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.