Question

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?

Answer 1

Answer: The second selection = Pd(O2CCH3)2

Answer 2

Answer:- molecular formula is
`PdC_4H_6O_4` .

Solution:- Percentages are given from which we will calculate the moles of each element.

moles of Pd =
`47.40gPd((1mol)/(106.42g))` = 0.445 mol Pd

moles of O =
`28.50gO((1mol)/(16g))` = 1.78 mol O

moles of C =
`21.40gC((1mol)/(12.01g))` = 1.78 mol C

moles of H =
`2.69gH((1mol)/(1.01g))` = 2.66 mol H

Now we divide the moles of each by the least one of them. The least one is 0.445. So, let's divide the moles of each by 0.445.

Pd =
`(0.445)/(0.445)` = 1

O =
`(1.78)/(0.445)` = 4

C =
`(1.78)/(0.445)` = 4

H =
`(2.66)/(0.445)` = 6

So, the molecular formula of the compound is
`PdC_4H_6O_4` .