Question

Hemoglobin is a large protein molecule that is responsible for carrying oxygen through the blood. Iron ions are a relatively small component of hemoglobin. There are four Fe2+ions that are part of the much larger hemoglobin structure. In a single red blood cell there are 2.50x108molecules of hemoglobin. If a single Fe2+ion has an atomic radius of 75.1 pm and a redblood cell has a volume of 95 μm3, what percentage of the total red blood cell volume is taken up by Fe2+ions?

Answer 1

The percentage of volume taken by iron(II) in the blood cell is 0.0019%.

Step-by-step explanation:

Radius of iron(II) ions ,r= 75.1 pm =
`7.51* 10^(-5) \mu m`

1 pm = 10^{-6} μm

Volume of sphere =
`(4)/(3)\pi r^3`

Volume of single iron(II) ion = V

`V=(4)/(3)* 3.14* (7.51* 10^(-5) \mu m)^3`

`V=1.7742* 10^(-12) \mu m^3`

Number of iron(II) ions in one hemoglobin structure = 4

Number of hemoglobin structure in blood cell =
`2.50* 10^8` molecules

Then number of iron (II) ions in
`2.50* 10^8` molecules of hemoglobin:

`4* 2.50* 10^8 ions=10^9 ions`

Volume of
`10^9` ions of iron =
`V* 10^9`

Volume of the hemoglobin structure,V' =
`95 \mu m^3`

Percentage volume of iron (II) ions in a single blood cell:

`(10^9* V)/(V')* 100`

`(10^9* 1.7742* 10^(-12) \mu m^3)/(95 \mu m^3)* 100`

=
`0.0019\%`