# Question 29a4c

Sep 25, 2014

$7.31 \times {10}^{- 23}$ $\text{g}$

#### Explanation:

In order to determine the mass of one molecule of ${\text{CO}}_{2}$, the process involves using mole and mass conversion factors to convert from molecules to grams by way of moles.

molecule $\to$ mole $\to$ grams

The conversion factors include:

$6.02 \times {10}^{23}$ $\text{molecules = 1 mole" " }$ (this is Avogadro's constant)

and

$\text{1 mole = the molar mass of the molecule}$

The molar mass of carbon dioxide, ${\text{CO}}_{2}$, can be determined by combining the mass of one carbon atom and 2 oxygen atoms.

• $C = 1 \times \text{12.01 g/mol" = "12.01 g/mol}$
• $O = 2 \times \text{16.0 g/mol" = "32.0 g/mol}$

So

$\text{12.01 g/mol + 32.0 g/mol = 44.01 g/mol}$

The conversion is completed as follows:

1 cancel("molecule CO"_2) xx (1 cancel("mole CO"_2)) / (6.02 xx 10^23 cancel("molecules")) xx "44.01 g"/(1 cancel("mole CO"_2))#

$= 7.31 \times {10}^{- 23}$ ${\text{g CO}}_{2}$

Molecules cancel with molecules and moles cancel with moles
the outcome is the mass in grams.