 2 2 1 Equations to balance: (Three equations and 4 unknowns) a = c =1 From the first equation: Second equation: d = 2a = 2Third equation: 3 Equations to Solve 1⋅ a + 0 ⋅ b = 1⋅ c + 0 ⋅ d 4⋅ a + 0⋅b = 0⋅c + 2⋅ d 0 ⋅ a + 2 ⋅ b = 2 ⋅ c + 1⋅ d b = c + 12 d = 2 Verify the Solution 1 ⋅1 + 0 ⋅ 2 = 1 ⋅1 + 0 ⋅ 2 4 ⋅1 + 0 ⋅ 2 = 0 ⋅1 + 2 ⋅ 2 0 ⋅1 + 2 ⋅ 2 = 2 ⋅1 + 1 ⋅ 2 Mass Percent Composition of Na2SO4 Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses (g) 2 x Na = 2 x 22.99 = 45.98 1xS= 4xO= Check % Na + % S + % O = 100% Percent of each Element % Na = Mass Na / Total mass x 100% % Na = % S = Mass S / Total mass x 100% %S= %O= %O= Mass Percent Composition of Na2SO4 Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses (g) 2 x Na = 2 x 22.99 = 45.98 1 x S = 1 x 32.07 = 32.07 4 x O = 4 x 16.00 = 64.00 142.05 Check Percent of each Element % Na = Mass Na / Total mass x 100% % Na = (45.98 / 142.05) x 100% =32.37% % S = Mass S / Total mass x 100% % S = (32.07 / 142.05) x 100% = 22.58% % O = Mass O / Total mass x 100% % O = (64.00 / 142.05) x 100% = 45.05% % Na + % S + % O = 100% 32.37% + 22.58% + 45.05% = 100.00% Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Multiply by MW (X in g / mol) Mass (g) of X in one mole of compound Divide by mass (g) MW of one mo of compound Mass fraction of X Multiply by 100 % Mass % of X Adrenaline is a very Important Compound in the Body - I • Analysis gives : • C = 56.8 % • H = 6.50 % • O = 28.4 % • N = 8.28 % • Calculate the Empirical Formula ! Adrenaline - II • Assume 100g! • C= • H= • O= • N= Divide by smallest • • • • C= H= O= N= => Adrenaline - II • • • • • • • • • • Assume 100g! C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N Divide by smallest (0.591) => C = 8.00 mol C = 8.0 mol C or H = 10.9 mol H = 11.0 mol H O = 3.01 mol O = 3.0 mol O C8H11O3N N = 1.00 mol N = 1.0 mol N