# Why do we use coefficients for RFM (relative formula mass) in atom economy? [closed]

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 CusterMay 31 at 15:49

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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.

• "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? – Ubaid Hassan May 31 at 15:54