One of the compounds used to increase the octane rating of gasoline is toluene (pictured). suppose 8.30 ml of toluene (d = 0.867 g/ml) is consumed when a sample of gasoline burns in air.

Missing question: How many grams of oxygen are needed for complete combustion of the toluene?
Chemical reaction: C₇H₈ + 9O₂ → 7CO₂ + 4H₂O.
V(C₇H₈) = 8,30 ml.
d(C₇H₈) = 0,867 g/ml.
n(C₇H₈) = V(C₇H₈) · d(C₇H₈).
m(C₇H₈) = 8,3 ml · 0,867 g/ml.
m(C₇H₈) = 7,19 g.
n(C₇H₈) = m(C₇H₈) ÷ M(C₇H₈).
n(C₇H₈) = 7,19 g ÷ 92,14 g/mol = 0,078 mol.
From chemical reaction: n(C₇H₈) : n(O₂) = 1 : 9.
n(O₂) = 9 · 0,078 mol = 0,702 mol.
m(O₂) = 0,702 mol · 32 g/mol = 22,5 g.