a gaseous compound has the following composition by mass: c, 17.7%; h, 0.7%; f, 69.8%; o, 11.8%. its…

Explanation:

Step1: Calculate moles of each element

Assume 100 g of the compound. Then mass of C = 17.7 g, mass of H = 0.7 g, mass of F = 69.8 g, mass of O = 11.8 g. Moles of C: $n_{C}=\frac{17.7\ g}{12.01\ g/mol}\approx1.47\ mol$ Moles of H: $n_{H}=\frac{0.7\ g}{1.01\ g/mol}\approx0.69\ mol$ Moles of F: $n_{F}=\frac{69.8\ g}{19.00\ g/mol}\approx3.67\ mol$ Moles of O: $n_{O}=\frac{11.8\ g}{16.00\ g/mol}\approx0.74\ mol$

Step2: Find the mole - ratio

Divide each number of moles by the smallest number of moles (0.69 mol). Ratio of C: $\frac{1.47\ mol}{0.69\ mol}\approx2$ Ratio of H: $\frac{0.69\ mol}{0.69\ mol}=1$ Ratio of F: $\frac{3.67\ mol}{0.69\ mol}\approx5$ Ratio of O: $\frac{0.74\ mol}{0.69\ mol}\approx1$ Empirical formula is $C_{2}HFO_{5}$. Empirical - formula weight: $2\times12.01 + 1.01+19.00 + 5\times16.00=24.02 + 1.01+19.00 + 80.00 = 124.03\ g/mol$

Step3: Find the molecular formula

$n=\frac{\text{Molecular weight}}{\text{Empirical - formula weight}}=\frac{136.0\ g/mol}{124.03\ g/mol}\approx1$ (rounding to the nearest whole - number). Molecular formula is the same as empirical formula $C_{2}HFO_{5}$. Analyze the Lewis - structures based on the formula: The correct Lewis formula is the one that satisfies the valency of each atom. Carbon has a valency of 4, hydrogen has a valency of 1, fluorine has a valency of 1 and oxygen has a valency of 2. The structure $\begin{array}{ccc} &&\ddot{O}:\ &&|\ :F:&\ddot{C}&-\ddot{O}-H\ |&&|\ F&-\ddot{C}&-F\ |&&\ F&& \end{array}$ is consistent with the formula $C_{2}HFO_{5}$

Answer: The third structure (from the top) $\begin{array}{ccc}

&&\ddot{O}:\ &&|\ :F:&\ddot{C}&-\ddot{O}-H\ |&&|\ F&-\ddot{C}&-F\ |&&\ F&& \end{array}$