How many nitrate ions, #"NO"_3^(-)#, and how many oxygen atoms are present in #1.00# #mu"g"# of magnesium nitrate, #"Mg" ("NO"_3)_2#?

Also, what mass (in grams) of oxygen is present in #1.00# #mu"g"# of magnesium nitrate?

1 Answer
Jun 15, 2017

Answer:

Here's what I got.

Explanation:

Start by converting the mass of magnesium nitrate

#1.00color(red)(cancel(color(black)(mu"g"))) * "1 g"/(10^6color(red)(cancel(color(black)(mu"g")))) = 1.00 * 10^(-6)# #"g"#

to moles by using the molar mass of the compound.

#1.00 * 10^(-6) color(red)(cancel(color(black)("g"))) * ("1 mole Mg"("NO"_3)_2)/(148.3color(red)(cancel(color(black)("g")))) = 6.743 * 10^(-9)# #"moles Mg"("NO"_3)_2#

Next, convert the number of moles of formula units of magnesium nitrate by using Avogadro's constant

#6.743 * 10^(-9) color(red)(cancel(color(black)("moles Mg"("NO"_3)_2))) * (6.022 * 10^(23)color(white)(.)"f. units Mg"("NO"_3)_2)/(1color(red)(cancel(color(black)("mole Mg"("NO"_3)_2))))#

# = 4.061 * 10^(15)# #"f. units Mg"("NO"_3)_2#

Now, you know that one formula unit of magnesium nitrate contains

  • one magnesium cation, #1 xx "Mg"^(2+)#
  • two nitrate anions, #2 xx "NO"_3^(-)#

This means that your sample will contain

#4.061 * 10^(15) color(red)(cancel(color(black)("f. units Mg"("NO"_3)_2))) * ("2 NO"_3^(-)color(white)(.)"anions")/(1color(red)(cancel(color(black)("f. unit Mg"("NO"_3)_2))))#

# = color(darkgreen)(ul(color(black)(8.12 * 10^(15)color(white)(.)"NO"_3^(-)color(white)(.)"anions")))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of magnesium nitrate.

Now, you know that each nitrate anion contains

  • one nitrogen atom, #1 xx "N"#
  • three oxygen atoms, #3 xx "O"#

This means that your sample contains

#8.12 * 10^(15)color(red)(cancel(color(black)("NO"_3^(-)"anions"))) * "3 atoms of O"/(1color(red)(cancel(color(black)("NO"_3^(-)"anion")))) #

# = color(darkgreen)(ul(color(black)(2.44 * 10^(16)color(white)(.)"atoms of O")))#

The answer is rounded to three sig figs.

Finally, to find the mass of oxygen present in the sample, use the fact that each mole of magnesium nitrate contains

#2 xx 3 = "6 moles of O"#

This means that you sample contains

#6.743 * 10^(-9) color(red)(cancel(color(black)("moles Mg"("NO"_3)_2))) * "6 moles O"/(1color(red)(cancel(color(black)("mole Mg"("NO"_3)_2))))#

# = 4.046 * 10^(-8)# #"moles O"#

Use the molar mass of oxygen to convert this to grams

#4.046 * 10^(-8) color(red)(cancel(color(black)("moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O"))))#

#= color(darkgreen)(ul(color(black)(6.47 * 10^(-7)color(white)(.)"g")))#

Once again, the answer is rounded to three sig figs.