Question

The organic compound Y contains C, H and O only. When Y is subjected to complete combustion, CO2 and H2O obtained are in the mole ratio of 2:1, respectively. The accurate relative molecular mass of Y is equal to 152. The percentage of O in Y is less than 40%, by weight. Determine the molecular formula of Y.
The relative atomic masses are as follows, C=12.0; H=1.0; O=16.0

Answer 1

Question #60427, Chemistry / Organic Chemistry |

The organic compound Y contains C, H and O only. When Y is subjected to complete combustion, CO2 and H2O obtained are in the mole ratio of 2:1, respectively. The accurate relative molecular mass of Y is equal to 152. The percentage of O in Y is less than 40%, by weight. Determine the molecular formula of Y.

The relative atomic masses are as follows, C=12.0; H=1.0; O=16.0

Answer:

Let's assume that the compound Y has formula $C_{\mathrm{a}}H_{\mathrm{b}}O_{\mathrm{k}}$.

Thus, its combustion can be described by the equation:

$$C _ {a} H _ {b} O _ {k} + f O _ {2} \rightarrow a C O _ {2} + (b / 2) H _ {2} O$$

As given $v(CO2) / v(H2O) = 2/1$, then

$$a / (0.5b) = 2,$$ $$a = b$$

The general formula can be written as $C_{\mathrm{a}}H_{\mathrm{a}}O_{\mathrm{k}}$ and the equation for molecular weight is:

$$12a + a + 16k = 152$$ $$13a + 16k = 152$$

The percentage of O gives the number of its atoms:

$$w(O) = m(O)/152 = [16k/152] < 0.4$$ $$16k < 60.8$$ $$k < 3.8$$

Thus, the possible number of oxygen can be $k = 1, 2, 3$.

The substitution of the parameter k into 1, 2, 3 in the equation $13a + 16k = 152$ results in:

$$k = 1, 13a + 16 = 152, a = 10.46$$ $$k = 2, 13a + 32 = 152, a = 9.23$$ $$k = 3, 13a + 48 = 152, a = 8.$$

The right combination is $k = 3$ and $a = 8$ and the formula is $C_8H_8O_3$.

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