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

2 Answers
P dilip_k
Mar 5, 2016

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 of1.98*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

Meave60
Mar 5, 2016

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

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