\(MX=3,375.32=172g/mol\)
\(nX=\dfrac{3,44}{172}=0,02\left(mol\right)\)
\(mCO_2+mH_2O=31,52-22,32=9,2\left(g\right)\)
\(nCO_2=n_{BaCO_3}=\dfrac{31,52}{197}=0,16mol\)
\(\Rightarrow nH_2O=\dfrac{9,2-0,16.44}{18}=0,12\left(mol\right)\)
\(C_X=\dfrac{0,16}{0,02}=8\)
\(H_X=\dfrac{0,12.2}{0,02}=12\)
\(O_X=\dfrac{172-8.12-12}{16}=4\)
\(\Rightarrow CTPT_X=C_8H_{12}O_4\)