Question

A certain alcohol fuel analysis indicates 64.9% carbon, 13.5% hydrogen and 21.6% oxygen by mass. Determine the equivalent molecular formula and the stoichiometric air fuel ratio by mass for this fuel.

Answer 1

Let us assume that 100g of the sample is present. Then the masses of O, C and H are as follows.

C = 64.9 g

H = 13.5 g

O = 21.6 g

Now calculate the number of moles of each substance.

C = 64.9 g/ 12 g/mol = 5.4 mol

H = 13.5 g/ 1 g/mol = 13.5 mol

O = 21.6 g/ 16 g/mol = 1.35 mol

To get the smallest whole number ratio, divide all by the smallest number in this case 1.35.

C = 5.4 mol/ 1.35 mol = 4

H = 13.5 mol/ 1.35 mol = 10

O = 1.35 mol/ 1.35 mol = 1

The empherical formula/ equivalent molecular formula is C₄H₁₀O.

The reaction for complete combustion of the uel is as follows.

C₄H₁₀O + 6 O₂ -------> 4 CO₂ + 5 H₂O

Calculate the mass of the reactants

C₄H₁₀O = (4X12)+(10X1)+(1X16)= 74 g

O₂ = 16X2 = 32 g

The oxygen fuel mass ratio = 32 g X 6

74 g X 1

The oxygen fuel mass ratio = 2.59