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 6.0 10 (Avogadro s Nuber, N A ) Exaples: 1 ol of carbo (C) 6.0 10 carbo atos 1.01 g (sae as the au) 1 ol of fluorie (F ) 6.0 10 fluorie olecules 38.00 g (iclude subscripts with au) Exaple 1: Fid the uber of olecules i 3.50 oles of platiu. 1 ol Pt 6.0 10 Pt atos 10 atos 3.50 ol.11 10 4 atos of Pt 1ol Exaple : How ay oles of sucrose are there if there are 3.54 10 5 olecules of C 1 H O 11? 3.54 10 5 olecules 1 ol C 1 H O 11 6.0 10 C 1 H O 11 olecules 1ol 10 olecules Exaple 3: Fid the uber of Br io i 0.65 ol of gbr. 58.8 ol of C 1 H O 11 1 ol gbr 6.0 10 gbr forula uits 0.65 ol 10 uits 1ol Br io 7.8 10 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 39.10 g/ol P 4 4 30.97 g/ol P 4 1.88 g/ol C 6 H 1 O 6 (6 1.01) + (1 1.01) + (6 16.00) C 6 H 1 O 6 180.18 g/ol d. sodiu chroate Na CrO 4 e. iro (III) itrite Fe(NO ) 3 Na CrO 4 (.99) + 5.00 + (4 16.00) Na CrO 4 /ol Fe(NO ) 3 55.85 + (3 14.01) + (6 16.00) Fe(NO ) 3 193.88 g/ol Page 7. Copyrighted by Gabriel Tag B.Ed., B.Sc.
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 7.50 10 olecules of sulphur (S 8 ). S 8 8 3.07 g/ol 56.56 g/ol 56.56 g of S 8 1 ol S 8 6.0 10 S 8 olecules 7.50 10 56.56 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 4 4 30.97 g/ol 1.88 g/ol 1.88 g of P 4 1 ol P 4 6.0 10 P 4 olecules There are four phosphorus atos i oe olecular uit of P 4. 60.0 g 10 olecules 1.88 g 4 P atos 1P olecule 4 1.17 10 4 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) + 30.97 + (4 16.00) 163.94 g/ol 163.94 g of Na 3 PO 4 1 ol Na 3 PO 4 6.0 10 Na 3 PO 4 forula uits There are three Na + ios i oe forula uit of Na 3 PO 4. 9.3 g 10 uits 163.94 g + 3 Na ios 1 Na PO uits 3 4 1.0 10 4 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) + 16.00 46.08 g/ol 0.0 g 0.434 ol 46.08 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.
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 1.67 10 7 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 91.754% ad.401% respectively, ad the rest is 54 Fe, calculate the average atoic ass of iro? % abudace of 54 Fe 100% 91.754%.401% 5.845% Average au of Iro (0.05845)(54) + (0.91754)(56) + (0.0401)(57) 5.845% of ass Nuber 56 91.754% of ass Nuber 56.401% of ass Nuber 57 Average au of Iro 55.91 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 13 7. pg. 9 40 #1 to 4 (Practice); pg. 6 #1 ad (Practice); pg. 40 #1, 3 to 1, 14 to 16 7.3: 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.
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 8.38605 % % Na 8.39 % % Na ( )( ) 100% 1ol 5.00 g/ol 3.107873 % % Cr 3.10 % % Cr ( )( ) 100% 4 ol 16.00 g/ol 39.51105075 % % O 39.51 % % 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 116.1 g/ol. What is the olecular forula of the copoud? Actual olar ass 116.1 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 18.95 g/ol 0.0084176004 ol Co(NO 3 ) 0. 0504994451 ol H. ass of H O released.45 g 1.54 g 0.91 g 0.91g 18.0 g/ol HO Co(NO ) 0 0084176004 ol Co(NO3) 3 olecular Forula C 4 H 1 N 4 O 0.0504994451 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.
Uit 4: The ole ad Cheical Copositio Cheistry Exaple 4: Vitai C has a olar ass of 176.14 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% 40.91 g O 100 g 54.50% 54.50 g H 100 g 4.59% 4.59 g C O C O 40.91g 1.01g/ol 54.50 g 16.00 g/ol 3.4063806 ol C H 3.4065 ol O 4.59 g 1.008 g/ol 4.55357149 ol H 3.4063806 ol C 1ol C 4.55357149 ol H 1.33 3.4065 ol O 1ol O O 3.4065 ol O C : O 1 : 1 Cobie Ratios H : O 4 : 3 H 4 ol H 3 ol O Actual olar ass 176.14 g/ol Epirical Forula C 3 H 4 O 3 (88.06 g/ol) Eprical olar ass 88.06 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. 40.91% 176.14 g 4.59% 176.14 g C 6.00 ol C H 8.00 ol H 1.01g/ol 1.01g/ol 54.50% 176.14 g O 6.00 ol O olecular Forula (C 6 H 8 O 6 ) will be foud first, the 16.00 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. 51 54 #19 to 66 Page 76. Copyrighted by Gabriel Tag B.Ed., B.Sc.