# Lecture 4: (Lec 3B) Stoichiometry

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2 Lecture 4: (Lec 3B) Stoichiometry Sections (Zumdahl 6 th Edition) Outline: Mass Composition of Molecules Conservation of Atoms (Modern version of the law of definite proportions) Balancing Chemical Reactions; Meaning Ch 3 Problems: (at the end of Ch 3 in Text) (What is different about 3.56b and 3.56g?) Discussion Problem: 3.9

3 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).

4 Review: The RYP molecule Atomic Masses given on each atom What is the molecular Formula: R 3 Y 3 P 2 Red, Yellow, Purple From the molecular formula and Atomic Masses, we can compute the mass fraction of each element in the molecule.

5 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.

6 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.

7 Some Examples of Compounds with the same Elemental Ratio s Empirical Formula Molecular Formula CH 2 (unsaturated Hydrocarbons) C 2 H 4, C 3 H 6, C 4 H 8 OH or HO H 2 O 2 S S 8 P P 4 Cl Cl 2 CH 2 O (carbohydrates) C 6 H 12 O 6

8 Empirical and Molecular Formulas Name Molecular Empirical water H 2 O hydrogen H 2 O 2 peroxide ethane C 2 H 6 sulfur S 8 acetic acid CH 3 COOH

9 Empirical and Molecular Formulas Name Molecular Empirical water H 2 O H 2 O hydrogen H 2 O 2 HO peroxide ethane C 2 H 6 CH 3 sulfur S 8 S acetic acid CH 3 COOH COH 2

10 FIGURE 3.6: Molecules with Different Empirical and Molecular Formulas

11 Ball-and-Stick Model: P 4 O 10 Various Names Molecular Empirical Empirical Weight Molecular Weight Ken O'Donoghue

12 Space Filling: Caffeine Molecule Name Formula Molar Mass C=black; N=blue; O=red; H=green

13 Steps to Determine Empirical Formulas from masses Mass (g) of Element in sample Or a relative mass A W (g/mol ) for that element Moles of Element Use no. of moles as subscripts. Preliminary Formula Empirical Formula Change to integer subscripts: smallest, conv. to whole #.

14 Determining Empirical Formulas from Measured Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, g Cr, and 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.

15 Determining Empirical Formulas from Measured Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, g Cr, and 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 ( ) ( ) ( ) 5.678g Na 5.678g Na 1mol = = = mol Na A Na 22.99g w ( ) 1 mol Cr n(cr) = g Cr x = mol Cr g Cr 1 mol O n(o) = g O x = mol O g O

16 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:

17 Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na Cr O Converting to integer subscripts (dividing all by smallest subscript): Na 1.99 Cr 1.00 O 4.02 Rounding off to whole numbers: Na 2 CrO 4 Sodium Chromate

18 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 = g/mol), elemental analysis shows that it contains mass % C, mass % H, and 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 =

19 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 = g/mol), elemental analysis shows that it contains mass % C, mass % H, and 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% = g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = g O g Cpd

20 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:

21 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 = moles C g C 1 mol H Moles of H = Mass of H x = moles H g H Moles of O = Mass of O x 1 mol O = moles O 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 / 3.33 O 3.33 / 3.33 = C 1 H 2 O 1 = CH 2 O

22 Alternative to finding Empirical Formula Begin with relative masses: mass % C, mass % H, and mass % O. Preliminary-Empirical-formula is C H O = C H O Empirical Formula C H O = C H O or CH O 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.

23 Compounds with empirical formula CH 2 O

24 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:

25 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 x x = g/mol M W s= Whole-number multiple = of the compound = empirical formula mass g (Mf) Mf s = = 6.00 = g (Ef) Ef Therefore the Molecular Formula is: C 1 x 6 H 2 x 6 O 1 x 6 = C 6 H 12 O 6

26 Obtain the Molecular Formula with no guessing Eg. A hydrocarbon sample is analyzed, to contain g C, g H, and g O. The Molecular Mass is g (from Mass Spec). What is the molecular formula? Strategy --- Scale the masses first to give the correct masses for the actual molecular mass. M s = W = = M E. F ( ) Therefore the new masses are g C, g H, g O. = s gc, s gh, s go C H O = C H O

