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Phases in 0.35% carbon steel

Short answer

By lever rule: cementite = 0.35/6.67 ≈ 5.25%; total ferrite = 100 − 5.25 = 94.75%; pearlite = 0.35/0.80 = 43.75%; free proeutectoid ferrite = 56.25%. Therefore: cementite 5.25%, ferrite 94.75% (56.25% free), pearlite 43.75%.

If a steel contains 0.35% of carbon in the form of iron carbide, what are the percentages of ferrite, cementite and pearlite present in the steel?

The carbon is present as iron carbide (Fe3CFe_3C, cementite). Pearlite is the eutectoid mixture of ferrite and cementite containing 0.80% C, and the steel is hypo-eutectoid (C<0.80%C < 0.80\%), so it has free ferrite plus pearlite.

Data: C=0.35%C = 0.35\%; cementite has 6.67% C; pearlite has 0.80% C.

Cementite

Cementite=0.356.67×100=5.25%\text{Cementite} = \frac{0.35}{6.67} \times 100 = 5.25\%

Ferrite (total, free + in pearlite)

Ferrite=100−5.25=94.75%\text{Ferrite} = 100 - 5.25 = 94.75\%

Pearlite

Pearlite=0.350.80×100=43.75%\text{Pearlite} = \frac{0.35}{0.80} \times 100 = 43.75\%

The rest, 100−43.75=56.25%100 - 43.75 = 56.25\%, is free (proeutectoid) ferrite. The ferrite and cementite inside the pearlite are in the ratio 88.01 : 11.99.

Answer: Cementite = 5.25%; total ferrite = 94.75% (of which free ferrite = 56.25%); pearlite = 43.75%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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