Question

Question #5d8ea

Answer 1

(3) `4` `"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 `"amu"` of one atom of the element, as well as the mass in `"g"` of one mole of the element.

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

• `"He"`: `4.00"g"/"mol"`

• `"N"`: `14.01"g"/"mol"` (`"N"_2 = 28.02"g"/"mol"`)

• `"H"`: `1.01"g"/"mol"` (`"H"_2 = 2.02"g"/"mol"`)

• `"O"`: `16.00"g"/"mol"` (`"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` `"mol He"`

• `28cancel("g N"_2)((1color(white)(l)"mol N"_2)/(28.02cancel("g N"_2))) = 0.999` `"mol N"_2`

• `4cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 1.98` `"mol H"_2`

• `48cancel("g O"_2)((1color(white)(l)"mol O"_2)/(32.00cancel("g O"_2))) = 1.50` `"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 xx 10^23` `"atoms He"`

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

`= 6.02xx10^23` `"molecules N"_2`

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

`= 1.19xx10^24` `"molecules H"_2`

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

`=9.03xx10^23` `"molecules O"_2`

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

• `9.03xx10^23` `"atoms He"`

• `6.02xx10^23` `"molecules N"_2 xx 2 = 1.20xx10^24` `"atoms N"`

• `1.19xx10^24` `"molecules H"_2 xx 2 = 2.38xx10^24` `"atoms H"`

• `9.03xx10^23` `"molecules O"_2 xx 2 = 1.81xx10^24` `"atoms O"`

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