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yanalaym [24]
3 years ago
15

Hemoglobin is a large protein molecule that is responsible for carrying oxygen through the blood. Iron ions are a relatively sma

ll 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?
Chemistry
1 answer:
gogolik [260]3 years ago
5 0

Answer:

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

Explanation:

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

1 pm = 10^{-6} μm

Volume of sphere  =\frac{4}{3}\pi r^3

Volume of single iron(II) ion = V

V=\frac{4}{3}\times 3.14\times (7.51\times 10^{-5} \mu m)^3

V=1.7742\times 10^{-12} \mu m^3

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

Number of hemoglobin structure in blood cell = 2.50\times 10^8 molecules

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

4\times 2.50\times 10^8 ions=10^9 ions

Volume of 10^9 ions of iron = V\times 10^9

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

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

\frac{10^9\times V}{V'}\times 100

\frac{10^9\times 1.7742\times 10^{-12} \mu m^3}{95 \mu m^3}\times 100

=0.0019\%

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