# 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?

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

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

"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

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

"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 $\text{O atoms}$

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