The empirical formula :
C₁₀H₁₆N₄SO₇
<h3>Further explanation</h3>
Given
6.4 g sample
Required
The empirical formula
Solution
mass C :
= 12/44 x 8.37 g
= 2.28
mass H :
= 2/18 x 2.75 g
= 0.305
mass N = 1.06
mass S :
= 32/64 x 1.23
= 0.615
mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g
Mol ratio :
= C : H : N : S : O
= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16
= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019
= 10 : 16 : 4 : 1 : 7
The empirical formula :
C₁₀H₁₆N₄SO₇
Elements occur naturally in different forms called isotopes. These are the same atoms with the same number of protons but has different number of neutrons. Almost all elements exist in nature as such. The amounts of these isotopes are expressed in terms of percent abundances and since it is in a percentage units, the sum of all the abundances should equate to 100 percent or else the calculation is wrong. This is not true onlt for magnesiun but with all of the elements. The atomic mass of these elements also depends on the percent abundances of the isotopes since it is a weighted average value of the individual masses of the isotopes.
