Question

What is the formula weight of silver chromate ($A{{g}_{2}}Cr{{O}_{4}}$) in amu?

Answer 1

To solve this question we first need to know what formula weight is. The sum of atomic weights of all the atoms according to the empirical formula of a compound is known as the formula weight of that compound. "amu" is the atomic mass unit of a compound.

Complete answer:
Now, to find out the formula weight of silver chromate ($A{{g}_{2}}Cr{{O}_{4}}$), we first need to determine its empirical formula.
The simplest integer ratio of the atoms present in the compound gives the empirical formula of the compound.
Now, in silver chromate, the atoms of silver, chromium, and oxygen are in the ratio of
Ag : Cr : O = 2 : 1 : 4
Since the atoms are already in the simplest ratio, the empirical formula of silver chromate is the same as the molecular formula of silver chromate i.e., $A{{g}_{2}}Cr{{O}_{4}}$.
Hence the formula weight of silver chromate is equal to the molecular weight of silver chromate.
Now, the molecular mass of one mole of a compound can be calculated by taking the sum of atomic masses of all the elements present in the molecule.
The atomic weight or atomic mass (amu) of an element can be calculated by taking the weighted average of all the isotopes of that element that exist in nature, considering their abundance.
The atomic masses of atoms in silver chromate are
Ag = 107.8682 u
Cr = 51.99610 u
O = 15.99977 u
So, the molecular weight or the formula weight of silver chromate ($A{{g}_{2}}Cr{{O}_{4}}$) is

& {{M}_{A{{g}_{2}}Cr{{O}_{4}}}}=2\times {{M}_{Ag}}+{{M}_{Cr}}+4\times {{M}_{O}} \\\ & {{M}_{A{{g}_{2}}Cr{{O}_{4}}}}=2\times 107.8682+51.99610+4\times 15.99977 \\\ & {{M}_{A{{g}_{2}}Cr{{O}_{4}}}}=331.73158\text{ u} \\\ \end{aligned}$$ **The formula weight of silver chromate ($A{{g}_{2}}Cr{{O}_{4}}$) is 331.73158 amu.** **Note:** It must be noted that in 2019, the SI base unit of molar mass was redefined. According to the new definition, the molar mass constant is $${{M}_{u}}=0.99999999965\times {{10}^{-3}}kg/mol$$ And not $1\times {{10}^{-3}}kg/mol$. But for practical purposes, the molar mass of an element is still considered to be equivalent to the atomic mass of the element since the change is so insignificant.