[chaneliman]

An impure sample of iron(II), weighing 1.545g, was treated to produce a precipitate of Fe2O3. If the mass of the dried precipitate was .315g, calculate the percentage of iron in the sample. Thanks
The answer is 14.3% but i keep getting 19%

[Borek]

Show your work, answer given (14.3) is OK.

[Alpha-Omega]

He is right you need to show your work....I will help you set it up....you need to do the math:

1.  Total sample = 1.545 g
2.  Parts = 0.315 g Fe2O3

Do not let the wording fool you just pay attention to the numbers Fe2O3...is Fe(III)...it is a trick....pay attention to what is asked...they want TOTAL Iron in the sample...total iron can be Fe(II) and Fe(III).

You start with what you have....

Fe2O3:
2 moles of Fe = 2 (55.85 g/mol) = 111.7 g
3 moles of O = 3 (16.00 g/mol) = 48.0 g
Fe2O3 = 159.7 g formula weight


0.315 g x (1 mol Fe2O3/159.7 g) x (2 mole Fe/1 mol Fe2O3) x (55.85 g Fe/1 mol Fe) = 0.2203 g Fe

Now it is just Parts over Total Parts x 100 to get your percentage...and remember significant figures...to get you 14.3%

Parts = 0.2203 g Fe
Total Parts = 1.545 g Sample

Now do the math

Your Pa