Calculate the mass percent of methanol, ethanol, and MTBE each that must be present in gasoline to yield a mixture of 2.9 % oxygen by weight. Hint given: Calculate percent oxygen in each compound.
I don't have the density of the gasoline in the question or in the textbook. However, all I can find relevant in the textbook is that the octane number of Methanol is 116, ethanol is 112, and MTBE is 116 and that gasoline produced simply by distillation of crude oil has an octane number of about 50.
An example to help solve for mass percent:
To get the mass percents (or the more-fundamental mass fractions), just multiply the atomic weights of each of the elements by their coefficients in the formula for beryl, add, and divide the result for each elements by the total.
The term for Be is 3xABe = 3x9 = 27 u.
For Al it is 2xAAl = 2x27 = 54 u.
For Si it is 6xASi = 6x28 = 168 u.
For O it is 18xAO = 18x16 = 288 u.
The total wt. is 27 + 54 + 168 + 288 = 537 u (actually 537.54 u).
Be's fraction is 27/537 = 0.0503 = 5.03%. (b) 1.00 kg of beryl would contain 0.0503x1000 g = 50.3 g of Be.
The three organic compounds methanol (CH3OH), ethanol (CH3CH2OH), and MTBE (CH3OC(CH3)3) by increasing mass fraction oxygen, you just ...
Detailed calculation with explanation.