# How can I determine the amount of cementite formed (in g) per 100 g of steel?

c(O) = 0.40 wt% C

c($$\alpha$$) = 0.022 wt% C

c(FeC3) = 6.70 wt% C3

For the Fe-C alloy with 99.6 wt% Fe-0.40 wt% C, determine (at the temperature immediately below the eutectoid) determine the amount of cementite formed (in g) per 100 g of steel

Solution:

$$\frac{Fe_3C}{Fe_3C+\alpha} =\frac{C_0 - C_{\alpha}}{C_{Fe_3C}-C_{\alpha}}.100 = \frac{0.4-0.022}{6.7-0.022}.100 = 5.7g$$

• $$Fe_3C = 5.7g$$
• $$\alpha = 94.3g$$

but i though that since its 100g steel we got that alpha=(0.022 * 100) and Fe3C=(6.7 * 100)g . why is this wrong?

• +1 Nice first question!. Next time can you use MathJaX rather than copying the images directly. It would make the question searchable for other users. – Anoop A Nair Jan 16 at 6:20