# How many molecules of sulfur trioxide are in 78.0 grams?

Apr 6, 2016

Almost $6.022 \times {10}^{23}$ sulfur trioxide molecules.
By definition there are 6.022140857(74)×10^23 individual $S {O}_{3}$ molecules in $80.066$ $g$ of this material. This represents the MOLAR MASS of sulfur trioxide.
To find the number of $S {O}_{3}$ molecules, I simply divide the mass by the molar mass, to get a quantity in moles, and mulitply this number by "Avogadro's number, "6.022140857(74)×10^23, which is the number of molecules that have a $80.066 \cdot g$ mass:
$\frac{78.00 \cdot \cancel{g}}{80.0666 \cdot \cancel{g} \cdot \cancel{m o {l}^{-} 1}}$ $\times$ 6.022140857(74)×10^23*cancel(mol^-1) $=$ ??