Question #76a7e

1 Answer
Jan 29, 2018

Answer:

Here's what I got.

Explanation:

Interestingly enough, the value you cited for the percent composition of iron in hemoglobin is off by an order of magnitude. The actual percent composition of iron in hemoglobin is approximately #0.335%#, so I think that I'll use #0.372%# for this example.

#color(white)(a/a)#
The problem provides you with the percent composition of iron in hemoglobin, so the first thing that you can do here is use this information to find the mass of iron present in #"2 g"# of hemoglobin.

#0.372color(red)(%) quad "Fe" => "0.372 g of Fe for every" quad color(red)("100 g") quad "of hemoglobin"#

You can thus say that your sample contains

#2 color(red)(cancel(color(black)("g hemoglobin"))) * "0.372 g Fe"/(100color(red)(cancel(color(black)("g hemoglobin")))) = "0.00744 g Fe"#

Now, in order to find the number of atoms of iron present in the sample, you must use the fact that #1# mole of iron has a mass of #"55.845 g"#.

The molar mass of iron, i.e. the mass of exactly #1# mole of iron, essentially tells you the mass of #6.022 * 10^(23)# atoms of iron because #1# mole of iron must contain #6.022 * 10^(23)# atoms of iron #-># think Avogadro's constant here.

#overbrace("55.845 g"/"1 mol Fe")^(color(blue)("molar mass of Fe")) stackrel(color(white)(acolor(red)("1 mole Fe" = 6.022 * 10^(23) quad "atoms Fe")aaa))(->) "55.845 g"/(6.022 * 10^(23) quad "atoms Fe")#

This means that your sample contains

#0.00744 color(red)(cancel(color(black)("g"))) * (6.022 * 10^(23) quad "atoms Fe")/(55.845color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(8 * 10^(29) quad "atoms Fe")))#

The answer is rounded to one significant figure, the number of sig figs you have for the mass of hemoglobin.