The molecule contains: P, N , Cl
Given data:
P = 26.73%
N = 12.09%
O = 100 -(26.73 + 12.09) = 61.18%
If the mass of the sample is 100 g then,
P = 27.73 g; N = 12.09 g and O = 61.18 g
Atomic mass of P = 31 g/mol
Atomic mass of N = 14 g/mol
Atomic mass of O = 16 g/mol
# moles of P = 27.73 g/31 g.mol-1 = 0.8945 moles
# moles of N = 12.09/14 = 0.8636 moles
# moles of O = 61.18/16 = 3.8238 moles
Divide by the smallest # moles:
C = 0.8945/0.8636 = 1.04 = 1
N = 0.8636/0.8636 = 1
O = 3.8238/0.8636 = 4.42 = 4
Empirical formula = CNO4
Empirical formula mass = 12 + 14 + 4*16 = 90 g/mol
Molar mass = 463.5 g/mol
Ratio of molar mass/empirical mass = 463.5/90= 5.15 = 5
Molecular formula = 5(empirical formula) = 5(CNO4)
Molecular formula = C5N5O20
Answer: The molecular formula will be=[tex]P_4N_4Cl_8[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of P = 26.73 g
Mass of N = 12.09 g
Mass of Cl = (100-12.09) g = 61.18 g
Step 1 : convert given masses into moles.
Moles of P =[tex]\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{26.73g}{31g/mole}=0.86moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{12.09g}{14g/mole}=0.86moles[/tex]
Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{61.18g}{35.5g/mole}=1.7moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For P = [tex]\frac{0.86}{0.86}=1[/tex]
For N =[tex]\frac{0.86}{0.86}=1[/tex]
For Cl =[tex]\frac{1.7}{0.86}=2[/tex]
The ratio of P: N: Cl = 1: 1: 2
Hence the empirical formula is [tex]PNCl_2[/tex]
The empirical weight of [tex]PNCl_2[/tex] = 1(31)+1(14)+2(35.5)= 116 g.
The molecular weight = 463.5 g/mole
Now we have to calculate the molecular formula:
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{463.5}{116}=4[/tex]
The molecular formula will be=[tex]4\times PNCl_2=P_4N_4Cl_8[/tex]
