How many carbon atoms in a #5.85*g# mass of #CBr_4#?

1 Answer
Aug 26, 2017

Answer:

Approx. #1xx10^22*"carbon atoms"#.

Explanation:

We got #5.85*g# of #CBr_4#.

This represents a molar quantity of .................

#(5.85*g)/(331.63*g*mol^-1)=0.0176*mol#

And of course in such a molar quantity, there are #4xx0.0176*mol# of bromine, and #1xx0.0176*mol# of carbon. Are you with me?

And so we take the product of the molar quantity and Avogadro's number, to get.......

#0.0176*molxx6.022xx10^23*mol^-1=1.06xx10^22# individual carbon atoms.....