Question

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

Answer 1

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.....