# Question b3926

Jul 4, 2017

$\text{0.5 gram-atoms}$

#### Explanation:

The trick here is to realize that you're actually looking for the number of moles of atoms present in, presumably, $\text{7 g}$ of nitrogen gas, ${\text{N}}_{2}$.

For starters, a gram-atom is simply Avogadro's constant of atoms, regardless if those atoms are combined in molecules or formula units or floating around on their own.

This is the important distinction between a mole and a gram-atom: a mole of a given substance contains Avogadro's constant of molecules, formula units, or atoms of said substance, while a gram-atom of a given substance contains Avogadro's constant of atoms, regardless of the nature of the substance (molecular, ionic).

In your case, a mole of nitrogen gas contains $6.022 \cdot {10}^{23}$ molecules of nitrogen gas, ${\text{N}}_{2}$ and $2$ moles of nitrogen atoms, $\text{N}$.

You can thus say that a mole of nitrogen gas is equivalent to $2$ gram-atoms of nitrogen, or that $1$ gram-atom of nitrogen gas is equivalent to $\frac{1}{2}$ moles of nitrogen gas.

Now, a mole of nitrogen gas has a mass of approximately $\text{28 g}$ as given by the molar mass of nitrogen gas, which is equal to ${\text{28 g mol}}^{- 1}$.

This means that you sample contains

7 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28color(red)(cancel(color(black)("g")))) = "0.25 moles N"_2

You can thus say that the sample contains

$0.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles N"_2))) * "2 gram-atoms N"_2/(1color(red)(cancel(color(black)("mole N"_2)))) = color(darkgreen)(ul(color(black)("0.5 gram-atoms N}}_{2}}}}$

The answer is rounded to one significant figure.

Notice that you get the same answer is you assume that you're dealing with $\text{7 g}$ of elemental nitrogen, $\text{N}$.

The molar mass of elemental nitrogen is approximately ${\text{14 g mol}}^{- 1}$, which means that $\text{7 g}$ will contain

7 color(red)(cancel(color(black)("g"))) * "1 mole N"/(14color(red)(cancel(color(black)("g")))) = "0.5 moles N"#

Now, a gram-atom of nitrogen atoms simply means Avogadro's number of nitrogen atoms, so you can say that $1$ mole of nitrogen atoms is equivalent to $1$ gram-atom of nitrogen atoms.

This implies that you once again have

$0.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles N"))) * "1 gram-atom N"_2/(1color(red)(cancel(color(black)("mole N")))) = color(darkgreen)(ul(color(black)("0.5 gram-atoms N}}_{2}}}}$