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

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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 = = 0.05 mole

moles of water = = 0.05 mole

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

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

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

grams of C = = 0.6 g

grams of H = = 0.1g

grams of O = 1.5 - (0.6 + 0.1)

grams of O = 1.5 - 0.7 = 0.9 g

moles of O = = 0.056 moles

Let's calculate the mole ratio now:

= 1

= 2

= 1

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

definition

                    entropy is a measure of the randomness or the extent of disorder of a chemical process.

                      enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

                        entropy is measured in joule per kelvin (jk⁻¹)   whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

                            entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

                              in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.

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