Question

A sample of compound containing \[C,O,\] and silver \[\left( {Ag} \right)\] weighed \[1.372g\] . On analysis it was found to contain \[0.288g\] of \[O\] and \[0.974g\] of \[Ag\] . The molar mass of the compound is \[303.8gmo{l^{ - 1}}\] . What is the empirical and molecular formula of the compound?

Answer 1

Hint: The mass of the atoms was given. From the molar mass and mass and atoms the number of moles can be calculated. Divide the number of moles of each atom with the atom containing the smallest number of moles gives the numbers. The ratio gives the number of atoms in the empirical formula. From the molar mass of the empirical formula and molar mass of the molecular formula the molecular formula can be calculated.

Complete answer:
Given that a sample of compound containing \[C,O,\] and silver \[\left( {Ag} \right)\] weighed \[1.372g\]
The mass of \[O\] is \[0.288g\] and the molar mass is \[16\] . Thus, number of moles of oxygen will be \[\dfrac{{0.288}}{{16}} = 0.018moles\]
The mass of \[Ag\] is \[0.974g\] and the molar mass is \[107.9\] . Thus, number of moles of silver will be \[\dfrac{{0.974}}{{107.9}} = 0.0090268moles\]
Given that the sample is \[1.372g\] from the mass of oxygen atom and mass of silver atom, the mass of carbon atom will be \[1.372 - 0.288 - 0.974 = 0.11\]
The mass of \[C\] is \[0.11g\] and the molar mass of \[C\] is \[12\] . Thus, number of moles of carbon will be
\[\dfrac{{0.11}}{{12}} = 0.009166moles\]
Now, divide the number of each moles of atom with the number of moles of atom with small number of moles to obtain mole ratio of atoms
Mole ratio of \[O\] is \[\dfrac{{0.018}}{{0.0090268}} = 1.99 \approx 2\]
Mole ratio of \[Ag\] is \[\dfrac{{0.0090268}}{{0.0090268}} = 1\]
Mole ratio of \[C\] is \[\dfrac{{0.009166}}{{0.0090268}} = 1.015 \approx 1\]
Mole ratio of \[Ag:C:O\] is \[1:1:2\]
Thus, the empirical formula is \[AgC{O_2}\]
The molar mass of the above empirical formula is \[147.95gmo{l^{ - 1}}\]
Given that the molar mass of the compound is \[303.8gmo{l^{ - 1}}\]
By dividing the molar mass of the compound and molar mass of the above empirical formula will get the units of the molecular formula as \[\dfrac{{303.8}}{{147.95}} = 2\]
Thus, the molecular formula of the compound is \[2 \times \] empirical formula
Molecular formula is \[A{g_2}{C_2}{O_4}\]

Note:
The molar masses of the atoms should be taken correctly as all the atoms will not have molar mass to the double of the atomic number. The moles obtained must divide with only the smallest moles to obtain a mole ratio. The mole ratio of atoms is the empirical formula but not the molecular formula.