Question #5d8ea

1 Answer
Jun 19, 2017

Answer:

(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.