Answer to Question #23613 in Inorganic Chemistry for john shane

Question

I want to accurately finished solution with precise octane values. What are the equations required to accomplish this task. For example, 1 solvent is 155 octane the other, 94. I want to finish off with 116 octane in a 205 quantity.

Answer 1

Derive a formula for a start:
octane value = octane value1*(% component1) + octane value2*(% component2)
octane value1 = 94
octane value2 = 155
% component1 = x
% component2 = 1-x
octane value = 116

116 = 94x + 155*(1-x)
116 = 94x + 155 - 155x
61x = 39
x = 64 % - % component1
36 % - % component2
If you need 205, then:
component1 = 205*0.64 = 131
component2 = 205*0.36 = 74