Question

What is the formula mass of $ethyl{\text{ }}alcohol$, ${C_2}{H_5}OH$ ?

Answer 1

To find the formula mass one must know the atomic weight of each of the atoms that are present in the molecule. Once we know that then we will multiply each of the elements corresponding to its moles in the compound.

Complete answer:
Formula mass also known as the formula weight is the sum of the atomic weights of the atoms in the empirical formula of the compound. Formula weight is given in atomic mass units $(amu)$
One must be familiar with the relative formula mass as well. This calculation is performed using relative atomic weights values of the elements, which is based on the natural isotopic ratio of elements found in earth’s atmosphere and crust.
Mole – The mole is an amount unit similar to familiar units like pair, dozen, gross etc. it provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules and ions) as the number of atoms in a sample of pure $^{12}C$ weighing exactly $12g$.
Now that we are familiar to these concepts, let us solve the question
For that first we must know the atomic weight of all the elements present in the $ethyl{\text{ }}alcohol$
$Carbon = 12.01g/mol$
$Hydrogen = 1.01g/mol$
$Oxygen = 16.00g/mol$
Now we will see the number of moles present in the $ethyl{\text{ }}alcohol$ and multiply it with their corresponding atomic number.
There are $2{\text{ }}mol$ of $Carbon$ and $6{\text{ }}mol$ of $Hydrogen$ and $1{\text{ }}mol$ of $Oxygen$
Formula mass =
=$12.01 \times 2 + 16.00 \times 1 + 1.01 \times 6$
$= 24.02 + 16.00 + 6.06$=$46.08g/mol$
Therefore our answer is $46.08g/mol$.

Note:
The number of entities composing a mole has been experimentally determined to be $6.02214179 \times {10^{23}}$ , a fundamental constant named Avogadro’s number (${N_A}$). This constant is properly reported with an explicit unit of “per mole” a conveniently rounded version being $6.022 \times {10^{23}}/mol$.