Question

“Analysis of a compound of phosphorus nitrogen and chlorine show that it is 26.73% P and 12.09% N with Cl accounting for the remainder. In a separate experiment, the molar mass of the compound was found to be 463.5 g/mol. Determine the molecular formula of the compound.” PLEASE HELP!

Answer 1

Answer: The molecular formula will be=
`P_4N_4Cl_8`

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 =
`\frac{\text{ given mass of P}}{\text{ molar mass of P}}= (26.73g)/(31g/mole)=0.86moles`

Moles of N =
`\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (12.09g)/(14g/mole)=0.86moles`

Moles of Cl =
`\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= (61.18g)/(35.5g/mole)=1.7moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P =
`(0.86)/(0.86)=1`

For N =
`(0.86)/(0.86)=1`

For Cl =
`(1.7)/(0.86)=2`

The ratio of P: N: Cl = 1: 1: 2

Hence the empirical formula is
`PNCl_2`

The empirical weight of
`PNCl_2` = 1(31)+1(14)+2(35.5)= 116 g.

The molecular weight = 463.5 g/mole

Now we have to calculate the molecular formula:

`n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=(463.5)/(116)=4`

The molecular formula will be=
`4* PNCl_2=P_4N_4Cl_8`

Answer 2

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