Question

Determine the molecular mass of the following compounds:

(a)

(b) \({\rm{HC}} \equiv {\rm{CH}}\)

(c)

(d)

Short answer

The total molecular or formula mass of each compound will be:

(a)99 amu

(b)26 amu

(c)173.8 amu

(d)98 amu

Step-by-step solution

Determine the molecular mass of part (a)

The molecular mass of a compound is equivalent to the sum of the atomic mass of each element present in that compound.

In (Cl)₂CO, there are 2 chlorine, 1 carbon, and 1 oxygen atom, and the atomic mass of 1 chlorine, 1 carbon, and 1 oxygen atoms is 35.5 amu, 12 amu, and 16 amu, respectively. Therefore, the molecular mass of (Cl)₂CO will be \({\rm{35}}{\rm{.5 }} \times {\rm{ 2 + 12 }} \times {\rm{ 1 + 16 }} \times {\rm{ 1 = 99 amu}}\)

Determine the molecular mass of part (b)

The molecular mass of a compound is equivalent to the sum of the atomic mass of each element present in that compound.

In \({\rm{HC}} \equiv {\rm{CH}}\), there are 2 hydrogen and 2 carbon atoms, and the atomic mass of 1 hydrogen and 1 carbon atom is 1 amu and 12 amu, respectively. Therefore, the molecular mass of \({\rm{HC}} \equiv {\rm{CH}}\) will be \({\rm{12 }} \times {\rm{ 2 + 2 }} \times {\rm{ 1 = 26 amu}}\)

Determine the molecular mass of part (c)

The molecular mass of a compound is equivalent to the sum of the atomic mass of each element present in that compound.

In CH₂Br₂, there are 1 carbon, 2 hydrogen, and 2 bromine atoms, and the atomic mass of 1 carbon, 1 hydrogen, and 1 bromine atom is 12 amu, 1 amu, and 79.9 amu, respectively. Therefore, the molecular mass of CH₂Br₂ will be \[{\rm{12 }} \times {\rm{ 1 + 1 }} \times {\rm{ 2 + 79}}{\rm{.9 + 2 = 173}}{\rm{.8 amu}}\]

Determine the molecular mass of part (d)

The molecular mass of a compound is equivalent to the sum of the atomic mass of each element present in that compound.

In SH₂O₄, there are 1 sulfur, 2 hydrogen, and 4 oxygen atoms, and the atomic mass of 1 sulfur, hydrogen, and oxygen atom is 32 amu, 1 amu, and 16 amu, respectively. Therefore, the molecular mass of SH₂O₄ will be \({\rm{32 }} \times {\rm{ 1 + 2 }} \times {\rm{ 1 + 16 }} \times {\rm{ 4 = 98 amu}}\)