What is the empirical compound having the following composition: A 36 g sample of iron and oxygen that is 78% iron?

1 Answer
Junaid Mirza
Nov 22, 2017

#"Fe"_2"O"#

Explanation:

  • Mass of iron in given sample#= 78/100 × "36 g" = "28.08 g"#

Moles of iron#= "28.08 g"/"55.8 g/mol" ≈ "0.50 mol"#

  • Mass of oxygen in sample#= "32 g" - "28.08 g" = "3.92 g"#

Moles of oxygen#= "3.92 g"/"16 g/mol" ≈ "0.25 mol"#

Now, ratio of their moles is

#n_"Iron"/n_"Oxygen" = "0.50 mol"/"0.25 mol" = 2.0#

Empirical formula#= "Fe"_2"O"#

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