Average weight of oxalate in percentage mk3fec2o433h2onumber...

Average weight of oxalate in percentage- m(K3(Fe(C2O4)3).3H2O)/(Number of attempts/titrations)=(0.118g + 0.102g + 0.0987g)/3 = 0.106g Mass of C2O42-/(Number of attempts/titrations) = (0.01796g + 0.01828g + 0.01768g)/3 = 0.01797g Average weight of oxalate in percentage = (0.01797g/0.106g) x 100 = 16.95% Theoretical percentage of oxalate in potassium ferric oxalate Theoretical percentage of oxalate: potassium ferric oxalate trihydrate = oxalate 491.25 g/mol = 88.018 100 g/mol = 'x' x = 88.018 x 100/491.25 Theoretical % of oxalate = 17.92% % Error = (practical % of oxalate - theoretical % of oxalate) x 100 The concentration of KMnO4- 0.02M Using the chemical formula, calculate the theoretical percentage of oxalate in potassium ferric oxalate. Calculate the percentage error in your analysis. Comment on the results. Specifically, is your analysis in error? Or, is any error you observe associated with the quality of the sample?

Transcript

00:01 Hello student we have an question that is asked here we calculate the percentage of error so here we write the theoretical percentage of oxalate potassium ferric oxalate so here you write the molar mass first theory calculated molar mass of what potassium ferriboxolate it's formula is k3 then f f f e c 204 whole 3 so which is multiplied by 3 mole of water so which is equal to 491 25 9 .1 .25 gram per mole is the what molar mass of potassium ferric oxalate so here you write the molar mass here right what molar mass of what c2o4 minus which is oxalate so which is equal to here right 88 .018 here here what there is a what moles of c2 over 2 minus which is present in one mole of compounds so here right here what the moles of here right what the moles of you write of minus are present here are present in there what a one mole of compound here right what a mole of compound here right what one mole of very right one mole of compound so its mass so here write its mass which is equal to here right its mass so which is equal to here write 80 .018 which is multiplied by three so which is equal to 264 .054 gram so here you write you write so 491 .255 gram of a compound here right of the compound the compound which contains you're right which contains oxalate he wrote which contains oxalate he wrote which is equal to you write what which is equal to 264 .0554 gram so we know one gram for what one gram he's right one gram of the compound of the compound he wrote one gram of the compound which consists here which consists what consists of xallate he related consists of xallate so which is equal to here write it 264 .054 which is divided by 491 .25 so when here it is a hundred and of compound oxalate which consists here you write 100 gram of the what compound 100 gram of the compound which consists you write out which consists of xallate here write what oxa oxalate so you write you write you write oxalate so which is equal to here you write 264 point a 0554 so which is debil divided by 491 .25 so which is multiplied by 100 so the theoretical percentage of oxalate when we calculate it here right word theoretical percentage of oxalate you write right word oxalate which is equal to here we calculate it then you get 50 .08 08 here 50 point what 08 percentage so now we calculate the percentage error so you'll write percentage error which is equal to theoretical here we write its formula theoretical value minus experimental here i don't experimental experimental value then which is divided by what? which is divided by what? theoretical value you are right word theoretical what? value here is right where theoretical in the book some spelling is mistake so you write theoretical you're right theoretical…

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