Answer:
Option 3
[tex]MgCO_{3}[/tex]
Explanation:
These are the steps for empirical forumla
1) If given percentage, assume everything in grams
2) Convert those grams to moles (using specific molar mass of each unique element)
3) Divide all the moles by the smallest mole you discovered.
4) Answer will be close to a whole number for each element and that's the answer.
First step, 28.6 g Mg, 14.3 g C, 57.1 g O
Second step:
28.6/24.305 = 1.1767 Mg
14.3/12.01= 1.1907 C
57.1/16.00 = 3.5688 O
Third step:
1.1767/1.1767 = 1 Mg
1.1907//1.1767 ~ 1 C
3.5688/1.1767 ~ 3 O
Put everything together you get!
[tex]MgCO_{3}[/tex]
The empirical formula for the compound is:
C. [tex]MgCO_3[/tex].
A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.
Given:
28.6 g Mg, 14.3 g C, 57.1 g O
In order to determine empirical formula we will divide mass of each given element with the molar mass of the respective element.
[tex]\frac{28.6}{24.305} = 1.1767 Mg[/tex]
[tex]\frac{14.3}{12.01}= 1.1907 C[/tex]
[tex]\frac{57.1}{16.00} = 3.5688 O[/tex]
Dividing each mass with the lowest mass:
[tex]\frac{1.1767}{1.1767 } = 1 Mg[/tex]
[tex]\frac{1.1907}{1.1767 } = 1 C[/tex]
[tex]\frac{3.5688}{1.1767} =3 O[/tex]
Thus, the empirical formula for the compound is [tex]MgCO_3[/tex].
Therefore, option C is correct.
Find more information about Empirical formula here:
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