Question

Complete combustion of 1.5 g of fructose a sugar that contains carbon, hydrogen, and oxygen yields 2.2 g of carbon dioxide and .9 g of water. determine the empirical formula of fructose

Answer 1

Answer:-
`CH_2O` .

Solution:- From given grams of carbon dioxide and water we could calculate their moles. using mole ratio, moles of C and H are calculated. These moles of C and H are converted to grams and then we subtract the sum of grams of C and H from given mass of compound to calculate the grams of oxygen. Grams of oxygen are converted to moles and then we find out the mole ratio of C, H and O that gives us empirical formula.

The calculations are as follows:

moles of carbon dioxide =
`2.2g((1mole)/(44g))` = 0.05 mole

moles of water =
`0.9g((1mole)/(18g))` = 0.05 mole

`CO_2` has one C means the mole ratio of
`CO_2` to C is 1:1. So, moles of C would also be 0.05.

`H_2O` has two H means the mole ratio of
`H_2O` to H is 1:2. So, moles of H would be 2 times the moles of
`H_2O` which is 0.10.

Let's convert moles of C and H to their grams as:

grams of C =
`0.05mole((12g)/(1mole))` = 0.6 g

grams of H =
`0.10mole((1g)/(1mole))` = 0.1g

grams of O = 1.5 - (0.6 + 0.1)

grams of O = 1.5 - 0.7 = 0.9 g

moles of O =
`0.9g((1mole)/(16g))` = 0.056 moles

Let's calculate the mole ratio now:

`C=(0.05)/(0.05)` = 1

`H=(0.1)/(0.05)` = 2

`O=(0.056)/(0.05)` = 1

From the mole ratio, the empirical formula of fructose is
`CH_2O` .