Question

How many grams are in `1.98 * 10^21` atoms of Fe?

Answer 1

The molar mass of Fe is `55.845 gmol^-1`
So it is the mass of 1 mole atom of Fe i.e. `6.023*10^23` atoms of Fe
Hence the mass of`1.98*10^21` atoms of Fe will be = `55.845*(1.98*10^21)/(6.023*10^23)=0.55845*1.98/6.023g~~0.1836g`

Answer 2

`1.98xx10^21"atoms Fe"` has a mass of `"0.184 g Fe"`.

Explanation:

`"1 mol atoms"=6.022xx10^23"atoms"`

`"Molar mass Fe"="55.845 g/mol"`

You need to make the following conversions:

`"atoms Fe"` `rarr` `"mol Fe"` `rarr` `"mass Fe"`

Divide the given atoms of iron by `6.022xx10^23"atoms Fe"`, then multiply by its molar mass.

`1.98xx10^21cancel"atoms Fe"xx(1cancel"mol Fe")/(6.022xx10^23cancel"atoms Fe")xx(55.845"g Fe")/(1cancel"mol Fe")="0.184 g Fe"` rounded to three significant figures