How many nitrate ions, NO3-, and how many oxygen atoms are present in 1.00 microgram of magnesium nitrate, Mg (NO3)2? Also what mass of oxygen is present in 1.00 microgram of magnesium nitrate?

1 Answer
Ernest Z.
Jun 13, 2017

8.12 × 10^15 color(white)(l)"NO"_3^"-"color(white)(l) "ions"; 2.44 × 10^16color(white)(l) "O atoms"; "0.647 µg O"

Explanation:

Number of nitrate ions

"Moles of Mg"("NO"_3)_2 = 1.00 × 10^"-6" color(red)(cancel(color(black)("g Mg"("NO"_3)_2))) × ("1 mol Mg"("NO"_3)_2)/(148.31 color(red)(cancel(color(black)("g Mg"("NO"_3)_2)))) = 6.743 × 10^"-9"color(white)(l) "mol Mg"("NO"_3)_2

"Moles of NO"_3^"-" = 6.743 × 10^"-9" color(red)(cancel(color(black)("mol Mg"("NO"_3)_2))) × ("2 mol NO"_3^"-")/(1 color(red)(cancel(color(black)("mol Mg"("NO"_3)_2)))) = 1.349 × 10^"-8"color(white)(l) "mol NO"_3^"-"

"NO"_3^"-" color(white)(l)"ions" = 1.349 × 10^"-8" color(red)(cancel(color(black)("mol NO"_3^"-"))) × (6.022 × 10^23 color(white)(l)"ions NO"_3^"-")/(1 color(red)(cancel(color(black)("mol NO"_3^"-"))))

= 8.12 × 10^15 color(white)(l)"ions NO"_3^"-"

Number of oxygen atoms

"Moles of O atoms" = 6.743 × 10^"-9" color(red)(cancel(color(black)("mol Mg"("NO"_3)_2))) × "6 mol O atoms"/(1 color(red)(cancel(color(black)("mol Mg"("NO"_3)_2)))) = 4.046 × 10^"-8"color(white)(l) "mol O atoms"

"O atoms" = 4.046 × 10^"-8" color(red)(cancel(color(black)("mol O atoms"))) × (6.022 × 10^23color(white)(l) "O atoms")/(1 color(red)(cancel(color(black)("mol O atoms"))))

= 2.44 × 10^16color(white)(l) "O atoms"

Mass of "O atoms"

"Mass of O" = 4.046 × 10^"-8" color(red)(cancel(color(black)("mol O atoms"))) × "16.00 g O"/(1 color(red)(cancel(color(black)("mol O atoms")))) = 6.47 × 10^"-7"color(white)(l) "g O" = "0.647 µg O"

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