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2-A gas sample is found to contain 39.10% carbon, 7.67% hydrogen, 26.11%
oxygen, 16.82% phosphorus, and 10.30% fluorine. If the molecular mass is 184.1
g/mol, what is the molecular formula?
C6H1403PF

Respuesta :

ardni313

The molecular formula :

C₆H₁₄O₃PF

Further explanation

Given

39.10% carbon, 7.67% hydrogen, 26.11%  oxygen, 16.82% phosphorus, and 10.30% fluorine.

Required

The molecular formula

Solution

mol ratio :

C = 39.1 : 12 = 3.258

H = 7.67 : 1 = 7.67

O = 26.11 : 16 = 1.632

P = 16.82 : 31 = 0.543

F = 10.3 : 19 = 0.542

Divide by 0.542

C = 6

H : 14

O = 3

P = 1

F = 1

The empirical formula :

C₆H₁₄O₃PF

(The empirical formula)n = the molecular formula

(C₆H₁₄O₃PF)=184.1

(6.12+14.1+3.16+31+19)n=184.1

(184)n=184.1

n = 1

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