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

Answer 1

Just to retire this question....

#"the empirical formula is the simplest whole ratio..."#

...#"the empirical formula is the simplest whole ratio"# #"that defines constituent elements in a species..."#

And so we got a monosaccharide, #C_nH_(2n)O_n#...and CLEARLY the empirical formula of this beast is #CH_2O# given the definition....

Additionally, a condensation reaction between two monosaccharides yields a disaccharide as well as WATER.

#2C_nH_(2n)O_n rarr C_(2n)H_(2n-2)O_(n-1)+H_2O#

And to use the obvious example, we could take glucose, #C_6H_12O_6#, whose dissacharide is sucrose, #C_12H_22O_11#...

#C_12H_22O_11-={2xxC_6H_12O_6}-H_2O#

For example, we assume that water is LOST during the condensation reaction and that the empirical formula needs to be changed to match the molecular formula.

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Answer 2

Owen Ambrose

Since monosaccharides already have the simplest whole-number ratio, doubling the empirical formula would no longer accurately represent this ratio. Instead, it represents the simplest whole-number ratio of elements in a compound.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.