Balancing Chemical Reactions. CHAPTER 3: Quantitative Relationships in Chemical Reactions. Zn + HCl ZnCl 2 + H 2. reactant atoms product atoms

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1 CHAPTER 3: Quantitative Relationships in Chemical Reactions Stoichiometry: Greek for measure elements Stoichiometry involves calculations based on chemical formulas and chemical equations (reactions) quantitative. Law of conservation of mass is central to stoichiometry. Atoms are neither created nor destroyed in a chemical reaction. Atoms are simply rearranged to yield new substances. Zn + HCl ZnCl + H Reactants 1 Zn 1 H 1 Cl Products 1 Zn H Cl Procedure for balancing a reaction: Balance the elements that appear in only one reactant and one product first. The following is a common teaching laboratory reaction: Zn + HCl ZnCl + H Reactants This reaction is not balanced: Products reactant atoms product atoms Zn - Already balanced Balance H and Cl; need H atoms on each side of the reaction Put a coefficient i of in front of the HCl molecule on the reactant side. Also results in balancing Cl's. Balance this reaction by the method of inspection. Balancing Chemical Reactions Formulas of reactants and products can t be changed (i.e., cannot change subscripts) Only amounts of reactants & products can be changed to reach atom balance put coefficients (multipliers) in front of the substances in the reaction Zn + HCl ZnCl + H Now, the reaction is balanced: Reactants Products 1 Zn 1 Zn H H Cl Cl reactant atoms = product atoms 1

2 Another example: C 4 H 10 + O CO + H O Balance C's and H's first: only appear in one reactant and one product each O's balanced last; appears in two products Balance C's, multiply the CO product by 4. C 4 H 10 + O 4 CO + H O Often, the physical state of each substance is specified. Designate physical states of each species: (g) gas (l) - liquid (s) - solid (aq) - aqueous (water) solution Chemical equation with physical states specified: C 4 H 10 (g)+13o (g) 8 CO (g) +10H O(g) Next, balance H's: multiply the HO product by 5 C 4 H 10 + O 4 CO + 5 H O 13 O s Balance the O's; multiply the O reactant by 13/ to give 13 O atoms: C 4 H / O 4 CO + 5 H O Reactants Products 4 C 4 C 10 H 10 H 13 O 13 O Combustion reactions: Involve O as a reactant, typically form CO and H O as products. Examples: CH 4 +O CO + H O C H +5O 4 CO + H O C 8 H 18 +5O 16 CO +18H O BALANCING CHEMICAL REACTIONS Typically, want the smallest whole-number coefficients in a balanced reaction. Hence, fractional coefficient is removed by multiplying the entire equation by. Examples: Balancing Reactions NO + Cl NOCl Fe O 3 + CO Fe + CO C 4 H O 8 CO + 10 H O Balanced chemical reactions are called chemical equations. C 3 H 6 + O CO + H O

3 ATOMIC AND MOLECULAR MASSES It is rather easy to determine the relative masses of atoms experimentally. For example, C has a mass 1 times that of H. O a mass that is 16 times that of H or 1.33 times that of C. Cl a mass that is 35.4 times that of H or.95 times that of C. Atoms have very, very small masses. The O-16 atom has a mass of only.6560 X 10-3 g. These very small numbers are awkward to use. Consequently, a new mass unit (atomic mass unit, amu) is used to express the mass of atoms/molecules/formula units. RELATIVE ATOMIC MASS (WEIGHT) SCALE H Derived from early experiments involving known substances; e.g. H (g) + Li(s) LiH(s) 1.00 g 6.88g 7.88g What is the atomic mass of Li relative to H? 6.88 to 1.00 Define the atomic mass unit (amu): The mass of one C-1 atom is assigned a value of 1 atomic mass units (amu), exactly! Mass of 1 C-1 atom = 1 amu 1 amu = X 10-4 g Some atomic masses: Hydrogen = amu Helium = amu Fluorine = amu RELATIVE ATOMIC MASS SCALE From the periodic table, the mass of C is given as amu, not amu. Developed In A Similar Manner For All Elements H C O Mg Why??? Carbon consists of two stable isotopes (C-1 & C-13) Masses relative to that for H. Isotope C-1 Mass (amu) % abundance C

Average Atomic Mass of C: Avg. Mass C = (1.00000)(0.98890) + (13.00335)(0.01110) Avg. Mass C = 1.01 amu (Sig.

