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Weight Percentage of MTBE (methyl tert-butylether) in Gasoline : Oxygen Content

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If gasoline had a hypothetical chemical formula of C8H17, and we knew the chemical formula of MTBE as CH3OC4H9, what weight (actual weight and percent by weight) of MTBE should be in the gasoline to achieve a final gasoline blend that has 2.7% oxygen by weight? Assume the densities of gasoline and MTBE equivalent at 0.75 g per cm3.

Also, what is the stoichiometric air to fuel ratio (lb to lb) for the above final mixture of gasoline? How is the a/f ratio calculated in this circumstance?

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Solution Summary

The mass of MTBE required to bring oxygen content to a specified level is calculated. Stoichiomtric air/fuel ratio is also calculated.

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Let the mole fraction of MTBE in the blend be equal to x
Molecular weight of MTBE (C5H12O) = 88
Molecular weight of Gasoline (C8H17) = 113
Average molecular weight of blend = 88 x + 113 (1-x) = 113-25 x

Weight of oxygen in one mole of blend = 16 x

% wt of oxygen ...

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