Chemistry Moles Calculation

vendredi 1 novembre 2013

1. The problem statement, all variables and given/known data

"1.79 g of a fuel was burned in the presence of oxygen. The fuel is a composite of 90% octane and 10% ethanol. Calculate the number of moles of this fuel that was burned."

2. Relevant equations

Moles = Mass/Molar Mass

3. The attempt at a solution

90% of 1.79 g is 1.61 g. This was the mass of octane burned.

10% of 1.79 g is 0.18 g. This was the mass of ethanol burned.

Moles of octane = Mass of octane burned/Molar mass of octane

Moles of octane = 1.61/(8*C + 18*H)

Moles of octane = 1.61/(8*12 + 18*1)

Moles of octane = 1.61/(96 + 18)

Moles of octane = 1.61/114

Moles of octane = 0.014 mol

Moles of ethanol = Mass of ethanol burned/Molar mass of ethanol

Moles of ethanol = 0.18/(2*C + 6*H + O)

Moles of ethanol = 0.18/(2*12 + 6*1 + 16)

Moles of ethanol = 0.18/(24 + 6 + 16)

Moles of ethanol = 0.18/46

Moles of ethanol = 0.0039 mol

Total moles burned = Moles of octane burned + Moles of ethanol burned

Total moles burned = 0.014 + 0.0039

Total moles burned = 0.018 mol

Therefore 0.018 moles of the fuel was burned.

Have I done this correctly?