What are the masses of the constituent atoms in a #6.3*g# mass of #"nitric acid"#?

1 Answer
anor277
Nov 2, 2017

We (i) calculate the molar quantity of nitric acid....

Explanation:

And this is #(6.3*g)/(63.01*g*mol^-1)=0.10*mol#....and note that this is a very contrived question in that we use nitric acid as a solution rather than as a weighed mass....

Now we know that #1*mol# of stuff specifies #6.022xx10^23# individual items of that stuff....and we use #N_A# to represent this number for convenience....

And so (ii) we calculate the number of individual atoms....

#"no. hydrogen atoms"-=0.10xxN_A# #"hydrogen atoms..."#

#"no. nitrogen atoms"-=0.10xxN_A# #"nitrogen atoms..."#

#"no. oxygen atoms"-=3xx0.10xxN_A# #"oxygen atoms..."#

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