relative molecular mass of tris[trinitratocerium(IV)]paraperiodate

tris[trinitratocerium(IV)]paraperiodate | relative molecular mass

Find the relative molecular mass, M_r, for tris[trinitratocerium(IV)]paraperiodate: M_r = sum _iN_im_i/m_u Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i and m_i using these items. • Finally, compute the mass. Write the chemical formula: [Ce(NO_3)_3]_3(H_2IO_6) Use the chemical formula, [Ce(NO_3)_3]_3(H_2IO_6), to count the number of atoms, N_i, for each element: | N_i Ce (cerium) | 3 H (hydrogen) | 2 I (iodine) | 1 N (nitrogen) | 9 O (oxygen) | 33 Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table. Since m_i is divided by the atomic mass constant, m_u, the result is a unitless relative atomic mass: | N_i | m_i/m_u Ce (cerium) | 3 | 140.116 H (hydrogen) | 2 | 1.008 I (iodine) | 1 | 126.90447 N (nitrogen) | 9 | 14.007 O (oxygen) | 33 | 15.999 Multiply N_i by m_i/m_u to compute the relative mass for each element. Then sum those values to compute the relative molecular mass, M_r: Answer: | | | N_i | m_i/m_u | relative mass Ce (cerium) | 3 | 140.116 | 3 × 140.116 = 420.348 H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016 I (iodine) | 1 | 126.90447 | 1 × 126.90447 = 126.90447 N (nitrogen) | 9 | 14.007 | 9 × 14.007 = 126.063 O (oxygen) | 33 | 15.999 | 33 × 15.999 = 527.967 M_r = 420.348 + 2.016 + 126.90447 + 126.063 + 527.967 = 1203.298

≈ 1.7 × relative molecular mass of fullerene ( ≈ 721 )

≈ 6.2 × relative molecular mass of caffeine ( ≈ 194 )

≈ 21 × relative molecular mass of sodium chloride ( ≈ 58 )

Molar mass M from M = M_uM_r: | 1.2 kg/mol (kilograms per mole)

Molecular mass m from m = M_rM_u/N_A: | 2×10^-21 grams | 2×10^-24 kg (kilograms)