(Campbell / Callis C142B) Chapter #3 : Stoichiometry -Mole - Mass Relationships in Chemical Systems 3.1: The Mole 3.2: Determining the Formula of an Unknown Compound 3.3: Writing and Balancing Chemical Equations 3.4: Calculating the amounts of Reactant and Product 3.5: Fundamentals of Solution Stoichiometry

MOLE The Mole is based upon the definition: The amount of substance that contains as many elementary particles (atoms, molecules, ions, or other?) as there are atoms in exactly 12 grams of carbon Mole = x particles (atoms, molecules, ions, electrons, or…) = N A particles ~100 million x 100 million x 100 million

Avogadro’s Number (N A ) N A = x = # of particles (atoms, molecules, ions, electrons, or…) in one mole of that thing.

Fig 3.1 (P 90) Counting objects of fixed relative mass 12 red 7g each = 84g 12 yellow each=48g 55.85g Fe = x atoms Fe 32.07g S = x atoms S

Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = amu 1 mole of H = g = x atoms 1 atom of Fe = amu 1 mole of Fe = g = x atoms 1 atom of S = amu 1 mole of S = g = atoms 1 atom of O = amu 1 mole of O = g = atoms Molecular mass: 1 molecule of O 2 = amu 1 mole of O 2 = g = molecules 1 molecule of S 8 = amu 1 mole of S 8 = g = molecules

Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = amu 1 mole of H = g = x atoms 1 atom of Fe = amu 1 mole of Fe = g = x atoms 1 atom of S = amu 1 mole of S = g = x atoms 1 atom of O = amu 1 mole of O = g = x atoms Molecular mass: 1 molecule of O 2 = x 2 = amu 1 mole of O 2 = g = x molecule 1 molecule of S 8 = x 8 = amu 1 mole of S 8 = g = x 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, called its molar mass. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( amu) + amu = amu Mass of one molecules of water = amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( g ) + g = g g H 2 O = x molecules of water = 1 mole H 2 O

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, called its molar mass. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( amu) amu = amu Mass of one molecules of water = amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( g ) g = g g H 2 O = x molecules of water = 1 mole H 2 O

Fig 3.2 (P 87) One mole of common sbustances CaCO g Oxygen g Copper g Water g (This balloon volume is not really big enough. Need ~10-20 liters, depending on pressure inside.)

Fig. 3.3

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 3680 o C. 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 = No. of W 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 3680 o C. 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 = mol 1.90 x mol NO. of W atoms = 1.90 x mol W x = = 1.15 x atoms of Tungsten 1 mol W g W x 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 Na 3 PO 4. Calculating the molar mass: M = Converting mass to moles: # Formula units =

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 Na 3 PO 4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x g/mol + 1 x g/mol + 4 x g/mol = g/mol g/mol g/mol = g/mol Converting mass to moles: Moles Na 3 PO 4 = 38.6 g Na 3 PO 4 x (1 mol Na 3 PO 4 ) g Na 3 PO 4 = mol Na 3 PO 4 Formula units = mol Na 3 PO 4 x x formula units 1 mol Na 3 PO 4 = 1.46 x formula units

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

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 of 1 mole sucrose = To find mass % of C = Mass Fraction of C = =

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 of 1 mole sucrose g Cpd/mol = To find mass % of C = x 100% = 42.10% Mass Fraction of C = =

Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = Mass % of O = x 100% = (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = mol H x M of H mass of 1 mol sucrose mol O x M of O mass of 1 mol sucrose

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% = % 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 mol H x M of H 22 x g H mass of 1 mol sucrose g mol O x M of O 11 x g O mass of 1 mol sucrose g g C 1 g sucrose

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

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 = x 100% = 35.00% 28.02g N g %H = x 100% = 5.037% 4.032g H g %O = x 100% = 59.96% 48.00g O g %

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 = x 100% = 2.06% H 2(1.008g H 2 ) 98.09g %S = x 100% = 32.69% S 1(32.07g S) 98.09g %O = x 100% = 65.25% O 4(16.00g O) g Check = %

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.

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

Steps to Determine Empirical Formulas Mass (g) of Element Moles of Element Preliminary Formula Empirical Formula ÷ M (g/mol ) Use no. of moles as subscripts. Change to integer subscripts: ÷ smallest, conv. to whole #.

Determining Empirical Formulas from 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 molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = Moles of Cr = Moles of O =

Determining Empirical Formulas from 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 molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = g Na x = mol Na Moles of Cr = g Cr x = mol Cr Moles of O = g O x = mol O 1 mol Na g Na 1 mol Cr g Cr 1 mol O g O

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: 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

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 =

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

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 = moles C Moles of H = Mass of H x = moles H Moles of O = Mass of O x = moles 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 = CH 2 O 1 mole C g C 1 mol H g H 1 mol O 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: Whole-number multiple = = M of Glucose 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 weight of the empirical formula is: 1 x C + 2 x H + 1 x O = 1 x x x = g/mol Whole-number multiple = = = = 6.00 = 6 M of Glucose empirical formula mass Therefore the Molecular Formula is: C 1 x 6 H 2 x 6 O 1 x 6 = C 6 H 12 O 6

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 (0.591) => 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/( 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