Answer
$C_{18}H_{18}O_3N_3As_3$ and $C_{30}H_{30}O_5N_5As_5$ respectively
Work Step by Step
From a 100.0 g sample:
C: 39.37 g, 12.011 g/mol --> 3.278 mols
H: 3.304 g, 1.008 g/mol --> 3.278 mols
O: 8.741 g, 15.999 g/mol --> 0.5463 mols
N: 7.652 g, 14.007 g/mol --> 0.5463 mols
As: 40.932 g, 74.922 g/mol --> 0.5463 mols
Normalizing by the smallest number of mols:
C, H: 3.278/0.5463 = 6.0
O, N, As: 0.5463/0.5463 = 1.0
The emprical formula of both compunds is: $C_6H_6ONAs$, which has a molar mass of $M_u=183.042\ g/mol$
For the first compound ($M=549\ g/mol$): $M/M_u=3.0$, so its molecular formula is:
$C_{18}H_{18}O_3N_3As_3$
For the second compound ($M=915\ g/mol$): $M/M_u=5.0$
$C_{30}H_{30}O_5N_5As_5$