4 Average Atomic Mass of C: Avg. Mass C = ( )( ) + ( )( ) Avg. Mass C = 1.01 amu (Sig. Figures) Average Atomic Mass of chlorine ( isotopes): Isotope Mass (amu) % abundance Cl Cl Avg. Mass Cl = (34.968)(0.7553) + (36.956)(0.447) Avg. Mass Cl = amu This number, 6.0 X 10 3, is given the special name mole X 10 3 Mole: SI base unit for the amount of a substance (mol). 1 mole of anything is 6.0 X 10 3 of whatever we a considering. Average atomic mass is also called atomic weight for an element. Example Problem Copper is a metal used in the production of cables, pennies, etc. The atomic masses of its two stable isotopes, Cu-63 (69.09 %) and Cu- 65 (30.91 %), are 6.93 amu and amu, respectively. Calculate the average atomic mass of Cu (63.55 amu). Mole: The amount of matter that contains as many elementary units (atoms, molecules, etc.) as there are atoms in exactly 1 g of C-1. 1 mole = 6.0 X 10 3 (Avogadro s number) 1 mole C-1 = 6.0 X 10 3 atoms = g KNOW!!!!! It is difficult to work with individual atoms-too small. Consider a very large number of C-1 atoms. One C-1 atom has a mass of 1 amu exactly. How many C-1 atoms are in exactly 1 g?? One C-1 atom has a mass of X 10-3 g. C -1 atom X 10 g 1 3 = - ( 1 g) 6.0 X 10 C atoms Molar Mass Molar mass (g) mass (g) of one mole of a substance Molar mass in grams is numerically equivalent to atomic mass in atomic mass units (amu). 1 C-1 atom = amu; 1 mol C-1 = g 1 H-1 atom = amu; 1 mol H-1 = g 4

We can use the result (5.81 X 10-3 g/cl-35 atom) to calculate the mass of the amu is terms of grams: gram 3 5.81 X 10 g -4 1Cl - 35 atom 1.66 X 10 g = = amu 1Cl 35 atom 34.968 amu amu 1 amu = 1.

011 amu amu/atom is numerically equivalent to grams/mole 1 mol = 6.0 X 10 3 These concepts apply to all elements!!! Example: What is the mass in grams of one Cl- 35 atom? Given: 1 mol Cl-35 = 34.

5 We can use the result (5.81 X 10-3 g/cl-35 atom) to calculate the mass of the amu is terms of grams: gram X 10 g -4 1Cl - 35 atom 1.66 X 10 g = = amu 1Cl 35 atom amu amu 1 amu = X 10-4 g 1 g = 6.0 X 10 3 amu Figure 3.1 Avogadro s number CONVERSION FACTORS: Molar Mass 1 mol C = g 1 avg. C atom = amu amu/atom is numerically equivalent to grams/mole 1 mol = 6.0 X 10 3 These concepts apply to all elements!!! Example: What is the mass in grams of one Cl- 35 atom? Given: 1 mol Cl-35 = g Desired: mass in g/cl-35 atom ANOTHER CALCULATION How many moles O are in g O? First look at given and desired units: g O ( g O) conversion factor 1mol O = g O moles O.31 mol O Solution to Cl-35 example g Cl mol Cl - 35 grams = X 10 1mol Cl X 10 atoms atom Cl (given) X (conversion factor) = desired Use the correct conversion factor to go from given to desired units. ANOTHER CALCULATION How many O atoms are in g O? g O moles O atoms O molar mass Avogadro s # ( g O) conversion factors 1 mol O g O X 10 O atoms 1 mol O = X 10 4 O atoms 5

6 Molar Mass: Figure 3. Mass, in g's, of one mole of a substance. Numerically equal to the formula mass (in amu) or molecular mass (in amu). Examples: Molar mass [Fe(NO 3 3) 3 ] = g Molar mass (NH 3 ) = g Another example: What is the molar mass of C H 5 OH? (46.07 g) Formula and Molecular Masses Formula Mass (FM): sum of the atomic masses of all atoms in the chemical formula (formula unit) Example: Fe(NO 3 ) 3 FM = 1(AM of Fe) + 3(AM of N) + 9(AM of O) FM = 1(55.85 amu) + 3(14.01 amu) + 9(16.00 amu) FM = amu Mass, Moles, Number of Particles molar mass Avogadro's # Grams Moles Molecules 1. What is the average mass in grams of one avg. chlorine atom? (5.89 X 10-3 g). What is the avg. mass in grams of one ethanol (C H 5 OH) molecule? (7.65 X 10-3 g) 3. How many moles of PbCrO 4 (Lead Chromate) are in 45.6 grams? (0.141 mol) 4. How many HCl (hydrogen chloride) molecules are in 46.0 grams? (7.60 X 10 3 molecules) Molecular Mass (MM): The sum of the atomic masses of all atoms of the molecular formula of a molecular substance. Example: NH 3 MM = 1(AM of N) + 3(AM of H) = 1(14.01 amu) + 3(1.008 amu) = amu Another example: What is the molecular mass of aspartame (nutrasweet), (C 14 H 18 N O 5 ), the artificial sweetener? (94.3 amu) Calculations: Percent Composition of elements by mass: The percent by mass of each element in a compound. Example: ammonia, NH 3 Divide the mass of each element in one mole of the compound by the molar mass of the compound, then multiply by 100 to get a percent. 6

