Respuesta :
Answer:- P = 43.63% and O = 56.37% and the empirical formula is [tex]P_2O_5[/tex]
Solution:- mass of Phosphorous = 30.98 g
mass of product(phosphorous oxide) = 71.00 g
mass of oxygen = 71.00 - 30.98 = 40.02 g
percentage of phosphorous in the compound = (30.98/71.00)100 = 43.63%
percentage of oxygen in the compound = (40.02/71.00)100 = 56.37%
moles of P = 30.98 g x ( 1mol/30.97 g) = 1.00 mol
moles of O = 40.02 g x (1 mol/ 16.00 g) = 2.50 mol
Ratio of moles of P to O is 1.00:2.50
The whole number ratio is 2:5
So, the empirical formula of the phosphorous oxide formed is [tex]P_2O_5[/tex]
Answer : The percent composition of P = 43.63 % and O = 53.37 % .
Empirical formula of Phosphorous oxide = P₂O₅
Part A : Percent composition :
It is percent of each element present in compound . It is given by formula :
[tex] Percent composition = \frac{mass of element}{ total mass of compound } * 100 [/tex]
Given : Mass of Phosphorous (P ) = 30.98 g
Mass of Compound Phosphorous oxide = 71.00 g
Mass of Oxygen (O) = mass of compound - mass of P
= 71.00 g - 30.98 g = 40.02 g
[tex] Percent composition of P = \frac{mass of P }{ mass of compound} [/tex][tex] = \frac{30.98 g }{71.00 g} * 100 [/tex]
Percent composition of P = 43.63 %
[tex] Percent composition of O = \frac{mass of Al}{Mass of compound } * 100
= \frac{40.02 g }{71.00 g} * 100 [/tex]
Percent composition of O = 53.37 %
Part B : Empirical formula of Phosphorous Oxide.
Empirical formula is formula which shows the proportion of element present in a compound . Following are the steps to calculate empirical formula of an compound :
Step 1 : Find masses of each element .
Mass of P = 30.98 g
Mass of O = 40.02 g
Step 2 : Conversion of masses of element to its mole .
[tex] Mole = \frac{mass }{molar mass } [/tex]
Given : Mass of P = 30.98 g [tex] Molar mass of P = 30.97 \frac{g}{ mol} [/tex]
[tex] Mole of P = \frac{30.98 g }{30.97\frac{g}{mol} } = 1 mol [/tex]
Given: Mass of O = 40.02 g [tex] Molar mass of O = 15.99\frac{g}{mol} [/tex]
[tex] Mole of O = \frac{40.02 g}{15.99\frac{g}{mol}} = 2.5 mol [/tex]
Step 3 : Finding Ratio of mole .
In this step ratio is found by dividing each mole by smallest mole .Since mole of P is smaller , so this will be used for division.
[tex] Ratio of mole of P= \frac{mole of P }{mole of P } = \frac{1 mol}{1 mol} = 1 [/tex]
[tex] Ratio of mole of O = \frac{ mole of O }{mole of P } = \frac{2.5 }{1 } = 2.5 [/tex]
Hence , Ratio of P : O = 1 : 2.5
Since the ratio is in fraction , it need to be converted to whole number . So we multiply the ratio by such a minimum number which gives us a whole number ratio. This step is skipped if the ratio already comes in whole number.
On multiplication the ratio by 2 :
Ratio of P : O = ( 1 : 2.5 ) * 2 = 1 * 2 : 2.5 * 2
Ratio of P : O = 2 : 5
Step 4 : Writing the empirical formula
The ratio of P and O is 2: 5 , which gives the empirical formula of compound as P₂O₅ .
