# Question 5d8ea

Jun 19, 2017

(3) $4$ ${\text{g H}}_{2}$

#### Explanation:

Given four different masses of substances, we're asked which substance contains the most atoms.

Most periodic tables list each element's molar/atomic mass beneath its symbol (for hydrogen, it should say $1.008$ or $1.01$). This number represents both the mass in $\text{amu}$ of one atom of the element, as well as the mass in $\text{g}$ of one mole of the element.

The molar masses of each of these elements (from a periodic table, or online) are

• $\text{He}$: $4.00 \text{g"/"mol}$

• $\text{N}$: $14.01 \text{g"/"mol}$ ($\text{N"_2 = 28.02"g"/"mol}$)

• $\text{H}$: $1.01 \text{g"/"mol}$ ($\text{H"_2 = 2.02"g"/"mol}$)

• $\text{O}$: $16.00 \text{g"/"mol}$ ($\text{O"_2 = 32.00"g"/"mol}$)

(each rounded to two decimal places)

The diatomic values were listed next to the diatomic elements, and are simply twice its molar mass.

Using dimensional analysis , we can convert each of the given masses to moles, using their molar masses:

• 6cancel("g He")((1color(white)(l)"mol He")/(4.00cancel("g He"))) = 1.50 $\text{mol He}$

• 28cancel("g N"_2)((1color(white)(l)"mol N"_2)/(28.02cancel("g N"_2))) = 0.999 ${\text{mol N}}_{2}$

• 4cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 1.98 ${\text{mol H}}_{2}$

• 48cancel("g O"_2)((1color(white)(l)"mol O"_2)/(32.00cancel("g O"_2))) = 1.50 ${\text{mol O}}_{2}$

Now what we can do is use Avogadro's number to calculate the number of particles of each substance:

• 1.50cancel("mol He")((6.022xx10^23color(white)(l)"atoms He")/(1cancel("mol He")))

$= 9.03 \times {10}^{23}$ $\text{atoms He}$

• 0.999cancel("mol N"_2)((6.022xx10^23color(white)(l)"molecules N"_2)/(1cancel("mol N"_2)))

$= 6.02 \times {10}^{23}$ ${\text{molecules N}}_{2}$

• 1.98cancel("mol H"_2)((6.022xx10^23color(white)(l)"molecules H"_2)/(1cancel("mol H"_2)))

$= 1.19 \times {10}^{24}$ ${\text{molecules H}}_{2}$

• 1.50cancel("mol O"_2)((6.022xx10^23color(white)(l)"molecules O"_2)/(1cancel("mol O"_2)))

$= 9.03 \times {10}^{23}$ ${\text{molecules O}}_{2}$

And since each diatomic species contains color(red)(2# atoms per molecule, the final quantities are

• $9.03 \times {10}^{23}$ $\text{atoms He}$

• $6.02 \times {10}^{23}$ ${\text{molecules N}}_{2} \times 2 = 1.20 \times {10}^{24}$ $\text{atoms N}$

• $1.19 \times {10}^{24}$ ${\text{molecules H}}_{2} \times 2 = 2.38 \times {10}^{24}$ $\text{atoms H}$

• $9.03 \times {10}^{23}$ ${\text{molecules O}}_{2} \times 2 = 1.81 \times {10}^{24}$ $\text{atoms O}$

The largest value is that of hydrogen, so option (3) is correct.