What is (i) the molar mass of #"lead nitrate"#, and (ii), the molar quantity of a #42.65*g# mass of this material?

2 Answers
Oct 1, 2017

We take the quotient, #"Mass of stuff"/"Molar mass of stuff"#.

Explanation:

The molar mass of #"lead nitrate"# is the sum of the individual atomic molar masses....

And so we gots...#(a)# #{207.2+2xx14.01+6xx15.999}*g*mol^-1=??*g*mol^-1#

From where did I get these atomic masses? Did I know them off the top of my head? From where will YOU get these masses if asked the question in an exam?

And so with respect to #"lead nitrate"# we take the quotient....

#(b)# #n_"lead nitrate"=(42.65*g)/(331.20*g*mol^-1)=0.1288*mol#

And how many moles of oxygen atoms are in this sample?

And please note the dimensional consistency of the answer; we wanted an answer with dimensions of moles. Did we get one?

i.e. #(g)/(g*mol^-1)=cancel(g)/(cancelg*mol^-1)=1/(1/(mol))=mol# as required. Capisce?

Oct 1, 2017

(a) & (b) done already, I will do
c) Formula units #=("moles of substance" * N_A)f.u.#

#=color(violet)( 0.129 "moles")*6.02*10^23#
#=color(orange)(7.75*10^22 f.u.#