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Answer: from the analysis results of C and 5, find the weight fraction of C and 5 in thiophene wt of carbon in 16.65 mg co₂: = (wit of co₂)* (at wit Carbon)/(mol wit (0₂) (16.65 mg) * (12)/(12+2 * 16) = (16.65 mg) 12/44 AED 1.541 mg. Now find the wt of sulfur in 4.31 mg of this phene: =(wit of Basa) ** (at. wt. sulfur)/(mol wit (1.96 mg * (32.06)/(137-34-+32-06 +47/ 50+). = = = 11.96 mg * 32.06/233.4. = 1.643 mg Conalysis, this sample had less thiophene than the We need to scale the weight as if. 7.96 mg of thisphere was analyzed: it's just the (7-96 mg/4-31 mg) * 1.643 mg = 3-034 mg. ratio the other component is hydrogen, Fince. be the remainder of the weight: 38 wt hydrogen = wt of thiophene - (wit of (+wit of s) ***** = 7.96 mg (4.541+ 3.034) Now Convert each element, #moles = 1 0.385 my wt into # of males of each (wit in grame)/(atomic we in ✓ #millimoles (= (4.541 mg C)/(12 milligramy/millimele) = 0.37-84 mmol #millimoler S = (3.034 mg S)/ (32 milligrams/millimole) = 0-0948 mmol #millimoles H = (0.385 mg H)/(1 milligrams/millimale) 0.385 mmol = Now divide by the smalley number to find the molas. rations of each element, In this case, S hey the smallest no. of millimoles to divide C and H by 5 and round to the neagest integer to get the ratios: 4 0.3784 mmol/0.0948 mmol mmol 0.0948 mmol/ 0.385 mmol/0.0948 mmol His ratio = 4 So the empirical formula for thisphene is C4 M4 S ( given this information thiophene molecular formula cibd be C:s ratio S.S ratio 0.0948 Caths, CBt8 S2+ etc. so long as (=H=4*S

Answer: from the analysis results of C and 5, find the weight fraction of C and 5 in thiophene wt of carbon in 16.65 mg co₂: = (wit of co₂)* (at wit Carbon)/(mol wit (0₂) (16.65 mg) * (12)/(12+2 * 16) = (16.65 mg) 12/44 AED 1.541 mg. Now find the wt of sulfur in 4.31 mg of this phene: =(wit of Basa) ** (at. wt. sulfur)/(mol wit (1.96 mg * (32.06)/(137-34-+32-06 +47/ 50+). = = = 11.96 mg * 32.06/233.4. = 1.643 mg Conalysis, this sample had less thiophene than the We need to scale the weight as if. 7.96 mg of thisphere was analyzed: it's just the (7-96 mg/4-31 mg) * 1.643 mg = 3-034 mg. ratio the other component is hydrogen, Fince. be the remainder of the weight: 38 wt hydrogen = wt of thiophene - (wit of (+wit of s) ***** = 7.96 mg (4.541+ 3.034) Now Convert each element, #moles = 1 0.385 my wt into # of males of each (wit in grame)/(atomic we in ✓ #millimoles (= (4.541 mg C)/(12 milligramy/millimele) = 0.37-84 mmol #millimoler S = (3.034 mg S)/ (32 milligrams/millimole) = 0-0948 mmol #millimoles H = (0.385 mg H)/(1 milligrams/millimale) 0.385 mmol = Now divide by the smalley number to find the molas. rations of each element, In this case, S hey the smallest no. of millimoles to divide C and H by 5 and round to the neagest integer to get the ratios: 4 0.3784 mmol/0.0948 mmol mmol 0.0948 mmol/ 0.385 mmol/0.0948 mmol His ratio = 4 So the empirical formula for thisphene is C4 M4 S ( given this information thiophene molecular formula cibd be C:s ratio S.S ratio 0.0948 Caths, CBt8 S2+ etc. so long as (=H=4*S

