a/ $d_{A/kk}=1,52→M_A=44,08(g/mol)$
Gọi CTHH là $C_xO_y$
$m_C=\dfrac{44,08.27,3}{100}=12(g)$
$→n_C=\dfrac{12}{12}=1(mol)$
$→x=1$
$m_O=\dfrac{44,08.72,7}{100}=32(g)$
$→n_O=\dfrac{32}{16}=2(mol)$
$→y=2$
Suy ra CTHH: $CO_2$
b/ $M_{R(H_2PO_4)_2}=M_R+4.M_H+2.M_P+8.M_O=234(g/mol)$
$↔M_R+4+62+128=234(g/mol)$
$↔M_R=40(g/mol)$
$→$ Nguyên tố R là Canxi (Ca)