Civil Engineers use mild steel for a wide variety of applications, including bri

Question

Civil Engineers use mild steel for a wide variety of applications, including bridge supports. A 1018 plain carbon steel contains 0.18 wt. % C. Compute the atomic % C in this steel. Assuming a density of 7.85 g/cc for this steel, how many C atoms would there be in cubic centimeter of this steel? Chromium is often added to steels to improve their oxidation resistance. If the oxide skin that offers corrosion protection contains Cr_2 O_3, determine the valence of chromium and specify the type of bonding between chromium and oxygen (support your answer with a calculation). Which are the bonding electrons for both chromium and oxygen (Specify one of these bonding electrons by its configuration and write its quantum number)?

Explanation / Answer

a) the weight percentage = 0.18

Let the weight of carbon steel taken = 100gram

so the mass of carbon = 0.18grams

Moles of carbon = mass / atomic weight = 0.18 / 12 = 0.015

it contains only carbon and iron

so weight of iron = 100-0.18 = 99.82grams

The moles of iron = mass / atomic weight = 99.82/55.85 = 1.787

Total moles = 1.787 + 0.015 = 1.802

Atomic % of carbon = Moles of carbon X 100 / total moles = 0.015 X 100 / 1.802 = 0.832%

b) the density = 7.85g / cubic centimeter

Mass of carbon in 7.85g = 0.18 X 7.85/ 100 = 0.0141 grams

the atomic weight of carbon = 12g / mole

in 12grams of carbon = 6.023 X 10^23 atoms are present

so in 0.0141 grams of carbon = 6.023 X 0.0141 X 10^23 / 12 atoms will be present = 7.077 X 10^20 atoms

c) the oxidation state of chromium or valency = x

2x + 3(-2) = 0

x = +3

the bonding present is ionic as it is made up of a metal and a nonmetal

d) the electronic configuration of chromium = [Ar]4s1 3d5

The electronic configuration of Cr+3 = [Ar] 3d3

so the bonding electrons are 3d electrons

the quantum number of 3d electrons will be

n = 3

l = 2

m = +2

s = +1/2

For oxygen

1s2 2s2 2p4

oxide

1s2 2s2 2p6

the bonding electrons are 2p6

n = 2

l = 1

m = +1 s = +1/2

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