Usually when calculating using moles = mass / RFM, the coefficient is not multiplied by a compound's RFM to calculate RFM.

But in calculations for atom economy, the (balanced) coefficient of a compound is used in calculating RFM. Why is this the case?


closed as unclear what you're asking by Karsten Theis, M.A.R., Mithoron, Tyberius, Jon Custer May 31 at 15:49

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Before we come to your actual question, we should clarify a few other problems related to quantities and units in your expression:

moles = mass / RFM

It is essential to distinguish between quantities and units. For example, density is defined as “mass per volume” and not “mass per litre”. In your expression “moles” is a unit name; but it should actually be the quantity “amount of substance”:

amount of substance = mass / RFM

Names of quantities (such as “amount of substance” and “mass”) shall not be arranged in the form of an equation. You should use the corresponding quantity symbols:

n = m / RFM

It is not permissible to use abbreviations (such as “RFM”) for unit symbols (“Mr”) or unit names (“relative molecular mass”).

n = m / Mr

There shall be a space on both sides of most operators but not for the solidus “/”.

n = m/Mr

And finally, your equation is wrong, since amount of substance is not given as mass per relative molecular mass. Remember that relative molecular mass is a dimensionless quantity. Actually, amount of substance is given as mass per molar mass:

n = m/M

Now coming to your actual question.

Atom economy is defined as the quotient of relative molecular mass of the desired product by relative molecular mass of all reactants.

Thus, for a simple reaction

$$\ce{A + B -> Y + W}$$

where $\ce{Y}$ is the desired product and $\ce{W}$ is a by-product that becomes waste, the atom economy would be

$$x=\frac{M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{A})+M_\mathrm r(\ce{B})}$$

However, for a slightly different reaction

$$\ce{C + D -> 2 Y + V}$$

you would get two molecules of the desired product $\ce{Y}$ for each molecule of the reactant $\ce{C}$ and for each molecule of the reactant $\ce{D}$. The yield of product per used amount of reactants is twice as high; therefore, the atom economy is twice as high:

$$x=\frac{2M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{C})+M_\mathrm r(\ce{D})}$$

In the same way, any other coefficients in the chemical equation would have to be taken into account.

  • $\begingroup$ "you would get two molecules of the desired product Y for each molecule of the reactant C" i thought the coefficient represented the number of moles? $\endgroup$ – Ubaid Hassan May 31 at 15:54

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