Calculate the empirical and molecular formula of a compound whose molecular weight is 120 and has the following percentage compositions:
Mg=19.68%; S=26.24%; O=52.48%.

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Solution

Given:
Molecular weight of compound=120.
Percentage composition: Mg=19.68%; S=26.24%; O=52.48%.
Step-1: Calculate the empirical formula
Empirical formula: Empirical formula is the simplest formula of the compound which gives the simplest whole number ratio of the atoms of the various elements present in one molecule of the compound.
Element % Of Element Atomic mass (u) Atomic ratio Simplest ratio
Magnesium (Mg) 19.68 24 $\frac{19.68}{24} = 0.82$ $\frac{0.82}{0.82} = 1$
Sulphur (S) 26.24 32 $\frac{26.24}{32} = 0.82$ $\frac{0.82}{0.82} = 1$
Oxygen (O) 52.48 16 $\frac{52.4}{16} = 3.275$ $\frac{3.275}{0.82} = 3.99 \approx 4$
As the simplest ratio of Magnesium is 1, Sulphur is 1 and Oxygen is 4, therefore the empirical formula comes out to be MgSO_4.
Step-2: Calculate the empirical and molecular mass
Empirical formula weight: Empirical formula weight is obtained by the addition of the atomic weight of the various atoms present in the empirical formula.
Atomic weight of Magnesium= 24 u
Atomic weight of Sulphur= 32 u
Atomic weight of Oxygen= 16 u
Empirical formula weight= Atomic weight of Magnesium + Atomic weight of Sulphur + 4(Atomic weight of Oxygen)
Empirical formula weight= 24 + 32 + 4(16) = 120 u
Molecular formula weight: Molecular formula weight is obtained by the addition of the atomic weight of the various atoms present in the molecular formula.
Molecular weight =120 u (given)
Step-3: Calculate the molecular formula
Molecular formula: Molecular formula is the actual formula of the compound which gives the actual number of the atoms present in the molecule of the compound.
Relationship between empirical formula and molecular formula is given as follows: Molecular formula = n × Empirical formula, where n is an integer such as 1,2,3, etc.
$n = \frac{Molecular weight of the compound}{Empirical formula mass}$
$n = \frac{120}{120} = 1$
As stated above, Molecular formula = n × Empirical formula, and empirical formula is MgSO₄
Therefore, the molecular formula is (MgSO₄)₁ or MgSO₄.

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Element	% Of Element	Atomic mass (u)	Atomic ratio	Simplest ratio
Magnesium (Mg)	19.68	24	$\frac{19.68}{24} = 0.82$	$\frac{0.82}{0.82} = 1$
Sulphur (S)	26.24	32	$\frac{26.24}{32} = 0.82$	$\frac{0.82}{0.82} = 1$
Oxygen (O)	52.48	16	$\frac{52.4}{16} = 3.275$	$\frac{3.275}{0.82} = 3.99 \approx 4$

Calculate the empirical and molecular formula of a compound whose molecular weight is 120 and has the following percentage compositions: Mg=19.68%; S=26.24%; O=52.48%.

Calculate the empirical and molecular formula of a compound whose molecular weight is 120 and has the following percentage compositions:
Mg=19.68%; S=26.24%; O=52.48%.