Question #819a0

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Question #819a0

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Answer 1 · 2017-08-31T19:10:26

The empirical formula is ${Fe}_{2} O_{3}$ $"Fe"_2"O"_3$ .

Explanation:

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

We must convert the masses of $Fe$ $"Fe"$ and $O$ $"O"$ to moles and then find their ratio.

$Mass of O = Mass of oxide - mass of Fe = 1000 g - 700 g = 300 g$ $"Mass of O" = "Mass of oxide - mass of Fe" = "1000 g - 700 g" = "300 g"$

$Moles of Fe = 700 g Fe \times \frac{1 mol Fe}{55.84 g Fe} = 12.5mol Fe$ $"Moles of Fe" = 700 color(red)(cancel(color(black)("g Fe"))) × "1 mol Fe"/(55.84 color(red)(cancel(color(black)("g Fe")))) = "12.5mol Fe"$

$Moles of O = 300 g O \times \frac{1 mol O}{16.00 g O} = 18.8mol O$ $"Moles of O" = 300 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "18.8mol O"$

From this point on, I like to summarize the calculations in a table.

$Element m Mass/g m m Moles m Ratio m \times 2 m m Integers - ------------------------------------------ -$ $bbul("Element"color(white)(m) "Mass/g"color(white)(mm) "Moles"color(white)(m) "Ratio"color(white)(m)×2color(white)(m)color(white)(m)"Integers")$
$m m Fe X X X m 700 X m m m 12.5 X m l 1 m m m l l 2 m m m m m l 2$ $color(white)(mm)"Fe" color(white)(XXXm)700 color(white)(Xmmm)12.5 color(white)(Xml)1color(white)(mmmll)2color(white)(mmmmml)2$
$m m O X X X X l 300 m m m m 18.8 X m l 1.50 m m l 3.00 m m m l l 3$ $color(white)(mm)"O" color(white)(XXXXl)300 color(white)(mmmm)18.8 color(white)(Xml)1.50 color(white)(mml)3.00color(white)(mmmll)3$

The molar ratio is $Fe:O = 2:3$ $"Fe:O = 2:3"$ .

The empirical formula is ${Fe}_{2} O_{3}$ $"Fe"_2"O"_3$ .

Here is a video that illustrates how to determine an empirical formula.

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Question #819a0

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