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Question: A sample of compound containing \[C,O,\] and silver \[\left( {Ag} \right)\] weighed \[1.372g\] . On ...

A sample of compound containing C,O,C,O, and silver (Ag)\left( {Ag} \right) weighed 1.372g1.372g . On analysis it was found to contain 0.288g0.288g of OO and 0.974g0.974g of AgAg . The molar mass of the compound is 303.8gmol1303.8gmo{l^{ - 1}} . What is the empirical and molecular formula of the compound?

Explanation

Solution

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,C,O, and silver (Ag)\left( {Ag} \right) weighed 1.372g1.372g
The mass of OO is 0.288g0.288g and the molar mass is 1616 . Thus, number of moles of oxygen will be 0.28816=0.018moles\dfrac{{0.288}}{{16}} = 0.018moles
The mass of AgAg is 0.974g0.974g and the molar mass is 107.9107.9 . Thus, number of moles of silver will be 0.974107.9=0.0090268moles\dfrac{{0.974}}{{107.9}} = 0.0090268moles
Given that the sample is 1.372g1.372g from the mass of oxygen atom and mass of silver atom, the mass of carbon atom will be 1.3720.2880.974=0.111.372 - 0.288 - 0.974 = 0.11
The mass of CC is 0.11g0.11g and the molar mass of CC is 1212 . Thus, number of moles of carbon will be
0.1112=0.009166moles\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 OO is 0.0180.0090268=1.992\dfrac{{0.018}}{{0.0090268}} = 1.99 \approx 2
Mole ratio of AgAg is 0.00902680.0090268=1\dfrac{{0.0090268}}{{0.0090268}} = 1
Mole ratio of CC is 0.0091660.0090268=1.0151\dfrac{{0.009166}}{{0.0090268}} = 1.015 \approx 1
Mole ratio of Ag:C:OAg:C:O is 1:1:21:1:2
Thus, the empirical formula is AgCO2AgC{O_2}
The molar mass of the above empirical formula is 147.95gmol1147.95gmo{l^{ - 1}}
Given that the molar mass of the compound is 303.8gmol1303.8gmo{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 303.8147.95=2\dfrac{{303.8}}{{147.95}} = 2
Thus, the molecular formula of the compound is 2×2 \times empirical formula
Molecular formula is Ag2C2O4A{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.