27 How Chemical Analysis is done: 1 2 eg : ( n+ m) O + CnHm mh O+ nco Mass Carbon = f C *Mass CO 2 Mass Hydrogen = f H *MassH 2 O f f C H = = = = = = = =

28 Chemical Analysis and empirical formula Know the mass of the sample From the chemical analysis: Obtain the mass of H 2 O and CO 2 Mass Carbon = f C *Mass CO 2 Mass Hydrogen = f H *MassH 2 O f f C H = = = = = = = = Our Goal: Determine the fraction (by mass) of each element in the original sample e.g. Mass Carbon = mass CO 2 *(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)

29 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 CO and mg H 2 O. Calculate its Empirical formula! Mass C: Mass H: Mass Oxygen =

30 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 CO and mg H 2 O. Calculate its Empirical formula! Mass C: 9.74 x10-3 g CO 2 x(12.01 g C/44.01 g CO 2 ) = 2.65 x 10-3 g C Mass H: 2.64 x10-3 g H 2 O x (2.016 g H/18.02 gh 2 O) = 2.95 x 10-4 g H Mass Oxygen = 6.49 mg mg mg = 3.54 mg O = 3.54x10-3 g O

31 Ascorbic acid combustion cont d Moles C = Moles H = Moles O = Divide each by smallest: Moles C = Moles H = Moles O =

32 Vitamin C combustion - II C = 2.65 x 10-3 g C / ( g C / mol C ) = = 2.21 x 10-4 mol C H = x 10-3 g H / ( g H / mol H ) = = 2.92 x 10-4 mol H O = 3.54 x 10-3 g O / ( 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 C 3 H 4 O 3

33 If the empirical formula of ascorbic acid is C 3 H 4 O 3, (empirical mass 88 g/mol) and the molecular mass of ascorbic acid is 176 g/mol, what is the molecular formula? 1. C 3 H 4 O 3 2. C 6 H 8 O 6 3. C 9 H 12 O 9 4. C 12 H 16 O None of the above

34 Change: The meaning of a chemical reaction A rxn represents a change. Burn Sugar: Sugar becomes CO 2 and Water. Sugar on the left (with O 2 ) 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 Δ X = New Initial

35 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 CH 4 (g) + O 2 (g) 1 CO 2 (g) + H 2 O (g) 1 CH 4 (g) + O 2 (g) 1 CO 2 (g) + 2 H 2 O(g) 1 CH 4 (g) + 2 O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) Make charges balance. (Remove spectator ions.) Ca 2+ (aq) + 2 OH - (aq) + Na + Ca(OH) 2 (s) + Na + DEMO: Methane Bubbles

36 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.

37 Calculate M W and % composition of NH 4 NO 3. 2 mol N x 4 mol H x 3 mol O x Molar mass = M = %N = g N x 100% = g cpd %H = g H x 100% = g cpd %O = x 100% = g O g cpd Check: 100% total?

38 Calculate M and % composition of NH 4 NO 3. 2 mol N x g/mol = g N 4 mol H x g/mol = g H 3 mol O x g/mol = g O g/mol %N = 28.02g N x 100% = 35.00% 80.05g %H = 4.032g H x 100% = 5.037% 80.05g %O = x 100% = 59.96% 48.00g O 80.05g %

39 Calculate the Percent Composition of Sulfuric Acid H 2 SO 4 Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = g/mol %H = 2(1.008g H) x 100% = 2.06% H 98.09g %S = 1(32.07g S) x 100% = 32.69% S 98.09g %O = 4(16.00g O) x 100% = 65.25% O g Check = %

40 Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. (a)what is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 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 = = = To find mass % of C = mass of 1 mole sucrose

41 Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. (a)what is the mass percent of each element in sucrose? ( b) How many grams of carbon are in g of sucrose? (a) Determining the mass percent of each element: mass of C per mole sucrose = 12 x g C/mol = g C/mol mass of H / mol = 22 x g H/mol = g H/mol mass of O / mol = 11 x g O/mol = g O/mol total mass per mole = g/mol Finding the mass fraction of C in Sucrose & % C : mass of C per mole g C/mol Mass Fraction of C = = mass of 1 mole sucrose g Cpd/mol = To find mass % of C = x 100% = 42.10%

42 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 =

43 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 g Hx 100% mass of 1 mol sucrose g = 6.479% H Mass % of O = mol O x M of O x 100% = 11 x g O x 100% mass of 1 mol sucrose g = % 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 = g sucrose x g C = g C 1 g sucrose

