Question

A 1.22 gram of unknown liquid is vaporized at 100 degrees Celsius and 102 kPa. the sample occupies 617 ml. If the percentage composition of this compound is 59.96% carbon, 13.42% hydrogen and 26.62% oxegyn, derive the molecular formula.

Answer 1

A 1.22 gram of unknown liquid is vaporized at 100 degrees Celsius and 102 kPa. the sample occupies 617 ml. If the percentage composition of this compound is 59.96% carbon, 13.42% hydrogen and 26.62% oxegyn, derive the molecular formula.

**Solution.** We derive the simplest formula of the compound, based on the percentage of the elements. Suppose there is 100 g of substance, then we find the molar ratio of the elements: $v(C):v(H):v(O) = \frac{59.96}{M(C)} : \frac{13.42}{M(H)} : \frac{26.62}{M(O)} = \frac{59.96}{12} : \frac{13.42}{1} : \frac{26.62}{16} = 5:13.42:1.66 = 3:8:1$.

So, the simplest formula for the desired compound is $C_3H_8O$.

Further, we have the following data: $m(\text{compound}) = 1.22$ grams, $t = 100$ degrees Celsius, or $100 + 273 = 373$ degrees Kelvin, $p = 102$ kPa, $V = 617$ ml, or $0.617$ L, then, according to the Mendeleev-Clapeyron equation, we can calculate the molar mass of the compound: $pV = \frac{m(\text{compound})}{M} RT$, $M = \frac{m(\text{compound})RT}{pV} = \frac{1.22 \times 8.31 \times 373}{102 \times 0.617} = 60 \frac{g}{\text{mol}}$. The molar mass of a simple substance, the formula of which we found above, is: $M(C_3H_8O) = 3 \times M(C) + 8 \times M(H) + M(O) = 36 + 8 + 16 = 60 \frac{g}{\text{mol}}$. The molar masses are the same, then the formula of the compound: $C_3H_8O$.

**Answer:** $C_3H_8O$.

Answer provided by www.AssignmentExpert.com