Unit 4: The Mole and Chemical Composition UNIT 4: THE MOLE AND CHEMICAL COMPOSITION
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1 Uit 4: The ole ad Cheical Copositio Cheistry UNIT 4: THE OLE AND CHEICAL COPOSITION Chapter 7: The ole ad Cheical Copositio 7.1 & 7.: Avogadro s Nuber ad olar Coversio & Relative Atoic ass ad Cheical Forulas ole (ol): - a group of atos or olecules ubered (Avogadro s Nuber, N A ) Exaples: 1 ol of carbo (C) carbo atos 1.01 g (sae as the au) 1 ol of fluorie (F ) fluorie olecules g (iclude subscripts with au) Exaple 1: Fid the uber of olecules i 3.50 oles of platiu. 1 ol Pt Pt atos 10 atos 3.50 ol atos of Pt 1ol Exaple : How ay oles of sucrose are there if there are olecules of C 1 H O 11? olecules 1 ol C 1 H O C 1 H O 11 olecules 1ol 10 olecules Exaple 3: Fid the uber of Br io i 0.65 ol of gbr ol of C 1 H O 11 1 ol gbr gbr forula uits 0.65 ol 10 uits 1ol Br io Br ios 1gBr uit olar ass (g/ol): - soeties refer to as forula ass, olecular weight, is the ass per oe ole of atos or olecules. - olar ass of a oo-atoic eleet is the sae as the atoic ass. - olar ass of a copoud, diatoic eleet, or polyatoic eleet is the sae as the cobie atoic asses of all atos i the olecule. (Be careful coutig uber of atos of a polyatoic io i parethesis.) Exaple 4: Fid the olar ass of the followig. a. potassiu b. phosphorus (P 4 ) c. glucose (C 6 H 1 O 6 ) K g/ol P g/ol P g/ol C 6 H 1 O 6 (6 1.01) + (1 1.01) + ( ) C 6 H 1 O g/ol d. sodiu chroate Na CrO 4 e. iro (III) itrite Fe(NO ) 3 Na CrO 4 (.99) ( ) Na CrO 4 /ol Fe(NO ) ( ) + ( ) Fe(NO ) g/ol Page 7. Copyrighted by Gabriel Tag B.Ed., B.Sc.
2 Cheistry Uit 4: The ole ad Cheical Copositio Covertig Betwee ass ad Nuber of Particles: 1. Fid the olar ass (i g/ol).. Set up a Coversio Factor betwee olar ass ad the Avogadro s Nuber ad Solve. Exaple 5: Calculate the ass of olecules of sulphur (S 8 ). S g/ol g/ol g of S 8 1 ol S S 8 olecules g olecules 30 g of S 8 10 olecules Exaple 6: Deterie the uber of phosphorus atos i 60.0 g of solid phosphorus (P 4 ). P g/ol 1.88 g/ol 1.88 g of P 4 1 ol P P 4 olecules There are four phosphorus atos i oe olecular uit of P g 10 olecules 1.88 g 4 P atos 1P olecule phosphorus atos Exaple 7: Deterie the uber of Na + io i 9.3 g of sodiu phosphate Na 3 PO 4. Na 3 PO 4 (3.99) ( ) g/ol g of Na 3 PO 4 1 ol Na 3 PO Na 3 PO 4 forula uits There are three Na + ios i oe forula uit of Na 3 PO g 10 uits g + 3 Na ios 1 Na PO uits Na + ios Covertig betwee ass ad oles: (eed to fid olar ass first!) oles (ol) ass (g) olar ass (g/ol) Exaple 8: Calculate the uber of oles for 0.0 g of ethaol. Exaple 9: Deterie the ass of 0.85 ol of aluiu carboate oles ass olar ass Ethaol C H 5 OH (1.01) + 6(1.01) g/ol 0.0 g ol g/ol Aluiu carboate Al (CO 3 ) 3 (6.98) + 3(1.01) + 9(16.00) 3.99 g/ol (0.85 ol)(3.99 g/ol) 199 g Copyrighted by Gabriel Tag B.Ed., B.Sc. Page 73.
