Chapter 3 Stoichiometry

Chapter 3 - Stoichiometry 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical equations 3.8 Stoichiometric Calculations: Amounts of Reactants and Products 3.9 Calculations Involving a Limiting Reactant

Reaction of zinc and sulfur.

Figure 3.1: Mass spectrometer

Chemists using a mass spectrometer to analyze for copper in blood plasma. Source: USDA Agricultural Research Service

A herd of savanna-dwelling elephants Source: Corbis

Figure 3.2: Relative intensities of the signals recorded when natural neon is injected into a mass spectrometer.

Figure 3.3: Mass spectrum of natural copper

Atomic Definitions II: AMU, Dalton, 12C Std. Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom on this scale Hydrogen has a mass of 1.008 AMU. Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12C has a mass of 12.00 daltons. Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12C the chosen standard. Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

Isotopes of Hydrogen 11H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu 21H (D) 1 Proton 1 Neutron 0.015 % 2.01410178 amu 31H (T) 1 Proton 2 Neutrons -------- ---------- The average mass of Hydrogen is 1.008 amu 3H is Radioactive with a half life of 12 years. H2O Normal water “light water “ mass = 18.0 g/mole , BP = 100.000000C D2O Heavy water mass = 20.0 g/mole , BP = 101.42 0C

Element #8 : Oxygen, Isotopes 168O 8 Protons 8 Neutrons 99.759% 15.99491462 amu 178O 8 Protons 9 Neutrons 0.037% 16.9997341 amu 188O 8 Protons 10 Neutrons 0.204 % 17.999160 amu

Calculating the “Average” Atomic Mass of an Element Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%). 24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu 26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu ___________ amu With Significant Digits = __________ amu

Calculate the Average Atomic Mass of Zirconium, Element #40 Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr. Isotope (% abd.) Mass (amu) (%) Fractional Mass 90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu 91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu 92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu 94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu 96Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu ____________ amu With Significant Digits = ___________ amu

Problem: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol. Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0 Solution: X(78.918336) + Y(80.91629) = 79.904 X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904 78.918336 - 78.918336 Y + 80.91629 Y = 79.904 1.997954 Y = 0.985664 or Y = 0.4933 X = 1.00 - Y = 1.00 - 0.4933 = 0.5067 %X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br

LIKE SAMPLE PROBLEM 3.2 During a perplexing dream one evening you come across 200 atoms of Einsteinium. What would be the total mass of this substance in grams? SOLUTION: 200 Atoms of Es X 252 AMU/Atom = 5.04 x 104 AMU 5.04 x 104 AMU x (1g / 6.022 x 1023 AMU) = ______________ g of Einsteinium

MOLE The Mole is based upon the following definition: The amount of substance that contains as many elementary parts (atoms, molecules, or other?) as there are atoms in exactly 12 grams of carbon -12. 1 Mole = 6.022045 x 1023 particles

Figure 3.4: One-mole samples of copper, sulfur, mercury, and carbon

One mole of common Substances CaCO3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g

Atoms Molecules Moles Molecular Formula Avogadro’s Number 6.022 x 1023

Mole - Mass Relationships of Elements Element Atom/Molecule Mass Mole Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms 1 molecule of O2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule 1 molecule of S8 = 256.56 amu 1 mole of S8 = _______ g = 6.022 x 1023 molecules

Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = __________ g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

LIKE SAMPLE PROBLEM 3.4 How many carbon atoms are present in a 2.0 g tablet of Sildenafil citrate (C28H38N6O11S) ? SOLUTION: MW of Sildenafil citrate = 28 X 12 amu (C) + 38 X 1 amu (H) + 6 X 14 amu (N) + 11 X 16 amu (O) + 1 X 32 amu (S) = 666 AMU

LIKE SAMPLE PROBLEM 3.4 (cont…) 2.0 g (C28H38N6O11S) X 1 mol/666g = 3.0 X 10-3 mol (C28H38N6O11S) 3.0 X 10-3 mol (C28H38N6O11S) X 6.022 X 1023 molecules / 1 mol (C28H38N6O11S) = 1.8 X 1021 molecules of C28H38N6O11S 1.8 X 1021 molecules of C28H38N6O11S X 28 atoms of C / 1 molecules of C28H38N6O11S = Carbon Atoms

Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = __________________ atoms of Tungsten 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

Calculating the Moles and Number of Formula Units in a given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.23545 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4 = ______________ formula units

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

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 = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd = 0.421046 To find mass % of C = 0.421046 x 100% = ______% Mass Fraction of C = =

Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 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 = C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose

Mol wt 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 28.02g N2 80.05g %N = x 100% = 35.00% N 4.032g H2 80.05g %H = x 100% = 5.037% H 48.00g O2 80.05g %O = x 100% = 59.96% O 99.997%

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 H2) 98.09g %H = x 100% = 2.06% H 1(32.07g S) 98.09g %S = x 100% = 32.69% S 4(16.00g O) 98.09 g %O = x 100% = 65.25% O Check = 100.00%

Penicillin is isolated from a mold Source: Getty Images

Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! 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.

Figure 3.6: Examples of substances whose empirical and molecular formulas differ.

Steps to Determine Empirical Formulas Mass (g) of Element M (g/mol ) Moles of Element use no. of moles as subscripts Preliminary Formula change to integer subscripts 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

Figure 3.7: Structural Formula of P4O10

Determining Empirical Formulas from 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 molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = _________ mol Na Moles of Cr = 6.420 g Cr x = ___________ mol Cr Moles of O = 7.902 g O x = ____________ mol O 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

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 = 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 = Mass of C x = 3.3306 moles C Moles of H = Mass of H x = 6.6657 moles H Moles of O = Mass of O x = 3.3294 moles O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, divide all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O 1 mole C 12.01 g C 1 mol H 1.008 g H 1 mol O 16.00 g O

Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula weight 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 M of Glucose empirical formula mass Whole-number multiple = = = = 6.00 = 6 180.16 30.03 Therefore the Molecular Formula is: C 1 x 6 H 2 x 6 O 1 x 6 = C6H12O6

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 = 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 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

Figure 3.5: Combustion device

Ascorbic acid ( Vitamin C ) - I contains 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 it’s Empirical formula! C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg 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 2.21 x 10-4 C = 1.00 Multiply each by 3 = 3.00 = 3.0 H = 1.32 = 3.96 = 4.0 O = 1.00 = 3.00 = 3.0 C3H4O3

Computer generated molecule: Caffeine, C8H10N4O2

Determining a Chemical Formula from Combustion Analysis - I Problem: Erthrose (M = 120 g/mol) is an important chemical compound 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 = = = = 0.2729 g C / 1 g CO2 Mass fraction of H in H2O = = = = 0.1119 g H / 1 g H2O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element mol C x M of C mass of 1 mol CO2 1 mol C x 12.01 g C/ 1 mol C 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 18.02 g H2O

Determining a Chemical Formula from Combustion Analysis - III 0.2729 g C 1 g CO2 Mass (g) of C = 1.027 g CO2 x = 0.2803 g C Mass (g) of H = 0.4194 g H2O x = 0.04693 g H 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 0.1119 g H 1 g H2O

Like Example 3.7 (P 65) - I Sucrose is the common sugar used in all homes, and chemical analysis tells us that the chemical composition is 42.14% carbon, 6.48% hydrogen and 51.46% oxygen. What is the molecular formula of sucrose if its molecular mass is approximately 340 g/mol? First determine the mass of each element in 1 mole (342.3g) of the compound, sucrose. 42.14g C 342.3g 144.24g C 100.0g sucrose mol mol 6.48g H 342.3g 22.18g H 100.0g sucrose mol mol 51.46g O 342.3g 176.15g O 100.0g sucrose mol mol X = X = X =

Like Example 3.7 (P 65) - II Now we convert to moles: 144.24g C 1 mol C 12.01g C mol sucrose 12.011g C mol sucrose 2 22.18g H 1 mol Hg 22.00g H mol sucrose 1.008g H mol sucrose 176.15g O 1 mol O 11.01g O Mol sucrose 16.00g O mol sucrose C: X = H: X = O: X = Divide by the smallest number: C: 1.09 Therefore Empirical Formula = CH2O since the molecular mass = 340g/mol, H: 2.00 we must divide the formula mass into the molecular mass or 340/30 = 11.3 = 11 O: 1.00 therefore the molecular formula is C11H22O11 !!!

