Answer
$x=0.268$
Work Step by Step
Final titration:
$0.1542\ mmol\ thiosulfate\rightarrow 0.0771\ mmol\ I_3^-$
Potassium iodate addition:
$10\ mL\times 0.70\ M=7.0\ mmol$, from stoichiometry, $1.4\ mmol\ I_3^-$ would be formed, so iodate is in excess.
Number of moles of $Cu^{2+}: 0.1542\ mmol$
Number of moles of the compound: $0.05214\ mmol$
Molar mass of the compound: $34.02\ mg/0.05214\ mmol=661.9\ g/mol$
$661.9=88.906+2\times 137.327+3\times 63.546+(7-x)\times 15.9994$
$x=0.268$
