Question

Find the molecular mass of:
$(a){H_2}S$
$(b)HCl$
$(c)N{H_3}$
$(d)C{l_2}$
$(e)C{H_3}COOH$
$(f)C{H_3}CHO$

Answer 1

Hint: Add the individual atomic masses of the atoms of various elements present in the given compounds. Refer to the periodic table if needed, for knowing the molecular masses of all the elements that are present in the compounds given to us. Proceed using this hint.

Complete answer:
The molecular mass or molecular weight of a compound is defined as the total mass of a compound. It is equal to the sum of the individual atomic masses of each atom in the molecule.
We can find the molecular mass of a compound using the following few steps -
First we need to determine the molecular formula of the molecule.
Then we need to use the periodic table to determine the atomic mass of each element that is present in the molecule.
Then we multiply each element's atomic mass by the number of atoms of that element present in the molecule. This number is often represented by the subscript next to the symbol of that element in the molecular formula.
At last we need to add these values together for each different atom in the molecule.
 The total weight we get will be the molecular mass of the compound.
 From the periodic table one can note that -
 The molecular mass H is 1 g.
The molecular mass of S is 32 g.
The molecular mass of Cl is 35.5 g.
The molecular mass of N is 14 g.
The molecular mass of C is 12 g.

(a)Therefore Molecular mass of ${H_2}S$ can be calculated as
$ \Rightarrow 2 \times 1 + 32 = 34g$ .

(b)Therefore Molecular mass of HCl can be calculated as
$ \Rightarrow 1 + 35.5 = 36.5g.$

(c)Therefore Molecular mass of $N{H_3}$​ can be calculated as
$ \Rightarrow 14 + 3 \times 1 = 17g.$

(d)Therefore Molecular mass of $C{l_2}$ ​ can be calculated as
$ \Rightarrow 2 \times 35.5 = 71g.$

(e)Therefore Molecular mass of $C{H_3}COOH$can be calculated as
$ \Rightarrow 2 \times 12 + 4 \times 1 + 2 \times 16 = 60g.$

(f)Therefore Molecular mass of $C{H_3}CHO$ can be calculated as
$ \Rightarrow 2 \times 12 + 16 + 4 \times 1 = 34g$ .

Note: Always remember that if there is no subscript given after an element symbol, it means there is only one atom present of that particular element in the compound.
Also, a subscript will always apply to the atom symbol it follows. We need to multiply the subscript by the atom's atomic weight.