Chemical Equations Reactants Products 2 H2 (g) + O2 (g) 2 H2O (g) Qualitative Information: Reactants Products Phases (States of Matter): (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H2 (g) + O2 (g) 2 H2O (g)

Balanced Equations mass balance (atom balance)- same number of each element (1) start with simplest element (2) progress to other elements (3) make all whole numbers (4) re-check atom balance charge balance (no “spectator” ions) 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) Ca2+ (aq) + 2 OH- (aq) Ca(OH)2 (s) + Na+ + Na+

Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy Molecules 2 molecules of C2H6 + 7 molecules of O2 = 4 molecules of CO2 + 6 molecules of H2O Amount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O Mass (amu) 60.14 amu C2H6 + 224.00 amu O2 = 176.04 amu CO2 + 108.10 amu H2O Mass (g) 60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O Total Mass (g) 284.14g = 284.14g

Flare in a natural gas field Source: Stock Boston

Figure 3.8: Methane/oxygen reaction

Table 3.2 (P 66) Information Conveyed by the Balanced Equation for the Combustion of Methane Reactants Products CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) 1 molecule CH4 + 1 molecule CO2 + 2 molecules of O2 2 molecules H2O 1 mol CH4 molecules + 1 mol CO2 molecules + 2 mol O2 molecules 2 mol H2O molecules 6.022 x 1023 CH4 molecules + 6.022 x 1023 CO2 molecules + 2 x (6.022 x 1023) O2 molecules 2 x (6.022 x 1023) H2O molecules 16g CH4 + 2 (32g) O2 44g CO2 + 2 (18g) H2O 80g reactants 80g products

Molecular model: Balanced equation C2H6O(aq) + 3 O2(g) 2 CO2 (g) + 3 H2O(g) + Energy

Decomposition of ammonium dichromate

Balancing Chemical Equations - I Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C6H14). Plan: Write the skeleton equation from the words into chemical compounds with blanks before each compound. begin the balance with the most complex compound first, and save oxygen until last! Solution: C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy Begin with one Hexane molecule which says that we will get 6 CO2’s! 1 6 C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy

Balancing Chemical Equations - II The H atoms in the hexane will end up as H2O, and we have 14 H atoms, and since each water molecule has two H atoms, we will get a total of 7 water molecules. C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy 1 6 7 Since oxygen atoms only come as diatomic molecules (two O atoms, O2),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore multiply the hexane by 2, giving a total of 12 CO2 molecules, and 14 H2O molecules. C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy 2 12 14 This now gives 12 O2 from the carbon dioxide, and 14 O atoms from the water, which will be another 7 O2 molecules for a total of 19 O2 ! 19 C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy 2 12 14

Chemical Equation Calc - I Atoms (Molecules) Avogadro’s Number 6.02 x 1023 Molecules Reactants Products

Chemical Equation Calc - II Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles

Sulfuric Acid Plant S(s) + O2 (g) SO2 (g) SO2 (g) + O2 (g) SO3 (g) SO3 (g) + H2O(l) H2SO4 (aq) Source: Southern States Chemical

The process for finding the mass of carbon dioxide produced from 96 The process for finding the mass of carbon dioxide produced from 96.1 grams of propane

Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I Problem: Given the following chemical reaction between Aluminum Sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles of water are required for the reaction? b) What mass of H2S & Al(OH)3 would be formed? Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g) Plan: Calculate moles of Aluminum Sulfide using it’s molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it’s molecular weight. Solution: a) molar mass of Aluminum Sulfide = 150.17 g / mol moles Al2S3 = = ____________ Al2S3 65.80 g Al2S3 150.17 g Al2S3/ mol Al2S3

