Question #5d8ea

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Question #5d8ea

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Answer 1 · 2017-06-19T17:08:34

(3) $4$ $4$ ${g H}_{2}$ $"g H"_2$

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 $1.008$ $1.008$ or $1.01$ $1.01$ ). This number represents both the mass in $amu$ $"amu"$ of one atom of the element, as well as the mass in $g$ $"g"$ of one mole of the element.

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

$He$ $"He"$ : $4.00 \frac{g}{mol}$ $4.00"g"/"mol"$
$N$ $"N"$ : $14.01 \frac{g}{mol}$ $14.01"g"/"mol"$ ( $N_{2} = 28.02 \frac{g}{mol}$ $"N"_2 = 28.02"g"/"mol"$ )
$H$ $"H"$ : $1.01 \frac{g}{mol}$ $1.01"g"/"mol"$ ( $H_{2} = 2.02 \frac{g}{mol}$ $"H"_2 = 2.02"g"/"mol"$ )
$O$ $"O"$ : $16.00 \frac{g}{mol}$ $16.00"g"/"mol"$ ( $O_{2} = 32.00 \frac{g}{mol}$ $"O"_2 = 32.00"g"/"mol"$ )

(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.

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Question #5d8ea

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