Question

An unknown sugar is found to have a molar mass of \[180.18gmo{l^{ - 1}}\] . The sugar contains \[40\] grams of C, \[6.7\] grams of H, and \[53.3\] grams of O. What is the molecular formula?

Answer 1

Hint: From the given mass of atoms, and molar mass the mole ratio of atoms can be determined. By dividing the moles of each atom with the mole of the atom containing the least number of moles, it gives the simple mole ratio which is nothing but an empirical formula. The ratio of empirical mass and molecular mass gives the molecular formula.

Complete answer:
Given that an unknown sugar is found to have a molar mass of \[180.18gmo{l^{ - 1}}\] .
This unknown sugar contains \[40\] grams of C, \[6.7\] grams of H, and \[53.3\] grams of O.
Molar mass of C is \[12gmo{l^{ - 1}}\] , moles will be \[\dfrac{{40}}{{12}} = 3.3mol\]
Molar mass of H is \[1.008gmo{l^{ - 1}}\] , moles will be \[\dfrac{{6.7}}{{1.008}} = 6.6mol\]
Molar mass of O is \[16gmo{l^{ - 1}}\] , moles will be \[\dfrac{{53.3}}{{16}} = 3.3mol\]
Divide the moles of each atom with moles of \[3.3\] , as the moles of carbon and oxygen are small to get the mole ratio.
\[C:H:O = \dfrac{{3.3}}{{3.3}}:\dfrac{{6.6}}{{3.3}}:\dfrac{{3.3}}{{3.3}} = 1:2:1\]
Thus, the empirical formula is \[C{H_2}O\] , the empirical mass would be \[1\left( {12} \right) + 2\left( 1 \right) + 1\left( {16} \right) = 30gmo{l^{ - 1}}\]
But given that the molecular mass would be \[180gmo{l^{ - 1}}\]
Thus, the ratio of empirical mass and molecular mass is \[\dfrac{{180}}{{30}} = 6\]
Multiply the empirical formula by \[6\] , to get the molecular formula which will be \[{C_6}{H_{12}}{O_6}\]
Thus, the unknown sugar has the molecular formula of \[{C_6}{H_{12}}{O_6}\]

Note:
The mass and molar mass should be taken exactly while calculating the moles. The atom with least moles must divide with the moles of all atoms is an important step to determine the mole ratio. By calculating the molar mass of all atoms in molecular formula, it should be equal to \[180gmo{l^{ - 1}}\]