Mole and Stoichiometry PDF

The Mole and the Stoichiometry of Formulas and Equations Photosynthesis 2  3: Stoichiometry of Formulas and Equations 3.1 The Mole 3.2 The Formula of Compounds 3.3 Writing and Bal...

The Mole and the Stoichiometry of Formulas and Equations Photosynthesis 2  3: Stoichiometry of Formulas and Equations 3.1 The Mole 3.2 The Formula of Compounds 3.3 Writing and Balancing Chemical Equations 3.4 Calculating Quantities of Reactant and Product 3.5 The Reversibility of Reactions and the Equilibrium State 3 ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. The Mole The mole (mol) is the amount of a substance that contains the same number of entities as the number of atoms in exactly 12 g of carbon-12 The term “entities” refers to atoms, ions, molecules, formula units, or electrons – in fact, any type of particle One mole (1 mol) contains 6.022x1023 entities (to four significant figures) This number is called Avogadro’s number and is abbreviated as NA 1 mol of carbon-12 contains 6.022x1023 carbon-12 atoms and has a mass of 12 g 4 One Mole of Some Familiar Substances The mass in atomic mass unit of one atom/molecule/formula unit of an element/compound is the same numerically as the mass in grams of 1 mole of atoms of the element/compound: → 1 atom of Cu has a mass of 63.55 amu, and 1 mol (6.022x1023 atoms) of Cu has a mass of 63.55 g → 1 molecule of H2O has a mass of 18.02 amu, and 1 mol (6.022x1023 molecules) of H2O has a mass of 18.02 g → 1 formula unit of NaCl has a mass of 58.44 amu, and 1 mol (6.022x1023 formula units) of NaCl has a mass of 58.44 g 5 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Molar Mass The molar mass (M) of a substance is the mass per mole of its entities (atoms, molecules, or formula units); its unit is g/mol For monatomic elements, the molar mass is the same as the atomic mass in grams per mole (the atomic mass is simply read from the Periodic Table) The molar mass of Ne = 20.18 g/mol 6 Molar Mass For molecular elements and for compounds, the formula is needed to determine the molar mass → the molar mass is the sum of the molar masses of the atoms in the formula The molar mass M of O2 = 2 x M of O = 2 x 16.00 g/mol = 32.00 g/mol The molar mass M of SO2 = 1 x M of S + 2 x M of O = [32.06 + 2 ∙ (16.00)] g/mol = 64.06 g/mol 7 Example: Chemical Formula of Glucose Subscripts in a formula refer to individual atoms (or ions) as well as to moles of atoms (or ions) Glucose: C6H12O6 Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of 6 atoms 12 atoms 6 atoms compound Moles of atoms/mole 6 mol of atoms 12 mol of atoms 6 mol of atoms of compound Atoms/mole of 6x(6.022x1023) 12x(6.022x1023) 6x(6.022x1023) compound atoms atoms atoms Mass/molecule of 6x(12.01 amu) 12x(1.008 amu) 6x(16.00 amu) compound = 72.06 amu = 12.10 amu = 96.00 amu Mass/mole of 72.06 g 12.10 g 96.00 g compound 8 Moles, Mass, and Number of Chemical Entities no. of grams Mass g = no. of moles × 1 mol 1 mol No. of moles = mass (g) × no. of grams 𝐦 𝐠 = 𝐧 𝐦𝐨𝐥 × M 𝐠Τ𝐦𝐨𝐥 6.022 × 1023 entities No. of entities = no. of moles × 1 mol 1 mol No. of moles = no. of entities × 6.022 × 1023 entities 𝐍 = 𝐧 𝐦𝐨𝐥 × 𝑵𝑨 𝟏Τ𝐦𝐨𝐥 9 Amount-Mass-Number Relationship Amount-mass-number relationships of elements: 10 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Sample Problem 1 Converting Between Mass and Amount of an Element PROBLEM: How many grams of Ag are in 0.0342 mol of Ag? 11 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Sample Problem 2 Converting Between Number of Entities and Amount of an Element PROBLEM: How many Ga atoms are in 2.85 x 10-3 mol of gallium? 12 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Amount-Mass-Number Relationship Amount-mass-number relationships of compounds: 13 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Sample Problem 3 Converting Between Number of Entities and Mass of Compound PROBLEM: How many molecules are in 8.92 g of nitrogen dioxide? 14 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Mass Percent from the Chemical Formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) × 100 (%) molecular (or formula) mass of compound (amu) Mass % of element X = moles of X in formula x molar mass of X (g/mol) × 100 (%) mass (g) of 1 mol of compound 15 Mass of an Element from its Mass Fraction Mass fraction can also be used to calculate the mass of a particular element in any mass of a compound Mass of any element in sample = mass of element in 1 mol of compound mass of compound x mass of 1 mol of compound 16 Sample Problem 4 Calculating the Mass of an Element in a Compound PROBLEM: Determine the mass (g) of nitrogen in 650. g of ammonium nitrate. 