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

- .

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 .

answer: i would say b.

explanation:

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.