Question

What is the Emirical formula of 69.25% gallium and 30.75% phosphorus?

Answer 1

The empirical formula of a compound with 69.25% gallium and 30.75% phosphorus is GaP (gallium phosphide).

Answer 2

`\qquad \:\:\:\:\:\:\star \bold{\underline{\boxed{\bold{Step\:1}}}}\:`

• Assume that the mass of a compound is 100 g.

As per question, we are given -

`\:\:\:\:\:\:\longrightarrow \sf \underline{Mass \:of \:Ga = 69.25}`

`\:\:\:\:\:\:\longrightarrow \sf \underline{Mass \:of\: P = 30.75}\\`

`\qquad\:\:\:\:\:\:\star \bold{\underline{\boxed{\bold{Step\:2}}}}\:`

• Calculate the number of moles of each element.

`\longrightarrow \sf No\:of \:moles, n= (Given\: Weight )/(Molar\: mass)\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf Ga_(n) = (69.25)/(69.73)\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf Ga_(n) = 0.9931.......\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{Ga_(n) = 0.993}\\`

`\longrightarrow \sf No\:of \:moles, n= (Given\: Weight )/(Molar\: mass)\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf P_(n) = (30.75)/(30.97)\\`

`\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf P_(n) = 0.9928…....\\`

`\:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{P_(n) = 0.993}\\`

`\qquad\:\:\:\:\:\:\star \bold{\underline{\boxed{\bold{Step\:3}}}}\:`

• Represent an empirical formula.

• 
  `\text{Empirical\: formula} = \bold{Ga_(x)P_(y)}`

`\qquad \:\:\:\:\:\:\star \bold{\underline{\boxed{\bold{Step\:4}}}}\:`

• Divide the number of moles of each element by the least number of moles to get the subscripts of each element.Since the smallest number of moles is 0.993 mol, we divide each number of moles by 0.993 mol.

`\:\:\:\:\:\:\longrightarrow \sf x = (0.993)/(0.993) = 1\\`

`\:\:\:\:\:\:\longrightarrow \sf y= (0.993)/(0.993) = 1\\`

`\qquad\:\:\:\:\:\:\star \bold{\underline{\boxed{\bold{Step\:5}}}}\:`

• Write the empirical formula.

• 
  `\boxed{\text{Empirical \: formula} = \text{GaP}}`