Chapter 12, Problem 3E

Calculate (to the nearest 0.1 u) the formula mass of these compounds.

1. (a) carbon dioxide, CO₂
2. (b) methane, CH₄
3. (c) sodium phosphate, Na₃PO₄

To determine

The formula mass of carbon dioxide $\left({\text{CO}}_{\text{2}}\right)$.

Answer to Problem 3E

The formula mass of carbon dioxide $\left({\text{CO}}_{\text{2}}\right)$ is $44.0\text{ u}$.

Explanation of Solution

Formula mass of any compound is given as the sum of the atomic masses of the constituent atoms given in the chemical formula of a substance. Formula mass and atomic mass are abbreviated as FM and AM respectively.

The formula mass of carbon dioxide $\left({\text{CO}}_{\text{2}}\right)$ is given as,

$\text{FM of }\left({\text{CO}}_{\text{2}}\right)=\left(\text{AM of carbon atom}\right)+\left(2\times \text{AM of oxygen atom}\right)$ (1)

Atomic mass (AM) of carbon atom is $12.0\text{ u}$.

Atomic mass (AM) of oxygen atom is $16.0\text{ u}$.

Substitute $12.0\text{ u}$ for $\text{AM of carbon atom}$ and $16.0\text{ u}$ for $\text{AM of oxygen atom}$ in

equation (1).

$\begin{array}{c}{\text{FM of CO}}_2=\left(12.0\text{ u}\right)+\left(2\times 16.0\text{ u}\right)\\ =44.0\text{ u}\end{array}$

Conclusion:

Therefore, the formula mass of carbon dioxide $\left({\text{CO}}_{\text{2}}\right)$ is $44.0\text{ u}$.

To determine

The formula mass of methane $\left({\text{CH}}_4\right)$.

Answer to Problem 3E

The formula mass of methane $\left({\text{CH}}_4\right)$ is $16.0\text{ u}$.

Explanation of Solution

Formula mass of any compound is given as the sum of the atomic masses of the constituent atoms given in the chemical formula of a substance. Formula mass and atomic mass are abbreviated as FM and AM respectively.

The formula mass of methane $\left({\text{CH}}_4\right)$ is given as,

$\text{FM of }\left({\text{CH}}_4\right)=\left(\text{AM of carbon atom}\right)+\left(4\times \text{AM of hydrogen atom}\right)$ (2)

Atomic mass (AM) of carbon atom is $12.0\text{ u}$.

Atomic mass (AM) of hydrogen atom is $1.0\text{ u}$.

Substitute $12.0\text{ u}$ for $\text{AM of carbon atom}$ and $1.0\text{ u}$ for $\text{AM of hydrogen atom}$ in

equation (2).

$\begin{array}{c}{\text{FM of CH}}_{\text{4}}=\left(12.0\text{ u}\right)+\left(4\times 1.0\text{ u}\right)\\ =16.0\text{ u}\end{array}$

Conclusion:

Therefore, the formula mass of methane $\left({\text{CH}}_4\right)$ is $16.0\text{ u}$.

To determine

The formula mass of sodium phosphate $\left({\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$.

Answer to Problem 3E

The formula mass of sodium phosphate $\left({\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$ is $164.0\text{ u}$.

Explanation of Solution

Formula mass of any compound is given as the sum of the atomic masses of the constituent atoms given in the chemical formula of a substance. Formula mass and atomic mass are abbreviated as FM and AM respectively.

The formula mass of sodium phosphate $\left({\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$ is given as,

$\text{FM of }\left({\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\right)\text{ =}\left[\begin{array}{c}\left(\text{3}\times \text{AM of sodium atom}\right)+\left(\text{AM of phosphorous atom}\right)+\\ \left(4\times \text{AM of oxygen atom}\right)\end{array}\right]$ (3)

Atomic mass (AM) of sodium atom is $23.0\text{ u}$.

Atomic mass (AM) of phosphorus atom is $31.0\text{ u}$.

Atomic mass (AM) of oxygen atom is $16.0\text{ u}$.

Substitute $23.0\text{ u}$ for $\text{AM of sodium atom}$ $31.0\text{ u}$ for $\text{AM of phosphorous atom}$ and $16.0\text{ u}$ for $\text{AM of oxygen atom}$ in equation (3).

$\begin{array}{c}{\text{FM of Na}}_{\text{3}}{\text{PO}}_{\text{4}}\text{ = }\left(3\times 23.0\text{ u}\right)+\left(31.0\text{u}\right)+\left(4\times 16.0\text{ u}\right)\\ =164.0\text{ u}\end{array}$

Conclusion:

Therefore, the formula mass of sodium phosphate $\left({\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$ is $164.0\text{ u}$.