Answer:
C2H3O3
Explanation:
Empirical formula is the simplest whole number ratio of moles of atoms that you can find in a molecule.
In combustion analysis all Carbon reacts producing CO2 and all hydrogen reacts producing H2O. With the differences in masses we can find the mass of oxygen and their moles:
<em>Moles CO2 = Moles C:</em>
14.08g * (1mol/44.01g) = 0.3199 moles C * (12.01g/mol) = 3.8423g C
<em>Moles H2O:</em>
4.32g H2O * (1mol/18.01g) = 0.2399 moles H2O * (2mol H / 1molH2O) = 0.4797moles H = 0.4797g H
<em>Mass O:</em>
12.01g = Mass O + 3.8423g C + 0.4797g H
Mass O = 7.688g O
<em>Moles O:</em>
7.688g O * (1mol/16g) = 0.48 moles O
The ratio of atoms (Dividing in the moles of C that are the lower number of moles):
O: 0.48moles O / 0.3199 moles C = 1.50
C: 0.3199 moles C / 0.3199 moles C = 1
H: 0.4797 moles H / 0.3199 moles C = 1.50
As empirical formula requires whole numbers:
O: 1.50* 2 = 3
C: 1*2 = 2
H: 1.50*2 = 3
The empirical formula is:
<h3>C2H3O3</h3>
Answer keys ? With out that I can't answer it Lol!!
The answer is Strontium(Sr). The reactive increase from right to left. And this element has two valence electrons. So Rb is not correct. Then the very reactive metal is Sr.