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:- molecular formula is $PdC_4H_6O_4$ .

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

moles of Pd = $47.40gPd(\frac{1mol}{106.42g})$ = 0.445 mol Pd

moles of O = $28.50gO(\frac{1mol}{16g})$ = 1.78 mol O

moles of C = $21.40gC(\frac{1mol}{12.01g})$ = 1.78 mol C

moles of H = $2.69gH(\frac{1mol}{1.01g})$ = 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 = $\frac{0.445}{0.445}$ = 1

O = $\frac{1.78}{0.445}$ = 4

C = $\frac{1.78}{0.445}$ = 4

H = $\frac{2.66}{0.445}$ = 6

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

Answer 2

Answer: The second selection = Pd(O2CCH3)2