# Question #957e1

Feb 29, 2016

A) $6.0 \cdot {10}^{22}$ molecules of $C {O}_{2}$
B) $6.0 \cdot {10}^{22}$ molecules ${O}_{2}$

#### Explanation:

You are given 4.4g of $C {O}_{2}$ gas. To calculate the molecules of $C {O}_{2}$ present, you need to first covert from grams of $C {O}_{2}$ to moles of $C {O}_{2}$.

4.4g $C {O}_{2}$ *$\frac{1 m o l}{44.1 g}$= .10 mol of $C {O}_{2}$

After you covert to moles of $C {O}_{2}$, you then covert to molecules of $C {O}_{2}$ with Avogadro's number; $6.022 \cdot {10}^{23}$,

.1 mol $C {O}_{2}$ $\frac{6.022 \cdot {10}^{23}}{1 m o l}$= $6.0 \cdot {10}^{22}$ molecules of $C {O}_{2}$

to do all your conversion in one step;

4.4g $C {O}_{2}$ $\frac{1 m o l}{44.1 g}$ $\frac{6.022 \cdot {10}^{23}}{1 m o l}$= $6.0 \cdot {10}^{22}$ molecules of $C {O}_{2}$

To determine the atoms of oxygen present;

You are given 4.4g $C {O}_{2}$, but this time we want atoms (or molecules) of Oxygen gas.

4.4g $C {O}_{2}$* $\frac{1 m o l}{44.1 g}$= .10 mol $C {O}_{2}$

For every mol of $C {O}_{2}$, there is 1 mol of oxygen $\left({O}_{2}\right)$

.10 mol $C {O}_{2}$ $\frac{1 m o l {O}_{2}}{1 m o l C {O}_{2}}$=.10 mol ${O}_{2}$

we use Avogadro's number again;

.10 mol ${O}_{2}$ $\frac{6.022 \cdot {10}^{23}}{1 m o l {O}_{2}}$= $6.0 \cdot {10}^{22}$ molecules/atoms ${O}_{2}$

All worked out;

4.4g $C {O}_{2}$ $\frac{1 m o l C {O}_{2}}{44.1 g C {O}_{2}}$ $\frac{1 m o l {O}_{2}}{1 m o l C {O}_{2}}$ $\frac{6.022 \cdot {10}^{23}}{1 m o l {O}_{2}}$= $6.0 \cdot {10}^{22}$ molecules of ${O}_{2}$