Chapter 3 : Stoichiometry

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Frederick Garrett
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Chapter : Stoichiometry 14 KMnO 4 + 4 C H 5 (OH) --> 7 K CO + 7 Mn O + 5 CO + 16 H O + HEAT Chemical changes : Why they occur? How fast?

1 Chapter : Stoichiometry 14 KMnO C H 5 (OH) --> 7 K CO + 7 Mn O + 5 CO + 16 H O + HEAT Chemical changes : Why they occur? How fast? => Need to know chemical stoicheometry Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions

Mass spectrometer - the most accurate method currently available for

2 Atomic Mass Atomic mass unit (amu) - 1 C is assigned a mass of exactly 1 amu, and the masses of all other atoms are given relative to this standard. Mass spectrometer - the most accurate method currently available for comparing the masses of atoms Mass Mass ( 1 ( 1 1 C) Mass ( C) C) amu

Atomic Masses => atomic mass (atomic weight) in amu. C => 1.011 but there is no such carbon with atomic mass 1.011?? Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 1 C 1.

3 Atomic Masses => atomic mass (atomic weight) in amu. C => but there is no such carbon with atomic mass 1.011?? Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 1 C 1.11% 1 C <0.01% 14 C Atomic mass = (0.9889)(1amu) + (0.0111)(1.004 amu) = 1.01 amu => Average atomic mass = Handful of carbons all together act like 1.01 C because we can't separate them easily.

4 Atomic Masses Ex) Average mass of an element Ne Ne 10 Ne 10 Ne Cu average atomic mass of copper (0.6909)(6.9 amu) (0.091)(64.9 amu) 6.55 amu 6.9 amu 64.9 amu

The Mole Mole : The number equal to the number of carbon atoms in exactly 1 grams of pure 1 C. = 6.

5 The Mole Mole : The number equal to the number of carbon atoms in exactly 1 grams of pure 1 C. = x 10 = Avogadro's number 1 e of anything = 6.0 x 10 units of that thing Cu 1amu ( atoms )( ) 1g atom amu 1g exact number I Al S Hg Fe 1 of element The e is defined such that a sample of a natural element with a mass equal to the element's atomic mass expressed in grams contains 1 e of atoms.

The Mole Ex) Mass of 6 Am atoms in gram (Am => 4 amu)? 4amu 1g mass 6 Amatoms atom 6.0 10 Ex) Number of Al atoms in 10.0 g of Al (Al => 6.98 amu)? 1.4 10 amu g 10.

6 The Mole Ex) Mass of 6 Am atoms in gram (Am => 4 amu)? 4amu 1g mass 6 Amatoms atom Ex) Number of Al atoms in 10.0 g of Al (Al => 6.98 amu)? amu g 10.0g 6.98g atoms. 10 atoms 10.0 g of Al Ex) Number of es and mass of 5.00 x 10 0 Co atoms (Co => 58.9)? atoms atoms g g 4

Molar Mass A substance s ar mass (ecular weight) is

ar mass of CH 4 => mass of 1 C = 1 x 1.01 g = 1.

04 g Ex) How many ecules of 1 mg (1 x 10-6 g) of

CH O odor of bananas or pears H C CH CH CH O C CH 1

7 Molar Mass A substance s ar mass (ecular weight) is the mass in grams of one e of the compound. ar mass of CH 4 => mass of 1 C = 1 x 1.01 g = 1.01 g mass of 4 H = 4 x g = 4.0 g g Ex) How many ecules of 1 mg (1 x 10-6 g) of isopentyl acetate? How many C atoms in it? CH O odor of bananas or pears H C CH CH CH O C CH 1 bee sting contains ca. 1 mg of isopentyl acetate call signal for attacking together

8 Molar Mass Ex) How many ecules of 1 mg (1 x 10-6 g) of isopentyl acetate? How many C atoms in it? H C CH CH CH CH O O C CH 1. 01g 7 C 84.07g C 1.008g 14 H 14.11g H 16.00g O.00g O ar mass = g g isopentyl acetate 10.18g isopentyl acetate C atoms isopentyl acetate 4 10 isopentyl acetate 16 C atoms

9 Percent Composition of Compounds mass of element Mass percent of an element in a compound = 100% mass of compound isopentyl acetate (C 7 H 14 O ) CH O CH CH C H C CH O CH 1. 01g 7 C 84.07g C 1.008g 14 H 14.11g H 16.00g O.00g O ar mass = g mass % of mass % of mass % of 84.07g C 100% 64.58% 10.18g 14.11g H 100% 10.84% 10.18g.00g O 100% 4.58% 10.18g

Determining Formula of a Compound Formulas ecular formula = (empirical formula) n ecular formula = C 6 H 6 = (CH) 6 empirical formula = CH [n = integer] Mass%of each element in compound 0.

