Question #9cc47

1 Answer
Jun 26, 2017

Answer:

The mass percentage of #"Fe"^"2+"# in #"Fe"_0.93"O"_1.00# is 64.93 %.

Explanation:

Assume that we have 100 formula units of #"Fe"_0.93"O"_1.00#.

Then we have 93 #"Fe"# atoms and 100 #"O"# atoms.

The #"Fe"# atoms are a mix of #"Fe"^"2+"# and #"Fe"^"3+"# ions, and the #"O"# atoms are present as #"O"^"2-"# ions.

The total charge of the #"O"^"2-"# ions is 200-, so the total charge of the #"Fe"^"2+"# and #"Fe"^"3+"# ions must be 200+.

Let #x = "number of Fe"^"2+"color(white)(l) "ions"#.
Then #"93 – x" = "number of Fe"^"3+"color(white)(l) "ions"#.

The total + charge of these ions is

#2x + 3(93 - x) = 200#

#2x + 279 - 3x = 200#

#x = 279 - 200 = 79#

So, 100 formula units of #"Fe"_0.93"O"_1.00# contain

#"79 Fe"^"2+"#
#"14 Fe"^"3+"#
#"100 O"^"2-"#

#"% Fe"^"2+" = (79 × 55.84 color(red)(cancel(color(black)("u"))))/((79 × 55.84 + 14 × 55.84 + 100 ×16.00) color(red)(cancel(color(black)("u")))) × 100 % = 4411/6793 × 100 % = 64.93 %#