Question

2.2 g of a compound of phosphorus and sulphur has 1.24 g of P in it. Determine its empirical formula.

Answer 1

The smallest ratio of atoms in the give organic compound called empirical formula.

For example,

Glucose molecular formula: ${ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }$

Empirical formula: ${ C }{ H }_{ 2 }O$

From the given compound, weight = 2.2 g

The element phosphorous weight = 1.24 g

The element sulphur weight = 2.2 g - 1.24 g = 0.96 g

Molar mass of phosphorous = 31 g

Molar mass of sulphur = 32 g

$P\quad :\quad S\quad =\quad \frac { 1.24 }{ 31 } \quad :\quad \frac { 0.96 }{ 32 } \quad \\ \\ \quad \quad \quad \quad \quad \quad =\quad 0.04\quad :\quad 0.03\\\\ \quad \quad \quad \quad \quad \quad =\quad 4\quad :\quad 3$

Therefore, Empirical formula = ${ P }_{ 4 }{ S }_{ 3 }$

Answer 2

Answer:

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