7 General Approach: (# moles of element)(mm element) % = MM of compound 3(1.008 g) % H = X 100 = 17.76% g X 100 First, calculate mass of each element: Consider a 100 g sample (any sample size will work). For a 100 g sample: % composition = mass of the element: 1(14.01 g) % N = g X 100 = 8.7 % g C g H Percent composition by mass sum of % composition must equal 100 % % % = % This small deviation from 100 % is due to round-off error. Another example: What is the percent composition of hydrogen in pentylacetate, CH 3 CO C 5 H 11 (responsible for the odor in bananas)? (10.84 %) Second, convert grams of each element into moles of each element: Conversion: Mass molar mass Moles ( g C) ( g H) 1 mol C = mol C 1.01 g C 1mol H = mol H g H Determine Empirical Formulas from a given Percent Composition by Mass (from experiment) What is the empirical formula of a compound that is % C and % H by mass? How to solve?? Third, determine empirical formula: Use the mole amounts as a subscript for each chemical symbol: C H % mass mass moles empirical formula Convert subscripts to smallest whole-number ratio; divide each by the smallest subscript. 7

7.130 C : = 1 7.130 14.56 H : = 1.9994 7.130 Molecular Formula: 8.05 g = 1.9993 empirical formula units 14.

Mass (from experiment) Example: What is the empirical formula of a compound that is 48.6% C, 8.% H, and 43.% O by mass? Calculate moles of each element in 100 g sample: 1mol C 1.01 g C ( 48.

8 7.130 C : = H : = Molecular Formula: 8.05 g = empirical formula units g The Empirical Formula is: The molecular formula is: (CH ) C H 4 CH (smallest whole-number ratio of atoms in the Chemical Formula) Determine Empirical Formulas from a given Percent Composition by Mass (from experiment) Example: What is the empirical formula of a compound that is 48.6% C, 8.% H, and 43.% O by mass? Calculate moles of each element in 100 g sample: 1mol C 1.01 g C ( 48.6 g C ) = mol C 1mol H g H ( 8. g H) = mol H 1mol O g O ( 43. g O) =.700 mol O Desire a Molecular Formula? To determine it we need to know the molar mass. Convert to small mole ratios by dividing each by.700: C 1.50 H 3.01 O 1 # of empirical formula Molar Mass - units in the molecular = Empirical Molar Mass formula Molar mass = 8.05 g (given) Empirical molar mass (CH ) = g (determined from the above example) Multiply mole ratios (subscripts) by an appropriate multiplier to get smallest whole number ratios. Multiply mole ratios by a multiply of to yield the empirical formula: C 3 H 6 O 8

Required Multipliers for Fractional Subscripts Subscripts 1.50 1.33 1.67 1.5 1.75 Multipliers 3 3 4 4 Whole # s 3 3.99 4 5.01 5 5 7 Examples: P 1 O.50 X P O 5 N 1 O 1.50 X N O 3 C 1 H.

33 X 3 C 3 H 4 Calculations Involving Chemical Equations Stoichiometric relationships in chemical equations: CH 4 +O CO + H O The coefficients in the equation give the ratios by molecules or moles of

9 Required Multipliers for Fractional Subscripts Subscripts Multipliers Whole # s Examples: P 1 O.50 X P O 5 N 1 O 1.50 X N O 3 C 1 H.67 X 3 C 3 H 8 C 1 H 1.33 X 3 C 3 H 4 Calculations Involving Chemical Equations Stoichiometric relationships in chemical equations: CH 4 +O CO + H O The coefficients in the equation give the ratios by molecules or moles of reactants and products. For stoichiometric calculations, the coefficients represent mole quantities. 1 mole methane reacts with moles of diatomic oxygen to form 1 mole carbon dioxide and moles water Mass % elements Molar mass Empirical formula Stoichiometric Relationships 100 g sample Grams each element Molecular formula Molar atomic mass Calculate mole ratio Moles each element Also called stoichiometric equivalents Show how reactants and products are related quantitatively (number of molecules or moles). How? Coefficients in the equation gives ratios by molecules or moles of reactants and products. 9

CH 4 + O CO + H O One Stoichiometric equivalent is: 1 mole CH moles 4 O One mole methane reacts with (is stoichiometrically equivalent to) two moles oxygen CH 4 + O CO + H O Stoichiometric

stoichiometric (quantitative) calculations. given g O CH 4 + O CO + H O conversion factor ( g O ) conversion factor conversion factor desired d g CO 1mol O 1 mol CO 44.01 g CO 10.0 3.