Calculating Reactants and Products in a Chemical Reaction - II a) cont. 0.4382 moles Al2S3 x = 2.629 moles H2O b) 0.4382 moles Al2S3 x = 1.314 moles H2S molar mass of H2S = 34.09 g / mol mass H2S = 1.314 moles H2S x = 44.81 g H2S 0.4382 moles Al2S3 x = 0.4764 moles Al(OH)3 molar mass of Al(OH)3 = 78.00 g / mol mass Al(OH)3 = 0.4764 moles Al(OH)3 x = = ________________ g Al(OH)3 6 moles H2O 1 mole Al2S3 3 moles H2S 1 mole Al2S3 34.09 g H2S 1 mole H2S 2 moles Al(OH)3 1 mole Al2S3 78.00 g Al(OH)3 1 mole Al(OH)3

Calculating the Amounts of Reactants and Products in a Reaction Sequence - I Problem: Calcium Phosphate could be prepared in the following reaction sequence: 4 P4 (s) + 10 KClO3 (s) 4 P4O10 (s) + 10 KCl (s) P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq) 2 H3PO4 (aq) + 3 Ca(OH)2 (aq) 6 H2O(aq) + Ca3(PO4)2 (s) Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass of Calcium Phosphate could be formed? Plan: (1) Calculate moles of P4. (2) Use molar ratios to get moles of Ca3(PO4)2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate.

Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: moles of Phosphorous = 15.50 g P4 x = 0.1251 mol P4 For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s) 4 P4O10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq) ] For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2 1 Ca3(PO4)2 + 6 H2O] 0.1251 moles P4 x x x = _______ moles Ca3(PO4)2 1 mole P4 123.88 g P4 4 moles P4O10 4 moles P4 4 moles H3PO4 1 mole P4O10 1 mole Ca3(PO4)2 2 moles H3PO4

Calculating the Amounts of Reactants and Products in a Reaction Sequence - III Molar mass of Ca3(PO4)2 = 310.18 g mole mass of product = 0.2502 moles Ca3(PO4)2 x = = g Ca3(PO4)2 310.18 g Ca3(PO4)2 1 mole Ca3(PO4)2

Hydrochloric acid reacts with solid sodium hydrogen carbonate

Two antacid tablets

Molecular model: N2 molecules require 3H2 molecules for the reaction N2 (g) + 3 H2 (g) 2 NH3 (g)

Figure 3.9: Hydrogen and Nitrogen reacting to form Ammonia, the Haber process

Limiting Reactant Problems a A + b B + c C d D + e E + f F Steps to solve 1) Identify it as a limiting Reactant problem - Information on the: mass, number of moles, number of molecules, volume and molarity of a solution is given for more than one reactant! 2) Calculate moles of each reactant! 3) Divide the moles of each reactant by the coefficient (a,b,c etc....)! 4) Which ever is smallest, that reactant is the limiting reactant! 5) Use the limiting reactant to calculate the moles of product desired then convert to the units needed (moles, mass, volume, number of atoms etc....)!

Limiting Reactant Problem: A Sample Problem Problem: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4). They ignite on contact ( hypergolic!) to form nitrogen gas and water vapor. How many grams of nitrogen gas form when exactly 1.00 x 102 g N2H4 and 2.00 x 102 g N2O4 are mixed? Plan: First write the balanced equation. Since amounts of both reactants are given, it is a limiting reactant problem. Calculate the moles of each reactant, and then divide by the equation coefficient to find which is limiting and use that one to calculate the moles of nitrogen gas, then calculate mass using the molecular weight of nitrogen gas. Solution: 2 N2H4 (l) + N2O4 (l) 3 N2 (g) + 4 H2O (g) + Energy

Sample Problem cont. molar mass N2H4 = ( 2 x 14.01 + 4 x 1.008 ) = 32.05 g/mol molar mass N2O4 = ( 2 x 14.01 + 4 x 16.00 ) = 92.02 g/mol 1.00 x 102 g 32.05 g/mol Moles N2H4 = = 3.12 moles N2H4 Moles N2O4 = = 2.17 moles N2O4 dividing by coefficients 3.12 mol / 2 = 1.56 mol N2H4 2.17 mol / 1 = 2.17 mol N2O4 Nitrogen yielded = 3.12 mol N2H4 = = 4.68 moles N2 Mass of Nitrogen = 4.68 moles N2 x 28.02 g N2 / mol = g N2 2.00 x 102 g 92.02 g/mol Limiting ! 3 mol N2 2 mol N2H4

Acid - Metal Limiting Reactant - I 2Al(s) + 6HCl(g) 2AlCl3(s) + 3H2(g) Given 30.0g Al and 20.0g HCl, how many moles of Aluminum Chloride will be formed? 30.0g Al / 26.98g Al/mol Al = 1.11 mol Al 1.11 mol Al / 2 = 0.555 20.0g HCl / 36.5gHCl/mol HCl = 0.548 mol HCl O.548 mol HCl / 6 = 0.0913 HCl is smaller therefore the Limiting reactant!

