Answer to Question #135195 in General Chemistry for Lucero

Question

Analysis revealed the composition of a substance to contain 48.6% carbon, 8.11% hydrogen and the remainder is oxygen. The molar mass is 222 g/mol. Find the molecular (true) formula.

Answer 1

oxygen:

100%-48.6%-8.11%=43.29

then:

ω (C) = 48,6%

ω (H) = 8,11%

ω (O) = 43,29%

M (C_xH_yO_z) = 222 g/mol

m (C_xH_yO_z) = 100 g;

$m (C) = ω (C)\times\frac{m (CxHyOz)}{100} = 48,6\times\frac{100}{100} = 48.6 g$

$n (C) = \frac {m(C)}{ M (C)} = \frac{48.6}{ 12} = 4.05mol;$

$m (H) = ω (H)\times\frac{m (CxHyOz)}{100} = 8.11\times\frac{100}{100} = 8.11g$

$n (H) = \frac {m(H)}{ M (H)} = \frac{8.11}{ 1} = 8,11 mol;$

$m (O) = ω (O)\times\frac{m (CxHyOz)}{100} = 43.29\times\frac{100}{100} = 43.29g$

$n (O) = \frac {m(O)}{ M (O)} = \frac{43.29}{ 16} =2.71 mol;$

x : y : z = n (C) : n (H) : n (O) = 4.05 : 8.11 : 2.71= 1.5 : 3 : 1;

simple formula - C_1,5H₃O;

$M (C2H4O) = Mr (C1.5H3O) = Ar (C)\times N (C) + Ar (H) \times N (H) + Ar (O) \times N (O) = 12 \times1.5 + 1\times3+ 16\times1 = 37g/mol$

$=\frac{M (CxHyOz) }{ M (C1,5H3O)} =\frac{ 222}{ 37}= 6;$

(C_1,5H₃O;)6 - C₉H₁₈O₆ - Acetone peroxide

C₉H₁₈O₆ - Acetone peroxide