Question

select all the correct answers. a compound has the percent composition 47.40% pd, 28.50% o, 21.40% c, and 2.69% h. based on this information, which molecular formulas could represent the compound? pdo₂c₂h₃ pd(o₂cch₃)₂ pd(o₂c₂h₃)₃ pdo₄c₂h₉ pd₂c₈h₁₂o₈

Explanation:

Step1: Assume 100 g of the compound

So, we have 47.40 g Pd, 28.50 g O, 21.40 g C, and 2.69 g H.

Step2: Calculate the number of moles of each element

Molar - mass of Pd = 106.42 g/mol, moles of Pd $n_{Pd}=\frac{47.40\ g}{106.42\ g/mol}\approx0.445\ mol$;
Molar - mass of O = 16.00 g/mol, moles of O $n_{O}=\frac{28.50\ g}{16.00\ g/mol}\approx1.781\ mol$;
Molar - mass of C = 12.01 g/mol, moles of C $n_{C}=\frac{21.40\ g}{12.01\ g/mol}\approx1.782\ mol$;
Molar - mass of H = 1.01 g/mol, moles of H $n_{H}=\frac{2.69\ g}{1.01\ g/mol}\approx2.663\ mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (0.445 mol).
Ratio of Pd : O : C : H = $\frac{0.445}{0.445}:\frac{1.781}{0.445}:\frac{1.782}{0.445}:\frac{2.663}{0.445}\approx1:4:4:6$.
The empirical formula is $PdO_{4}C_{4}H_{6}$.

Step4: Check each option

• For $PdO_{2}C_{2}H_{3}$: The ratio of Pd : O : C : H is 1:2:2:3, not consistent with the empirical - formula ratio.
• For $Pd(O_{2}CCH_{3})_{2}$: Expand it to $PdC_{4}H_{6}O_{4}$, the ratio of Pd : O : C : H is 1:4:4:6, consistent with the empirical - formula ratio.
• For $Pd(O_{2}C_{2}H_{3})_{3}$: Expand it to $PdC_{6}H_{9}O_{6}$, the ratio of Pd : O : C : H is 1:6:6:9, not consistent with the empirical - formula ratio.
• For $PdO_{4}C_{2}H_{9}$: The ratio of Pd : O : C : H is 1:4:2:9, not consistent with the empirical - formula ratio.
• For $Pd_{2}C_{8}H_{12}O_{8}$: Divide by 2, we get $PdC_{4}H_{6}O_{4}$, the ratio of Pd : O : C : H is 1:4:4:6, consistent with the empirical - formula ratio.

Answer:

Pd$(O_{2}CCH_{3})_{2}$, $Pd_{2}C_{8}H_{12}O_{8}$