# How many electrons in ONE MOLE of carbon dioxide?

Apr 8, 2017

first calculate moles of $C O 2$

#### Explanation:

first calculate the moles of $C O 2$ = 100/44
= 2.27 mol
now number of electrons in $C O 2$ are obtained by adding total electrons in each of the three atoms i.e. two O and one C = 6+8+8
= 22
thus one mole of $C O 2$ has $22 \cdot 6.022 \cdot {10}^{23}$ electrons and 2.27 mol has $2.27 \cdot 22 \cdot 6.022 \cdot {10}^{23}$ electrons = $301.1 \cdot {10}^{23}$ electrons

Apr 8, 2017

$3 \cdot {10}^{25}$

#### Explanation:

The first thing to do here is to calculate the number of moles of carbon dioxide present in your sample. To do that, use the compound's molar mass

100 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.0color(red)(cancel(color(black)("g")))) = "2.27 moles CO"_2

Next, use Avogadro's constant to figure out the number of molecules of carbon dioxide present in the sample.

2.27 color(red)(cancel(color(black)("moles CO"_2))) * (6.022 * 10^(23)color(white)(.)"molecules CO"_2)/(1color(red)(cancel(color(black)("mole CO"_2))))

$= 1.37 \cdot {10}^{24}$ ${\text{molecules CO}}_{2}$

Now, every molecule of carbon dioxide contains

• one atom of carbon, $1 \times \text{C}$
• two atoms of oxygen, $2 \times \text{O}$

This means that your sample contains

1.37 * 10^(24)color(red)(cancel(color(black)("molecules CO"_2))) * "1 atom C"/(1color(red)(cancel(color(black)("molecule CO"_2))))

$= 1.38 \cdot {10}^{24}$ $\text{atoms of C}$

and

1.37 * 10^(24) color(red)(cancel(color(black)("molecules CO"_2))) * "2 atoms O"/(1 color(red)(cancel(color(black)("molecule CO"_2))))

$= 2.74 \cdot {10}^{24}$ $\text{atoms of O}$

Next, grab a periodic Table and look for the atomic numbers of the two elements. You will find

$\text{For C: } Z = 6$

$\text{For O: } Z = 8$

As you know, a neutral atom has equal numbers of protons located inside its nucleus and electrons surrounding the nucleus.

Therefore, you can say that every atom of carbon will contain $6$ electrons and every atom of oxygen will contain $8$ electrons.

This means that you will have

"total no. of e"^(-) = overbrace(6 * 1.37 * 10^(24))^(color(blue)("coming from C atoms")) + overbrace(8 * 2.74 * 10^(24))^(color(purple)("coming from O atoms"))

${\text{total no. of e}}^{-} = \left(8.22 + 21.92\right) \cdot {10}^{24}$

which gets you

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{total no. of e}}^{-} = 3 \cdot {10}^{25}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of carbon dioxide.

Apr 9, 2017

Approx. $3 \times {10}^{25}$ $\text{electrons........}$

#### Explanation:

First, we calculate the number of electrons in ONE MOLECULE of $C {O}_{2}$. There is ONE CARBON ATOM, that is 6 electrons; and TWO OXYGEN ATOMS, that is 16 electrons, i.e. 22 electrons per molecule.

And then we calculate the number of carbon dioxide molecules in a mass of $100 \cdot g$ of gas. How do we do this? We use the mole as a counting unit, i.e. $6.022 \times {10}^{23}$ molecules of $C {O}_{2}$ have a mass of $\left(12.01 + 2 \times 15.999\right) \cdot g \cdot m o {l}^{-} 1 = 44.0 \cdot g \cdot m o {l}^{-} 1$

$\text{Moles of carbon dioxide}$ $=$ $\frac{100 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} = 2.27 \cdot m o l$.

And (finally) we solve the product:

$22 \times \frac{100 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot \text{electrons} \cdot m o {l}^{-} 1 =$

$\text{how many electrons...........?}$