44 Calculating the Mass of an Element in a Compound: Ammonium Nitrate How much Nitrogen is in 455 kg of Ammonium Nitrate? Ammonium Nitrate = NH 4 NO 3 The Formula Mass of Cpd is: 4 x H = 4 x = g Therefore mass fraction N: 2 x N = 2 X = g 3 x O = 3 x = g g Nitrogen g Cpd = g Mass N in sample =

45 Calculating the Mass of an Element in a Compound: Ammonium Nitrate How much Nitrogen is in 455 kg of Ammonium Nitrate? Ammonium Nitrate = NH 4 NO 3 The Formula Mass of Cpd is: 4 x H = 4 x = g Therefore mass fraction N: 2 x N = 2 X = g 3 x O = 3 x = g g Nitrogen g Cpd = g N / g Cpd g or: 455 kg x 1000g / kg = 455,000 g NH 4 NO 3 455,000 g Cpd x g N / g Cpd = 1.59 x 10 5 g Nitrogen kg Nitrogen 455 kg NH 4 NO 3 X kg NH4 NO 4 = 159 kg Nitrogen

46 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 mg sample yielded: g CO 2 and g H 2 O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H 2 O, and C in CO 2. 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.

47 Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO 2 = Mass fraction of H in H 2 O = Calculating masses of C and H: Mass of Element = mass of compound x mass fraction of element

48 Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: mol C xmof C Mass fraction of C in CO 2 = = mass of 1 mol CO 2 = 1 mol C x g C/ 1 mol C = g C / 1 g CO g CO 2 Mass fraction of H in H 2 O = mol H xmof H = mass of 1 mol H 2 O 2 mol H x g H / 1 mol H = = g H / 1 g H 2 O g H 2 O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element

49 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 =

50 Determining a Chemical Formula from Combustion Analysis - III Mass (g) of C = g CO 2 x g C 1 g CO g H Mass (g) of H = g H 2 O x 1 g H 2 O = g C = g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = g g C g H = g O Calculating moles of each element: C = g C / g C/ mol C = mol C H = g H / g H / mol H = mol H O = g O / g O / mol O = mol O C H O = CH 2 O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd => C 4 H 8 O 4

51 Chemical Equations Qualitative Information: Reactants Products States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H 2 (g) + O 2 (g) 2 H 2 O (g) But also Quantitative Information!

52 Using A Table to Balance Molecular Reaction: Molecular Species Coefficients C H O CH 4 a= O 2 b= ach bo cco dh O CO 2 c= H 2 O d=2 Equations to balance: (Three equations and 4 unknowns) From the first equation: d Second equation: Third equation: 1 a+ 0 b= 1 c+ 0 d 4 a+ 0 b= 0 c+ 2 d 0 a+ 2 b= 2 c+ 1 d a = c= 1 = 2a= 2 b= c+ 1 2 d = 2 3 Equations to Solve Verify the Solution = = =

53 Mass Percent Composition of Na 2 SO 4 Na 2 SO 4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses (g) 2 x Na = 2 x = x S = 4 x O = Percent of each Element % Na = Mass Na / Total mass x 100% % Na = % S = Mass S / Total mass x 100% % S = Check % Na + % S + % O = 100% % O = % O =

54 Mass Percent Composition of Na 2 SO 4 Na 2 SO 4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses (g) 2 x Na = 2 x = x S = 1 x = x O = 4 x = Check % Na + % S + % O = 100% Percent of each Element % Na = Mass Na / Total mass x 100% % Na = (45.98 / ) x 100% =32.37% % S = Mass S / Total mass x 100% % S = (32.07 / ) x 100% = 22.58% % O = Mass O / Total mass x 100% % O = (64.00 / ) x 100% = 45.05% 32.37% % % = %

55 Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Mass (g) of X in one mole of compound Multiply by M W (X in g / mol) Mass fraction of X Divide by mass (g) M W of one mo of compound Multiply by 100 % Mass % of X

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57 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!

58 Assume 100g! C = Adrenaline - II H = O = N = C = H = O = N = Divide by smallest =>

59 Adrenaline - II Assume 100g! C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C H = 6.50 g H/( 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) = 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 C 8 H 11 O 3 N N = 1.00 mol N = 1.0 mol N

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