A sample of calcium nitrate, Ca(NO_3)_2, with a formula weight of 164 g/mol, has 5.00 x 10^25 atoms of oxygen. How many kilograms of Ca(NO_3)_2 are present?

1 Answer
Ernest Z.
Sep 1, 2016

There are 2.27 kg of "Ca"("NO"_3)_2 present.

Explanation:

Step 1. Calculate the moles of "O" atoms.

"Moles of O" = 5.00 × 10^25 color(red)(cancel(color(black)("atoms O"))) × "1 mol O"/(6.022 × 10^23color(red)(cancel(color(black)("atoms O")))) = "83.03 mol O"

Step 2. Calculate the moles of "Ca"("NO"_3)_2.

The formula tells us that 1 mol of "Ca"("NO"_3)_2 contains 6 mol of "O" atoms.

∴ "Moles of Ca"("NO"_3)_2 = 83.03color(red)(cancel(color(black)( "mol O"))) × ("1 mol Ca"("NO"_3)_2)/(6 color(red)(cancel(color(black)("mol O")))) = "13.84 mol Ca"("NO"_3)_2

Step 3. Calculate the mass of "Ca"("NO"_3)_2

"Mass of Ca"("NO"_3)_2 = 13.84 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2))) × ("164 g Ca"("NO"_3)_2)/(1 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2)))) = "2270 g Ca"("NO"_3)_2 = "2.27 kg Ca"("NO"_3)_2

Related questions
See all questions in Determining Formula
Impact of this question
16340 views around the world
You can reuse this answer
Creative Commons License