17 Empirical and Molecular Formulas The empirical formula is the simplest formula for a compound that agrees with the elemental analysis; it shows the lowest whole number of moles and gives the relative number of atoms of each element present The empirical formula for hydrogen peroxide is HO The molecular formula shows the actual number of atoms of each element in a molecule of the compound The molecular formula for hydrogen peroxide is H2O2 The structural formula also shows the relative placement and connections of atoms in the molecule The structural formula for hydrogen peroxide is H–O–O–H 18 Determining the Empirical Formula To find the empirical formula: Determine the mass (g) of each component element Convert each mass (g) to amount (mol), and write a preliminary formula Convert the amounts (mol) mathematically to whole-number (integer) subscripts 19 Determining the Molecular Formula The molecular formula gives the actual numbers of moles of each element present in 1 mol of compound The molecular formula is a whole-number multiple of the empirical formula molar mass (g/mol) = whole−number multiple empirical formula mass (g/mol) 20 Sample Problem 5 Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: During excessive physical activity, lactic acid (ℳ = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound has 40.0 mass % C, 6.71 mass % H, and 53.3 mass % O. a) Determine the empirical formula of lactic acid. b) Determine its molecular formula. 21 Chemical Formulas and Molecular Structures The empirical formula tells nothing about molecular structure because it is based solely on mass analysis; different compounds can have the same empirical formula The molecular formula tells nothing about molecular structure because the atoms can bond in different arrangements Isomers are compounds with the same molecular formula, and thus molar mass, but different properties 22  Compounds with Empirical Formula CH2O Whole-Number Name Molecular Formula ℳ (g/mol) Use or Function Multiple Disinfectant; biological Formaldehyde CH2O 1 30.03 preservative Acetate polymers; vinegar Acetic acid C2H4O2 2 60.05 (5% solution) Causes milk to sour; forms in Lactic acid C3H6O3 3 90.08 muscles during exercise Forms during sugar Erythrose C4H8O4 4 120.10 metabolism Component of many nucleic Ribose C5H10O5 5 150.13 acids and vitamin B2 Major nutrient for energy in Glucose C6H12O6 6 180.16 cells 23 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Chemical Equations A chemical equation uses formulas to express the identities and quantities of substances involved in a physical or chemical change 24 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Balancing a Chemical Equation The equation must be balanced; the same number and type of each atom must appear on both sides A balanced chemical equation is adhered to the law of conservation of mass Balancing process: 1) Translating the statement: we first translate the chemical statement into a “skeleton” equation 25 Balancing a Chemical Equation 2) Balancing the atoms: we match the numbers of each type of atom on the left and the right of the yield arrow → we place a balancing (stoichiometric) coefficient 1 1 1 1 ½ 1 3) Adjusting the coefficients: - use the smallest whole-number coefficients - the coefficient of 1 is not written 2 1 2 2 2 26 Balancing a Chemical Equation 4) Checking: after balancing and adjusting the coefficients, we always check that the equation is balanced Reactants (2 Mg, 2 O) Products (2 Mg, 2 O) 5) Specifying the states of matter: the final equation also indicates the physical state of each substance or whether it is dissolved in water 2 Mg (s) + O2 (g) 2 MgO (s) 27 A Three-level View of a Reaction 28 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Balancing Process Chemical formulas cannot be altered Other reactants or products cannot be added A balanced equation remains balanced if you multiply all the coefficients by the same number (by convention, the smallest whole-number coefficients are used) Each side of the equation may not have the same number of molecules or moles 29 Sample Problem 6 Balancing a Chemical Equation PROBLEM: Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. 