3 Uit 4: The ole ad Cheical Copositio Cheistry Atoic ass: - soeties called atoic weight. - the ass of the ato i atoic ass uit (au). - 1 au exactly oe-twelfth the ass of oe carbo-1 ato kg. Average Atoic ass: - Average ass of a ato ad its isotopes after accoutig their proportios of abudace (as stated o the Periodic Table of Eleets). Relative Abudace: - the relative proportio of various isotopes of a eleet. Relative Percetage Abudace: - the relative proportio of various isotopes of a eleet i percetage. Average Atoic ass (Relative Abudace of Isotope A)(ass Nuber of Isotope A) + (Relative Abudace of Isotope B)(ass Nuber of Isotope B) + (Relative Abudace of Isotope C)(ass Nuber of Isotope C) + Exaple 10: Iro has three atural isotopes. If 56 Fe ad 57 Fe have relative percetage of abudace at % ad.401% respectively, ad the rest is 54 Fe, calculate the average atoic ass of iro? % abudace of 54 Fe 100% %.401% 5.845% Average au of Iro ( )(54) + ( )(56) + (0.0401)(57) 5.845% of ass Nuber % of ass Nuber % of ass Nuber 57 Average au of Iro au Assiget 7.1 pg. 9 #1 to 5 (Practice); pg. 1 #1 to 4 (Practice); pg. #1 to 3 (Practice); pg. 3 # 1 to pg #1 to 4 (Practice); pg. 6 #1 ad (Practice); pg. 40 #1, 3 to 1, 14 to : Forulas ad Percetage Copositio Percetage Copositio: - also called ass percet or percetage ass. - it is the ass percetage of each eleet i a copoud. For Copoud A x B y C z with its Total ass (), the ass Percetages are: %A 100% A %B B 100 % %C 100 % For Copoud A x B y C z with its olar ass (), the ass Percetages are: C x ( )( ) A %A 100% ( )( ) %B y ()( ) B 100% %C C 100 % z Page 74. Copyrighted by Gabriel Tag B.Ed., B.Sc.
4 Cheistry Uit 4: The ole ad Cheical Copositio Exaple 1: Calculate the ass percetage of sodiu chroate. Na CrO 4 /ol Assue we have (1 ole) of Na CrO 4, there are oles of Na, 1 ole of Cr ad 4 oles of O: ol.99 g/ol % % Na 8.39 % % Na ( )( ) 100% 1ol 5.00 g/ol % % Cr 3.10 % % Cr ( )( ) 100% 4 ol g/ol % % O % % O ( )( ) 100% Epirical Forula: - the siplest ratio betwee the eleets i a cheical forula. olecular Forula: - the actual cheical forula of a copoud. olecular Forula (Epirical Forula) where atural uber Exaple: C 6 H 1 O 6 CH O olecular Forula for Glucose Epirical Forula Note: Kowig the ass percetages of a copoud allow us to fid the epirical forula. To kow the olecular forula, we ust also kow the olar ass. Exaple : A copoud has a epirical forula of CH 3 N has a olar ass of g/ol. What is the olecular forula of the copoud? Actual olar ass g/ol Epirical Forula CH 3 N (9.05 g/ol) 4 Eprical olar ass 9.05 g/ol olecular Forula Epirical Forula 4 (CH 3 N) 4 Exaple 3: Cobalt (II) itrate is a hydrate with a cheical forula of Co(NO 3 ) xh O. Whe the.45 g of hydrate is heated, 1.54 g of residual is left behid. Deterie the uber of hydrate uit for cobalt (II) itrate. ass of Co(NO 3 ) 1.54 g 1.54 g g/ol ol Co(NO 3 ) ol H. ass of H O released.45 g 1.54 g 0.91 g 0.91g 18.0 g/ol HO Co(NO ) ol Co(NO3) 3 olecular Forula C 4 H 1 N 4 O ol H O 6 ol HO 1ol Co(NO ) 3 olecular Forula Co(NO 3 ) 6 H O Copyrighted by Gabriel Tag B.Ed., B.Sc. Page 75.
5 Uit 4: The ole ad Cheical Copositio Cheistry Exaple 4: Vitai C has a olar ass of g/ol ad cotais carbo, hydroge, ad oxyge atos. If the % ass of carbo ad oxyge are 40.91% ad 54.50% respectively, deterie the epirical ad olecular forula of vitai C. % C 40.91% % O 54.50% % H 100% 40.91% 54.50% 4.59% Assue 100 g of Vitai C. The, there are C 100 g 40.91% g O 100 g 54.50% g H 100 g 4.59% 4.59 g C O C O 40.91g 1.01g/ol g g/ol ol C H ol O 4.59 g g/ol ol H ol C 1ol C ol H ol O 1ol O O ol O C : O 1 : 1 Cobie Ratios H : O 4 : 3 H 4 ol H 3 ol O Actual olar ass g/ol Epirical Forula C 3 H 4 O 3 (88.06 g/ol) Eprical olar ass g/ol olecular Forula Epirical Forula olecular Forula C 6 H 8 O 6 OR Aother ethod ay be used where the Actual olar ass becoes the ass of Vitai used. The, the ole of each Ato is calculated to deterie the olecular Forula first % g 4.59% g C 6.00 ol C H 8.00 ol H 1.01g/ol 1.01g/ol 54.50% g O 6.00 ol O olecular Forula (C 6 H 8 O 6 ) will be foud first, the g/ol the Epirical Forula (C 3 H 4 O 3 ) will be stated. Assiget 7.3 pg. 43 #1 to 4 (Practice); pg. 5 #1 to 3 (Practice); pg. 48 #1 to 5 (Practice); pg. 48 #1 to 10 Chapter 7 Review pg #19 to 66 Page 76. Copyrighted by Gabriel Tag B.Ed., B.Sc.
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