Problem: We typically calculate the percent yield using the actual yield and theoretical yield in units of mass (grams or kilograms).Would the percent yield be different if the actual yield and theoretical yield were in units of amount (moles)?

Recall that percent yield is given by the following equation:

$\boxed{\mathbf{\%}\mathbf{yield}\mathbf{=}\frac{\mathbf{actual}\mathbf{ }\mathbf{yield}}{\mathbf{theoreticl}\mathbf{ }\mathbf{yield}}\mathbf{\times }\mathbf{100}}$

If calculated using moles:

$\mathbf{\%}\mathbf{yield}\mathbf{=}\frac{\mathbf{moles}\mathbf{ }\mathbf{\left(}\mathbf{actual}\mathbf{\right)}}{\mathbf{moles}\mathbf{ }\mathbf{\left(}\mathbf{theortical}\mathbf{\right)}}\mathbf{\times }\mathbf{100}$

But mass can be calculated from moles:

molar mass = g/mol

$\mathbf{mass}\mathbf{=}\mathbf{mole}\mathbf{\times }\frac{\mathbf{grams}}{\mathbf{mole}}$

If calculated using mass:

$\mathbf{\%}\mathbf{yield}\mathbf{=}\frac{{(\mathrm{moles}\times {\displaystyle \frac{\mathrm{grams}}{\mathrm{mole}}})}_{\mathbf{actual}}}{{\mathbf{(}\mathbf{moles}\mathbf{\times }\frac{\mathbf{grams}}{\mathbf{mole}}\mathbf{)}}_{\mathbf{theoretical}}}\mathbf{\times }\mathbf{100}$