calculate the empirical formula

[enas1970]

%iron =63.63% & sulpher =36.47
iron atomic mass= 55.8 and sulpher atomic mass =32

[chemistry girl]

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]

شكرا جزيلا للاسف كتاب المعاصر حلها بالغلط

[محمد عبد السلام عواد]

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]

المشاركة الأصلية كتبت بواسطة chemistry girl

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