Input interpretation
barium titanate | relative molecular mass
Result
Find the relative molecular mass, M_r, for barium titanate: 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: BaTiO_3 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Ba (barium) | 1 O (oxygen) | 3 Ti (titanium) | 1 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 Ba (barium) | 1 | 137.327 O (oxygen) | 3 | 15.999 Ti (titanium) | 1 | 47.867 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 Ba (barium) | 1 | 137.327 | 1 × 137.327 = 137.327 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 Ti (titanium) | 1 | 47.867 | 1 × 47.867 = 47.867 M_r = 137.327 + 47.997 + 47.867 = 233.191
Comparisons
≈ 0.32 × relative molecular mass of fullerene ( ≈ 721 )
≈ 1.2 × relative molecular mass of caffeine ( ≈ 194 )
≈ 4 × relative molecular mass of sodium chloride ( ≈ 58 )
Corresponding quantities
Molar mass M from M = M_uM_r: | 233 g/mol (grams per mole)
Molecular mass m from m = M_rM_u/N_A: | 3.9×10^-22 grams | 3.9×10^-25 kg (kilograms)