Respuesta :
Answer:
Percentage composition of phosphorus is 43.66%.
Percentage composition of oxygen is 56.66%.
[tex]P_2O_5[/tex]is the empirical formula for this compound.
Explanation:
Moles of phosphorus =[tex]\frac{35.07 g}{31 g/mol}=1.1312 mol[/tex]
Moles of phosphorus oxide =[tex]\frac{71.00 g}{31x+16 y g/mol}=z[/tex]
1.1312 moles of moles of phosphorus gives z moles of phosphorus oxide.
The from 1 mol phosphorus :
[tex]\frac{z}{1.1312 }[/tex] moles of phosphorus oxide...(1)
[tex]2xP+yO_2\rightarrow 2P_xO_y[/tex]
According to reaction, 2x moles of phosphorus gives 2 mol of phosphorus oxide.
The from 1 mol phosphorus :
[tex]\frac{2}{2x}[/tex] moles of phosphorus oxide...(2)
(1)=(2)
[tex]\frac{z}{1.1312 }=\frac{2}{2x}=\frac{71.00 g}{31x+16 y g/mol}[/tex]
[tex]x=\frac{2}{5}y[/tex]
The molecular formula of the phosphorus oxide :[tex]P_xO_y[/tex]:
[tex]P_{\frac{2y}{5}}O_{y}=P_2O_5[/tex]
The empirical formula is simplest ratio of elements in a compound.
[tex]P_2O_5[/tex]is the empirical formula for this compound.
Percentage composition of phosphorus =
[tex]P\%=\frac{2\times 31 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]
[tex]P\%=43.66\%[/tex]
Percentage composition of oxygen=
[tex]O\%=\frac{5\times 16 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]
[tex]O\%=56.33\%[/tex]
