The gasoline contains #5.40 × 10^27"atoms of H"#.

This is an interesting exercise in the use of conversion factors.

We could do it in one big chain calculation, but let's break it into smaller parts.

**Step 1.** Calculate the volume of gasoline.

#400 cancel("mi") × (1 cancel("gal"))/(20.0 cancel("mi")) × (3.7854 cancel("L"))/(1 cancel("gal")) × "1000 mL"/(1 cancel("L")) = 7.571 × 10^4" mL"#

**Step 2.** Calculate the mass of the gasoline.

#7.571 × 10^4 cancel("mL") × "0.752 g"/(1 cancel("mL")) = 5.693 × 10^4" g"#

**Step 3.** Calculate the moles of octane.

#5.693 × 10^4 cancel("g C₃H₁₈") × ("1 mol C"_8"H"_18)/(114.23 cancel("g C₈H₁₈")) = 498.4" mol C"_8"H"_18#

**Step 4.** Calculate the atoms of H.

#498.4 cancel("mol C₈H₁₈") × (18 cancel("mol H"))/(1 cancel("mol C₈H₁₈")) × (6.022 × 10^23"atoms H")/(1 cancel("mol H")) = 5.40 × 10^27"atoms H"#