Question

a 25.0g sample of iron metal is exposed to the air, reacting with oxygen. The product weighs 35.6g. What is the empirical formula of this compound? Name it.

What is the percent error in the weight of oxygen combined?

Answer 1

1)

we have mass of each elements as:

Fe: 25 g

O: 10.6 g

Divide by molar mass to get number of moles of each:

Fe: 25/55.85 = 0.447628

O: 10.6/16.0 = 0.6625

Divide by smallest:

Fe: 0.447628/0.447628 = 1

O: 0.6625/0.447628 = 1.50

multiply both by 2 to get simplest whole number ratio:

Fe. :1*2 = 2

O : 1.50*2 = 3

So empirical formula is:

Fe2O3

Its name is:

Iron(III) oxide

2)

from above mole of iron = 0.447628

In compound Fe2O3,

mole of O should be = (3/2)*moles of Fe

= (3/2)*0.447628

= 0.6714

But we have 0.6625 moles of O as calculate above

% error = (correct value - actual value) * 100 / correct value

= (0.6714 - 0.6625)*100/0.6714

= 1.33 %

Answer: 1.33 %