30 Visualizing a Reaction with a Molecular Scene 31 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Stoichiometric Calculations Stoichiometry is the chemical arithmetic needed for mole- mass conversions In a balanced equation, the amounts (mol) of substances are stoichiometrically equivalent to each other, which means that a specific amount of one substance is formed from, produces, or reacts with a specific amount of the other The quantitative relationships are expressed as stoichiometrically equivalent molar ratios that we use as conversion factors to calculate the amounts 32 Information Contained in a Balanced Equation Viewed in Reactants Products → Terms of C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) 1 molecule C3H8 Molecules → 3 molecules CO2 + 4 molecules H2O + 5 molecules O2 Amount (mol) 1 mol C3H8 + 5 mol O2 → 3 mol CO2 + 4 mol H2O 44.09 amu C3H8 Mass (amu) → 132.03 amu CO2 + 72.06 amu H2O + 160.00 amu O2 44.09 g C3H8 Mass (g) → 132.03 g CO2 + 72.06 g H2O + 160.00 g O2 Total mass (g) 204.09 g → 204.09 g 33 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Amount-Mass-Number Relationships 34 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Sample Problem 7 Calculating Quantities of Reactants and Products: Amount (mol) to Mass (g) PROBLEM: During the process of roasting (heating with oxygen gas) copper(I) sulfide, powdered copper(I) oxide and gaseous sulfur dioxide are formed. How many grams of sulfur dioxide form when 10.0 mol of copper(I) sulfide reacts? 35 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Reactions in Sequence For stoichiometric purposes, when the same substance forms in one reaction and reacts in the next (it is common to both reactions), we eliminate that substance in an overall (net) equation The steps in writing the overall equation are: 1) Write the sequence of balanced equations 2) Adjust the equations arithmetically to cancel the common substance(s) 3) Add the adjusted equations together to obtain the overall balanced equation 36 Sample Problem 8 Writing an Overall Equation for a Reaction Sequence PROBLEM: Roasting is the first step in extracting copper from chalcocite (copper(I) sulfide). In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two- step process. 37 Limiting Reactants One reactant may limit the amount of product that can form Limiting Reactant: the reactant that is present in limiting amount → the limiting reactant will be completely used up in the reaction, and the extent to which a chemical reaction takes place depends on it Excess Reactant: any of the other reactants still present after determination of the limiting reactant → some of this reactant will be left over 38 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Limiting Reactants – Example At a high temperature, ethylene oxide reacts with water to form ethylene glycol, which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C2H4O ( aq ) + H2O ( l ) → C2H6O2 ( I ) If 3 mol of ethylene oxide react with 5 mol of water: 39 Copyright ©Pearson Education, Inc. All rights reserved. Reaction Table The reaction table shows: Initial quantities of reactants and products before the reaction Change in the quantities of reactants and products during the reaction Final quantities of reactants and products after the reaction If 3 mol of ethylene oxide react with 5 mol of water: C2H4O ( aq ) + H2O ( l ) → C2H6O2 ( I ) Amount (mol) C2H4O (aq) + H2O (l) → C2H6O2 (l) Initial 3 5 0 Change -3 -3 +3 Final 0 2 3 40 Copyright ©Pearson Education, Inc. All rights reserved.  Sample Problem 9 Calculating Quantities in a Limiting-Reactant Problem: Amount to Amount PROBLEM: In a preparation of ClF3, 0.750 mol of Cl2 reacts with 3.00 mol of F2. (a) Find the limiting reactant. (b) Write a reaction table (reporting the initial, change, and final quantities of reactants and products). 41 Stoichiometric Relationships 42 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Reaction Yields The theoretical yield is the amount of product calculated using the molar ratios from the balanced equation The actual yield is the amount of product actually obtained; it is usually less than the theoretical yield Theoretical and actual yield are expressed in units of amount (moles) or mass (grams) Reactant mixtures often proceed through side reactions that form different products 43 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Percent Reaction Yield The percent yield is the actual yield expressed as a percentage of the theoretical yield: actual yield % yield = × 𝟏𝟎𝟎 theoretical yield Percent yield is always less than 100% 44 Sample Problem 10 Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is made by reacting sand (silicon dioxide, SiO2) with powdered carbon at high temperature. Carbon monoxide is also formed. What is the percent yield if 51.4 kg of SiC is recovered from processing 100.0 kg of sand? 45 The Equilibrium State Dynamic equilibrium: no further changes appear in the amounts of reactants or products 46 Copyright ©McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.