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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From the analysis results of ( and $₁ Answer: find the weight fraction of C and I in thiophene wt of carbon in 16.65 mg CO₂: - = (wt of CO₂) * (at. wit Carbon)/ (mol wt (0₂) (16.65 mg)* (12)/(12+2* 16). 4.541 mg. Now find the wot of sulfur is =(wt of Bas04)* (at. wt. sulfur)/(mol - = = (16.65 mg) 12/44 11.96 mg * (32.06)/(137-34+32-06 +4 06+ 4 * 16-0). 11.96 mg * 32:06/233.4 1.643 mg C Now element, #moles 4.31 mg of this phene: wt BaSO4). = this sample had less thiophene than the We need to scale the weight as if: 7.96 mg of thiophene, was analyzed; it's (7-96 mg/4-31mg) * 1.643 mg it's just the ratio 10110101 893 = 3-034 mg. since, the the other component is hydrogen, it must (136) remainder of the weight: be wt hydrogen = wt of thiophene - (wit. of (+wy of s). Canalysis, 0-385my (4-591 +3.034) = 7.96 mg Convert each wt into # of moles of each (wt in grame) / (atomic we in grams/mole) or equivalently # millimoles = (wt in milligrams) / (atomic we in milligramy/millimale) #millimoles ( = (4-541 mg C) / (12 milligramy/millimole) = 0.37-84 mmol #millimales S = (3.034 mg S) / (32 milligrams/millimole) 0-0948 mmol = #millimoles H = (0.385 mg H)/(1 milligrams/millimole) 0.385 mmol divide by the smallest number to find the molas. Now S hoy the smallest rations of each element, In this case, no. of millimoles so divide C and H by S and round to the neagest integer to get the ratios: 4 1 0.3784 mmol/0.0948 mmol 0-0948 mmol / 0.0948 mmol 0.385 mmol /0.0948 mmol empirical ( given this Could be t Now add So the C:s ratio Sis ratio His ratio = = 4 formula for thiophene is C4 H4 S information thiophene molecular formula as (=H=4*S Cg Hq S2¢ etc. so long C4H451 find mol. wt: 4* 12 + 4*1 +32 = 84 up to therefore, the molecular formula of this phone is CHAS.
Transcribed Image Text:From the analysis results of ( and $₁ Answer: find the weight fraction of C and I in thiophene wt of carbon in 16.65 mg CO₂: - = (wt of CO₂) * (at. wit Carbon)/ (mol wt (0₂) (16.65 mg)* (12)/(12+2* 16). 4.541 mg. Now find the wot of sulfur is =(wt of Bas04)* (at. wt. sulfur)/(mol - = = (16.65 mg) 12/44 11.96 mg * (32.06)/(137-34+32-06 +4 06+ 4 * 16-0). 11.96 mg * 32:06/233.4 1.643 mg C Now element, #moles 4.31 mg of this phene: wt BaSO4). = this sample had less thiophene than the We need to scale the weight as if: 7.96 mg of thiophene, was analyzed; it's (7-96 mg/4-31mg) * 1.643 mg it's just the ratio 10110101 893 = 3-034 mg. since, the the other component is hydrogen, it must (136) remainder of the weight: be wt hydrogen = wt of thiophene - (wit. of (+wy of s). Canalysis, 0-385my (4-591 +3.034) = 7.96 mg Convert each wt into # of moles of each (wt in grame) / (atomic we in grams/mole) or equivalently # millimoles = (wt in milligrams) / (atomic we in milligramy/millimale) #millimoles ( = (4-541 mg C) / (12 milligramy/millimole) = 0.37-84 mmol #millimales S = (3.034 mg S) / (32 milligrams/millimole) 0-0948 mmol = #millimoles H = (0.385 mg H)/(1 milligrams/millimole) 0.385 mmol divide by the smallest number to find the molas. Now S hoy the smallest rations of each element, In this case, no. of millimoles so divide C and H by S and round to the neagest integer to get the ratios: 4 1 0.3784 mmol/0.0948 mmol 0-0948 mmol / 0.0948 mmol 0.385 mmol /0.0948 mmol empirical ( given this Could be t Now add So the C:s ratio Sis ratio His ratio = = 4 formula for thiophene is C4 H4 S information thiophene molecular formula as (=H=4*S Cg Hq S2¢ etc. so long C4H451 find mol. wt: 4* 12 + 4*1 +32 = 84 up to therefore, the molecular formula of this phone is CHAS.
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