10 Determining Formula of a Compound Formulas ecular formula = (empirical formula) n ecular formula = C 6 H 6 = (CH) 6 empirical formula = CH [n = integer] Mass%of each element in compound g mass% of H 100% 16.% g g mass% of C 100% 8.67% g mass% of N ( )% 45.11% A g of compound A consisting of C, H, N Massesof H and C in compound A H O => g emprical fomula? Mass of H in g of H O g (mass% of H in H O) ar massof H atom g g g g ar massof H O 18.0g Mass of C in g of CO g (mass% of C in CO ) ar massof C atom 1.01g g g g ar massof CO 44.01g CO => g Moles of each element in 100 g of 16. g H g 8.67 g C g 45.11g N g compound A Whole-number ratio of atoms in A C:H:N=.0:16.09:.19 = 1:5:1 Empirical formula of A => CH 5 N

Determining Formula of a Compound 1. Base calculation on 100 grams of a compound. (mass % of each element in the compound. Determine es of each element in 100 grams of the compound.

=> Empirical formula => ar mass information => Molecular formula Ex) Caffeine contains 49.48% C, 5.15% H, 8.87 % N, and 16.9 % O by mass and has ar mass of 194. g/. => Empirical formula?

11 Determining Formula of a Compound 1. Base calculation on 100 grams of a compound. (mass % of each element in the compound. Determine es of each element in 100 grams of the compound.. Divide each value of es by the smallest of the values. 4. Multiply each number by an integer to obtain all whole numbers. => Empirical formula => ar mass information => Molecular formula Ex) Caffeine contains 49.48% C, 5.15% H, 8.87 % N, and 16.9 % O by mass and has ar mass of 194. g/. => Empirical formula? Molecular formula? Moles Massesof of H and each C in compound element A in 100 g of caffeine g C 4.10 Mass of H in g of Hg O g (mass% of H in HO) ar 1massof H atom g g H g g ar massof gh O 18.0g Mass of C in g 8.87 g N of CO g (mass% of C in CO) ar massof g C atom 1.01g g g g ar 1massof g O CO g 16.00g Mass%of each element in compound A Whole-number ratio gof atoms in caffeine mass% of H 100% 16.% g C:H:N:O=4.10:5.11:.061:1.06 = 4:5:: g mass% of C 100% 8.67% Empirical formula of g caffeine mass% of N ( )% 45.11% => C 4 H 5 N O Moles of each element in 100 g of compound A Molecular 1formular of caffeine 16. g H g C 4 H 5 N O 1=> empirical formula mass 8.67 g C.0 = (1.01 x g1.008 x x x 11) g/ = g/ 45.11g => ar N mass g of caffeine = 194. g/ => Molecular formula = empirical formula x Whole-number => C 8 H 10 N 4 O ratio of atoms in A C:H:N=.0:16.09:.19 = 1:5:1 Empirical formula of A => CH 5 N Look at the text book. It has a different way to solve.

12 Chemical Equations Chemical Reaction: a reorganization of the atoms in one or more substances. Ex) Methane (CH 4 ) burns in air to produce carbon dioxide (CO ) and water (H O) Chemical Equation : a representation of a chemical reaction with reactants and products CH 4 + O CO + H O reactants products In a chemical reaction, atoms are neither created or destroyed. => chemical equation must be balanced. CH 4 + O CO + H O reactants products

Liquid Gas Dissolved in water (in aqueous solution) Symbol (s) (l) (g) (aq) CH 4 (g)

13 Chemical Equations The Meaning of Chemical Equation : two types of information => the nature of the reactants and products and the relative numbers of each State Solid Liquid Gas Dissolved in water (in aqueous solution) Symbol (s) (l) (g) (aq) CH 4 (g) + O (g) CO (g) + H O(g) reactants products HCl(aq) + NaHCO (s) CO (g) + H O(l) +NaCl(aq)

Ex) Balance the chemical equation for the decomposition of ammonium dichromate. 1. (NH 4 ) Cr O 7 (s) Cr O (s) + N (g) + H O(g).

