Respuesta :
The empirical formula of the iron oxide is [tex]Fe_2O_3[/tex].
And the empirical formula is in an agreement with [tex]Fe_2O_3[/tex] that is iron(III) oxide.
Given:
A sample of iron oxide with 69.943 % of iron and rest with of the mass from oxygen.
To find:
The empirical formula of iron oxide & to identify the type of iron oxide.
Solution:
Consider in 100 grams iron oxide has 69.943 % of iron.
Percentage of oxygen in 100 grams of iron oxide :
= 100% - 69.943%=30.054%
- Mass of iron in 100 grams sample =[tex]100g\times \frac{69.943 }{100}=69.943 g[/tex]
- Mass of oxygen in 100 grams = [tex]100 g\times \frac{30.054}{100}=30.054 g[/tex]
- Moles of iron = [tex]\frac{69.943 g}{55.845 g/mol}=1.2524mol[/tex]
- Moles of oxygen = [tex]\frac{30.054 g}{15.999 g/mol}=1.8785 mol[/tex]
Let the empirical formula of iron oxide be [tex]Fe_xO_y[/tex]
For empirical formula divide all moles of elements with the least value of moles of an element in the compound:
[tex]x=\frac{1.2524mol}{1.2524mol}=1\\\\y=\frac{1.8785 mol}{1.2524mol}=1.5\\\\Fe_xO_y=Fe_1O_{1.5}\\\\=Fe_{10}O_{15}=Fe_2O_3[/tex]
The empirical formula of the iron oxide is [tex]Fe_2O_3[/tex].
And the empirical formula is in an agreement with [tex]Fe_2O_3[/tex] that is iron(III) oxide.
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