Input interpretation
ethyl acetate | relative molecular mass
Result
Find the relative molecular mass, M_r, for ethyl acetate: 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: CH_3COOC_2H_5 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 4 H (hydrogen) | 8 O (oxygen) | 2 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 C (carbon) | 4 | 12.011 H (hydrogen) | 8 | 1.008 O (oxygen) | 2 | 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 C (carbon) | 4 | 12.011 | 4 × 12.011 = 48.044 H (hydrogen) | 8 | 1.008 | 8 × 1.008 = 8.064 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 M_r = 48.044 + 8.064 + 31.998 = 88.106
Comparisons
≈ ( 0.12 ≈ 1/8 ) × relative molecular mass of fullerene ( ≈ 721 )
≈ 0.45 × relative molecular mass of caffeine ( ≈ 194 )
≈ 1.5 × relative molecular mass of sodium chloride ( ≈ 58 )
Corresponding quantities
Molar mass M from M = M_uM_r: | 88 g/mol (grams per mole)
Molecular mass m from m = M_rM_u/N_A: | 1.5×10^-22 grams | 1.5×10^-25 kg (kilograms)