# Why is the empirical formula not double that of the monosaccharides?

Mar 28, 2018

Just to retire this question....

$\text{the empirical formula is the simplest whole ratio...}$

#### Explanation:

...$\text{the empirical formula is the simplest whole ratio}$ $\text{that defines constituent elements in a species...}$

And so we got a monosaccharide, ${C}_{n} {H}_{2 n} {O}_{n}$...and CLEARLY the empirical formula of this beast is $C {H}_{2} O$ given the definition....

And a disaccharide results from the condensation reaction of two monosaccharides to the give the disaccharide and WATER....

$2 {C}_{n} {H}_{2 n} {O}_{n} \rightarrow {C}_{2 n} {H}_{2 n - 2} {O}_{n - 1} + {H}_{2} O$

And to use the obvious example, we could take glucose, ${C}_{6} {H}_{12} {O}_{6}$, whose dissacharide is sucrose, ${C}_{12} {H}_{22} {O}_{11}$...

${C}_{12} {H}_{22} {O}_{11} \equiv \left\{2 \times {C}_{6} {H}_{12} {O}_{6}\right\} - {H}_{2} O$

...i.e. we conceive that water is LOST in the condensation reaction...and the empirical formula must be altered to be the same as the molecular formula...