14 Balancing Chemical Equations In a chemical reaction, atoms are neither created or destroyed. => chemical equation must be balanced. 1. Determine what reation is occuring: reactants and products => unbalanced equation. Balance the equation by inspection starting with the most complicated ecule(s).. Check. Ex) Balance the chemical equation for the decomposition of ammonium dichromate. 1. (NH 4 ) Cr O 7 (s) Cr O (s) + N (g) + H O(g). (NH 4 ) Cr O 7 (s) Cr O (s) + N (g) + 4H O(g) Ex) Balance the chemical equation for the reaction of ammonia gas with oxygen to form gaseous nitric oxide and water vapor (1000 o C) => first step of Ostwald process (commertial production of nitric acid, HNO ) 1. NH (g) + O (g) NO(g) + H O(g). NH (g) + O (g) NO(g) + H O(g) NH (g) + O (g) NO(g) + H O(g) NH (g) +.5O (g) NO(g) + H O(g) (NH 4 ) Cr O 7 Cr O 4NH (g) + 5O (g) 4NO(g) + 6H O(g)

9 O 49g O Ex) What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide (LiOH)?

15 Stoichiometric Calculations: Amounts of Reactants and Products Ex) What mass of oxygen will react with 96.1 g of propane? C H 8 (g) + 5O (g) CO (g) + 4H O(g) g C H 44.1g 8.18 CH8 5 x.18 O = 10.9 O reacts..00g 10.9 O 49g O Ex) What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide (LiOH)? LiOH(s) + CO (g) Li CO (s) + H O(l) Columbia Apollo 1 Emergency Rig LiOH Unit LiOH(s) + CO (g) Li CO (g) + H O(l) 1.00kgLiOH 41.8 LiOH.95g 0.5 x 41.8 CO = 0.9 CO reacts. 44.0g 0.9 CO g CO

Calculations Involving a Limiting Reactants CH 4 (g) + H O(g) H (g) + CO(g) The limiting reactant is the reactant that is consumed first, limiting the amounts of products

06 NH 17.0g 90.4g CuO 79.55g N (g), H O(g) Cu(s) 1.14 CuO The excess reactant is the reactant that still remains after the reaction completes. 1.14 CuO reacts with 1.

16 Calculations Involving a Limiting Reactants CH 4 (g) + H O(g) H (g) + CO(g) The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. Ex) 18.1 g 90.4 g NH (g) CuO(s) After reaction, mass of each compound? NH (g) + CuO(s) N (g) + Cu(s) + H O(g) NH (g) + CuO(s) N (g) + Cu(s) + H O(g) 18.1g NH 1.06 NH 17.0g 90.4g CuO 79.55g N (g), H O(g) Cu(s) 1.14 CuO The excess reactant is the reactant that still remains after the reaction completes CuO reacts with 1.14 x / (=0.760 ) NH and produces 1.14 Cu, 1.14/ (=0.80 ) N, and 1.14 H O. 6.55g 1.14 Cu 7.4g Cu 8.0g 0.80 N 10.6g N 18.01g 1.14 HO 0.5g HO 17.0g ( ) NH 5.1g limiting NH

17 Calculations Involving a Limiting Reactants Theoretical yield : the amount of a product formed when the limiting reactant is completely comsumed g 90.4 g NH (g) CuO(s) N (g), H O(g) Cu(s) Theoretical yield of N? 10.6 g Actual yield Percent yield = 100% Theoretical yield Ex) Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. When 68.5 kg of CO is reacted with 8.76 kg of H,.57 x 10 4 g of methanol is actually produced. Percent yield of methanol? H (g) + CO(g) CH OH(l) g H.016g g CO 8.0g H CO Theoretical yield of limiting Percent yield CH OH.04g CHOH g g % 5.0% 4 g CH OH

18 Summary of Solving a Stoichiometry Problem 1. Balance the equation.. Convert masses to es.. Determine which reactant is limiting. 4. Use es of limiting reactant and e ratios to find es of desired product. 5. Convert from es to grams.

Chapter 5. Stoichiometry

Chapter 5. Stoichiometry Chapter 5 Stoichiometry Chapter 5 Table of Contents (5-1) Counting by weighing (5-2) Atomic masses (5-3) Learning to solve problems (5-4) The mole (5-5) Molar mass (5-6) Percent composition of compounds