Answer to Question #316262 in General Chemistry for mmbali

Question

A compound of molar mass 229 g/mol

g/mol contains only carbon, hydrogen, iodine, and sulfur. Analysis shows that a sample of the compound contains 6 times as much carbon as hydrogen, by mass.

Answer 1

M(C_aH_bS_cI_d) = 229 g/mol;

M(C) = 12 g/mol;

M(H) = 1 g/mol;

M(S) = 32 g/mol;

M(I) = 127 g/mol;

The ratio by mass of carbon to hydrogen is 6:1. 

So, (6 g C)/(1 g H);

If 1 mol C then m(C)/M(C) = 6/12 = 1/2;

If 1 mol H then m(H)/M(H) = 1/1 = 1;

Therefore n(C)/n(H) = 1/2.

Simple formula of C_aH_b is (CH₂)_x;

M(C_aH_b) = M(C_aH_bS_cI_d) - M(S) - M(I) = 229-127-32 = 70 g/mol;

M((CH₂)_x) = 70 g/mol;

x = M((CH₂)_x)/M(CH₂) = 70/14 = 5;

So, (CH₂)_x = (CH₂)₅ = C₅H₁₀;

Therefore, the compound is C₅H₁₀SI.

Answer: C₅H₁₀SI.