enas1970
27-09-2012, 02:30 PM
%iron =63.63% & sulpher =36.47
iron atomic mass= 55.8 and sulpher atomic mass =32
iron atomic mass= 55.8 and sulpher atomic mass =32
مشاهدة النسخة كاملة : calculate the empirical formula enas1970 27-09-2012, 02:30 PM %iron =63.63% & sulpher =36.47 iron atomic mass= 55.8 and sulpher atomic mass =32 chemistry girl 01-10-2012, 06:41 PM the iron = 63.63 the sulpher = 36.47 (in the case of giving you the percent you can just delete the percentage symbol and complete your problem ) then divide the percent of the elemnt on it's atomich mass therefor = iron = 63.63\558 = 0.114 sulpher = 36.47 \32 = 1.13 the two numbers resulted from the dividing of the percent of element on it's atomich mass is your number of moles then you get the smaller result which is the (0.114) and divide the both results on it and you will get the ratio of moles which is 0.114 \ 0.114 = 1 1.13 \ 0.114 = 9 so the empirical formula is FeS9 you can chech that with your teacher enas1970 02-10-2012, 04:57 PM شكرا جزيلا للاسف كتاب المعاصر حلها بالغلط محمد عبد السلام عواد 04-10-2012, 03:48 PM number of moles (iron) = 63.63\55.8 = 1.14 number of moles (sulpher) = 36.47 \32 = 1.14 the two numbers resulted from the dividing of the percent of element on it's atomich mass is your number of moles 1 : 1 so the empirical formula is FeS محمد عبد السلام عواد 04-10-2012, 03:53 PM the iron = 63.63 the sulpher = 36.47 (in the case of giving you the percent you can just delete the percentage symbol and complete your problem ) then divide the percent of the elemnt on it's atomich mass therefor = iron = 63.63\558 = 0.114 sulpher = 36.47 \32 = 1.13 the two numbers resulted from the dividing of the percent of element on it's atomich mass is your number of moles then you get the smaller result which is the (0.114) and divide the both results on it and you will get the ratio of moles which is 0.114 \ 0.114 = 1 1.13 \ 0.114 = 9 so the empirical formula is FeS9 you can chech that with your teacher atomic mass of iron = 55.8 not 558 |