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"#
