What is the empirical formula for fructose given its percent composition: 40.00% $C$ , 6.72% $H$ , and 53.29% $O$ ?

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Answer 1 · 2016-05-26T06:22:17

$CH_2O$

The empirical formula is the simplest whole number ratio defining constituent atoms in a species.

In all of these problems it is useful to assume $100$ $g$ of compund, and calculate the ATOMIC composition.

So, in $100*g$ fructose, there are:

$(40.00*g)/(12.01*g*mol^-1)$ $=$ $3.33*mol*C$ ,

$(6.72*g)/(1.00794*g*mol^-1)$ $=$ $6.66*mol*H$ ,

$(53.29*g)/(15.999*g*mol^-1)$ $=$ $3.33*mol*O$ ,

If we divide thru by the smallest molar quantity, we get a formula of $CH_2O$ .

The molecuar mass of fructose is $180.16*g*mol^-1$ . How do we use these data to provide a molecular formula?

What is the empirical formula for fructose given its percent composition: 40.00% CC, 6.72% HH, and 53.29% OO?