# Answer on Inorganic Chemistry Question for john shane

Question #23613

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.

Expert's answer

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

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

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