Respuesta :
Answer:- [tex]CH_2O[/tex] .
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 = [tex]2.2g(\frac{1mole}{44g})[/tex] = 0.05 mole
moles of water = [tex]0.9g(\frac{1mole}{18g})[/tex] = 0.05 mole
[tex]CO_2[/tex] has one C means the mole ratio of [tex]CO_2[/tex] to C is 1:1. So, moles of C would also be 0.05.
[tex]H_2O[/tex] has two H means the mole ratio of [tex]H_2O[/tex] to H is 1:2. So, moles of H would be 2 times the moles of [tex]H_2O[/tex] which is 0.10.
Let's convert moles of C and H to their grams as:
grams of C = [tex]0.05mole(\frac{12g}{1mole})[/tex] = 0.6 g
grams of H = [tex]0.10mole(\frac{1g}{1mole})[/tex] = 0.1g
grams of O = 1.5 - (0.6 + 0.1)
grams of O = 1.5 - 0.7 = 0.9 g
moles of O = [tex]0.9g(\frac{1mole}{16g})[/tex] = 0.056 moles
Let's calculate the mole ratio now:
[tex]C=\frac{0.05}{0.05}[/tex] = 1
[tex]H=\frac{0.1}{0.05}[/tex] = 2
[tex]O=\frac{0.056}{0.05}[/tex] = 1
From the mole ratio, the empirical formula of fructose is [tex]CH_2O[/tex] .
