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

2 Answers
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 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#

Mar 5, 2016

Answer:

#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