How many oxygen atoms are present in a #120*g# mass of #"nitric acid"#?

2 Answers
Feb 24, 2017

Answer:

Approx. #3.5xx10^24# #"oxygen atoms......"#

Explanation:

#"Moles of nitric acid"-=(120*g)/(63.01*g*mol^-1)=1.90*mol#.

And thus, given a formula of #HNO_3#, there are #3xx1.90*mol# #"oxygen atoms"# in the given mass of nitric acid.

And since #1*mol# specifies a quantity of #6.022xx10^23*mol^-1#, there are:

#3xx1.90*molxx6.022xx10^23*mol^-1# #"oxygen atoms"#

#=??# #"oxygen atoms............"#

Feb 24, 2017

Answer:

Depends on the concentration.....

Explanation:

If you are thinking of concentrated nitric acid, then it is typically 68% #HNO_3#, 32% water. Fuming nitric acid will have #HNO_3# content of 86%, and only 14% water.

Oxygen atoms are present in both components, so you'd need to work out the number of moles of both.......

If you take standard conc. nitric acid, then 120 g contains 81.6 g of #HNO_3# and 38.4 g of water. This gives you (81.6 / 63.01) = 1.295 moles of #HNO_3# and (38.4 / 18.02) = 2.131 moles of water.

Each mole of #HNO_3# contains 6.02 x #10^23# molecules each containing an oxygen atom. The same applies to water.

So you have (6.02 x #10^23# x 1.295) + (6.02 x #10^23# x 2.131) = 2.062 x #10^24# oxygen atoms in total.