3. Materials Balance The whole is equal to the sum of the parts.
At 250oC, the lead in the alloy = Pb in  + Lead in liquid
Let us use 600 g of 80 Pb – 20 Sn solder at 250oC.
There are 480 g of lead (and 120 g Sn).
There are A grams of  (88.5 Pb Sn) and L grams of liquid (63 Pb – 37 Sn).
A + L = 600 g

4. 63%Pb
88.5% Pb

5. Thus, on the basis of the chemical compositions of the 2 phases and of the total alloy, the lead balance is
0.885A L = 0.80 (A+L)
Or, in this case, A = 2L. This means that we have 200 g of liquid and 400 g of  - the same one third and two thirds values we calculated previously.
Genrally,
Cx(X) + Cy(Y) = C (X+Y)

6. Example 5-4.1
Consider a 90 Cu-10 Sn bronze. (a) What phases are present at 300oC What are their chemical compositions? What is the fraction of each? (b) Repeat part (a) for 600oC. (c) At what temperature is there one-third liquid? (d) At what temperature is there two-thirds liquid?
Procedure Refer to Fig , the Cu - Sn diagram. Read the phases and their compositions directly. Interpolate for the fractions, either with an mm scale or by using the abscissa values.

9. Phase Diagrams with Intermediate Phases
Eutectic systems that we have studied so far have only two solid phases (a and b) that exist near the ends of phase diagrams. These phases are called terminal solid solutions.
Some binary alloy systems have intermediate solid solution phases. In phase diagrams, these phases are separated from the composition extremes (0% and 100%).
Example: in Cu-Zn,  and  are terminal solid solutions, , ’, , ,  are intermediate solid solutions.

11. Phase Diagrams with Intermetallic Compounds
Besides solid solutions, intermetallic compounds, that have precise chemical compositions can exist in some systems.
When using the lever rules, intermetallic compounds are treated like any other phase, except they appear not as a wide region but as a vertical line.