Acid - Metal Limiting Reactant - II since 6 moles of HCl yield 2 moles of AlCl3 0.548 moles of HCl will yield: 0.548 mol HCl / 6 mol HCl x 2 moles of AlCl3 = ______________ mol of AlCl3

Ostwald Process Limiting Reactant Problem What mass of NO could be formed by the reaction 30.0g of Ammonia gas and 40.0g of Oxygen gas? 4NH3 (g) + 5 O2 (g) 4NO(g) + 6 H2O(g) 30.0g NH3 / 17.0g NH3/mol NH3 = 1.76 mol NH3 1.76 mol NH3 / 4 = 0.44 mol NH3 40.0g O2 / 32.0g O2 /mol O2 = 1.25 mol O2 1.25 mol O2 / 5 = 0.25 mol O2 Therefore Oxygen is the Limiting Reagent! 1.25 mol O2 x = 1.00 mol NO mass NO = 1.00 mol NO x = g NO 4 mol NO 5 mol O2 30.0 g NO 1 mol NO

Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation. Side Reactions: These form smaller amounts of different products that take away from the theoretical yield of the main product. Actual yield: The actual amount of product that is obtained. Percent yield: (%Yield) % Yield = x 100 Actual Yield Theoretical Yield

Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe3O4 and Hydrogen gas given below. If 4.55g of Iron is reacted with sufficient water to react all of the Iron to form rust, what is the percent yield if only 6.02g of the oxide are formed? Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield. Solution: 3 Fe(s) + 4 H2O(l) Fe3O4 (s) + 4 H2 (g) 4.55 g Fe 55.85 g Fe mol Fe = 0.081468 mol = 0.0815 mol 0.0815 mol Fe x = 0.0272 mol Fe3O4 1 mol Fe3O4 3 mol Fe 231.55 g Fe3O4 1 mol Fe3O4 0.0272 mol Fe3O4 x = 6.30 g Fe3O4 Percent Yield = x 100% = x 100% = ____ % Actual Yield Theoretical Yield 6.02 g Fe3O4 6.30 g Fe3O4

Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber Process using Nitrogen and Hydrogen Gas. If 85.90g of Nitrogen are reacted with 21.66 g Hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: Divide by coefficient to get limiting: 3.066 g N2 1 10.74 g H2 3 85.90 g N2 28.02 g N2 1 mole N2 moles N2 = = 3.066 mol N2 = 3.066 21.66 g H2 2.016 g H2 1 mole H2 moles H2 = = 10.74 mol H2 = 3.582

Percent Yield/Limiting Reactant Problem - II N2 (g) + 3 H2 (g) 2 NH3 (g) Solution Cont. We have 3.066 moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: 2 mol NH3 1 mol N2 3.066 mol N2 x = 6.132 mol NH3 (Theoretical Yield) 6.132 mol NH3 x = 104.427 g NH3 17.03 g NH3 1 mol NH3 Actual Yield Theoretical Yield Percent Yield = x 100% 98.67 g NH3 104.427 g NH3 Percent Yield = x 100% = %

Flowchart : Solving a stoichiometry problem involving masses of reactants

Mass Percent Composition of Na2SO4 Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses Percent of each Element 2 x Na = 2 x 22.99 = 45.98 % 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% 1 x S = 1 x 32.07 = 32.07 4 x O = 4 x 16.00 = 64.00 142.05 Check % Na + % S + % O = 100% 32.37% + 22.58% + 45.05% = 100.00%

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 3 x O = 3 x 16.00 = 48.00 g Therefore gm Nitrogen/ gm Cpd 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 28.02 kg Nitrogen or: 455 kg NH4NO3 X = 159 kg Nitrogen 80.052 kg NH4NO4