10 CH 4 + O CO + H O One Stoichiometric equivalent is: 1 mole CH moles 4 O One mole methane reacts with (is stoichiometrically equivalent to) two moles oxygen CH 4 + O CO + H O Stoichiometric Equivalents 1 mole CH 4 mole O 1 mole CH 4 1 mole CO 1 mole CH 4 mole H O mole O 1 mole CO mole O mole H O 1 mole CO mole H O Use these relationships to construct conversion factors for stoichiometric (quantitative) calculations. given g O CH 4 + O CO + H O conversion factor ( g O ) conversion factor conversion factor desired d g CO 1mol O 1 mol CO g CO g O mol O 1 mol CO = 6.88 g CO STOICHIOMETRIC CALCULATIONS Construct conversion factors for calculations from stoichiometric equivalents CH 4 + O CO + H O Example: How many grams of CO is produced from 10.0 g of O as a reactant? Another example: The smelting (refining) of chalcopyrite ore to produce copper metal. CuFeS +5O Cu + FeO + 4 SO Chalcopyrite ore given X conversion ratios = desired grams O & stoichiometric equivalents grams CO 10

An analogous part to this problem is to calculate how much of the pollutant, sulfur dioxide (SO ), is produced when 1.

5 g Cu ( 1.00 g CuFeS ) 183 g CuFeS mol CuFeS 1mol Cu = 0.347 g Cu How many g oxygen (O ) is required to react with 1.

00 g O 1.00 g CuFeS 183 g CuFeS mol CuFeS 1mol O = 0.

11 An analogous part to this problem is to calculate how much of the pollutant, sulfur dioxide (SO ), is produced when 1.00 g of raw ore is refined? g SO Chalcopyrite ore (CuFeS ) How many g Copper (Cu) can be produced from 1.00 g of CuFeS ore? Set up the conversion scheme: g CuFeS mol CuFeS mol Cu g Cu CuFeS + 5 O Cu + FeO + 4 SO 1mol CuFeS mol Cu 63.5 g Cu ( 1.00 g CuFeS ) 183 g CuFeS mol CuFeS 1mol Cu = g Cu How many g oxygen (O ) is required to react with 1.00 g of CuFeS ore? Set up the conversion scheme: g CuFeS mol CuFeS mol O g O CuFeS + 5 O Cu + FeO + 4 SO ( ) 1mol CuFeS 5 mol O 3.00 g O 1.00 g CuFeS 183 g CuFeS mol CuFeS 1mol O = g O Limiting Reactants Dictate how much product can be produced. Consider the following reaction: N +3H NH 3 1 mole N reacts with 3 moles H to yield moles NH 3 Stoichiometric amounts of reactants N :H mole ratio of 1:3 Both reactants are completely consumed in the reaction. 11

12 N + 3 H NH 3 What would happen if 3 moles N and 3 moles H are reacted? H is completely consumed, however, mol of N remain unreacted. Consider 3 mol N reacting with 3 mol H Calculate the amount of product produced from each reactant: mol NH N = NH 1 mol N 3 ( 3 mol N ) 6 mol 3 H Limiting reactant N Excess reactant Limiting reactant dictates (limits) how much product can be produced. mol NH H = NH 3 mol H 3 ( 3 mol H ) mol 3 H is the limiting reactant, N remains in excess, and moles of NH 3 is produced. N + 3 H NH 3 Reaction Yield Products Consider that 3 moles N and 3 moles H are reacted together. How many moles of N react with the H? moles H moles N 1mol N ( 3 mol H ) = 1mol N 3 mol H How many moles N remain unreacted? 3 mol N 1mol N = mol N unreacted TWO TYPES THEORETICAL YIELD PERCENT YIELD Theoretical Yield: Maximum amount of product that can be obtained from a given amount of reactants - based on the limiting reactant. For the N + H reaction, the theoretical yield of NH 3 is mol. Limiting Reactants in Chemical Reactions How is the limiting reactant determined? By calculating the amount of product that each reactant would produce if it reacted completely. The reactant yielding the least amount of product is the limiting reactant. Consider our previous reaction: N + 3 H NH 3 Typically, the actual yield of a reaction is somewhat less than the theoretical yield Percent Yield actual yield % yield = theoretical yield X 100 Consider that 1.5 mol of NH 3 was obtained. 1.5 mol % yield = X 100 = 75%.0 mol % yield < 100 % 1

Chapter 3. Mass Relationships in Chemical Reactions

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