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molecular mass of lead(IV) acetate

Input interpretation

lead(IV) acetate | molecular mass
lead(IV) acetate | molecular mass

Result

Find the molecular mass, m, for lead(IV) acetate: m = sum _iN_im_i 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: Pb(CH_3CO_2)_4 Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  C (carbon) | 8  H (hydrogen) | 12  O (oxygen) | 8  Pb (lead) | 1 Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table:  | N_i | m_i/u  C (carbon) | 8 | 12.011  H (hydrogen) | 12 | 1.008  O (oxygen) | 8 | 15.999  Pb (lead) | 1 | 207.2 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: Answer: |   | | N_i | m_i/u | mass/u  C (carbon) | 8 | 12.011 | 8 × 12.011 = 96.088  H (hydrogen) | 12 | 1.008 | 12 × 1.008 = 12.096  O (oxygen) | 8 | 15.999 | 8 × 15.999 = 127.992  Pb (lead) | 1 | 207.2 | 1 × 207.2 = 207.2  m = 96.088 u + 12.096 u + 127.992 u + 207.2 u = 443.4 u
Find the molecular mass, m, for lead(IV) acetate: m = sum _iN_im_i 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: Pb(CH_3CO_2)_4 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 8 H (hydrogen) | 12 O (oxygen) | 8 Pb (lead) | 1 Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | N_i | m_i/u C (carbon) | 8 | 12.011 H (hydrogen) | 12 | 1.008 O (oxygen) | 8 | 15.999 Pb (lead) | 1 | 207.2 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: Answer: | | | N_i | m_i/u | mass/u C (carbon) | 8 | 12.011 | 8 × 12.011 = 96.088 H (hydrogen) | 12 | 1.008 | 12 × 1.008 = 12.096 O (oxygen) | 8 | 15.999 | 8 × 15.999 = 127.992 Pb (lead) | 1 | 207.2 | 1 × 207.2 = 207.2 m = 96.088 u + 12.096 u + 127.992 u + 207.2 u = 443.4 u

Unit conversions

443.4 Da (daltons)
443.4 Da (daltons)
0.4434 kDa (kilodaltons)
0.4434 kDa (kilodaltons)
7.362×10^-22 grams
7.362×10^-22 grams
7.362×10^-25 kg (kilograms)
7.362×10^-25 kg (kilograms)
443.4 chemical atomic mass units  (unit officially deprecated)
443.4 chemical atomic mass units (unit officially deprecated)
443.5 physical atomic mass units  (unit officially deprecated)
443.5 physical atomic mass units (unit officially deprecated)

Comparisons as mass of molecule

 ≈ 0.62 × molecular mass of fullerene ( ≈ 721 u )
≈ 0.62 × molecular mass of fullerene ( ≈ 721 u )
 ≈ 2.3 × molecular mass of caffeine ( ≈ 194 u )
≈ 2.3 × molecular mass of caffeine ( ≈ 194 u )
 ≈ 7.6 × molecular mass of sodium chloride ( ≈ 58 u )
≈ 7.6 × molecular mass of sodium chloride ( ≈ 58 u )

Corresponding quantities

Relative atomic mass A_r from A_r = m_aN_A/M_u:  | 443
Relative atomic mass A_r from A_r = m_aN_A/M_u: | 443
Molar mass M from M = m_aN_A:  | 443 g/mol (grams per mole)
Molar mass M from M = m_aN_A: | 443 g/mol (grams per mole)
Relative molecular mass M_r from M_r = m_mN_A/M_u:  | 443
Relative molecular mass M_r from M_r = m_mN_A/M_u: | 443
Molar mass M from M = m_mN_A:  | 443 g/mol (grams per mole)
Molar mass M from M = m_mN_A: | 443 g/mol (grams per mole)