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molar mass of α-bromo-4-fluorophenylacetic acid

Input interpretation

α-bromo-4-fluorophenylacetic acid | molar mass
α-bromo-4-fluorophenylacetic acid | molar mass

Result

Find the molar mass, M, for α-bromo-4-fluorophenylacetic acid: 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: C_8H_6BrFO_2 Use the chemical formula, C_8H_6BrFO_2, to count the number of atoms, N_i, for each element:  | N_i  Br (bromine) | 1  C (carbon) | 8  F (fluorine) | 1  H (hydrogen) | 6  O (oxygen) | 2 Look up the atomic mass, m_i, in g·mol^(-1) for each element in the periodic table:  | N_i | m_i/g·mol^(-1)  Br (bromine) | 1 | 79.904  C (carbon) | 8 | 12.011  F (fluorine) | 1 | 18.998403163  H (hydrogen) | 6 | 1.008  O (oxygen) | 2 | 15.999 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molar mass, M: Answer: |   | | N_i | m_i/g·mol^(-1) | mass/g·mol^(-1)  Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904  C (carbon) | 8 | 12.011 | 8 × 12.011 = 96.088  F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163  H (hydrogen) | 6 | 1.008 | 6 × 1.008 = 6.048  O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998  M = 79.904 g/mol + 96.088 g/mol + 18.998403163 g/mol + 6.048 g/mol + 31.998 g/mol = 233.036 g/mol
Find the molar mass, M, for α-bromo-4-fluorophenylacetic acid: 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: C_8H_6BrFO_2 Use the chemical formula, C_8H_6BrFO_2, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 8 F (fluorine) | 1 H (hydrogen) | 6 O (oxygen) | 2 Look up the atomic mass, m_i, in g·mol^(-1) for each element in the periodic table: | N_i | m_i/g·mol^(-1) Br (bromine) | 1 | 79.904 C (carbon) | 8 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 6 | 1.008 O (oxygen) | 2 | 15.999 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molar mass, M: Answer: | | | N_i | m_i/g·mol^(-1) | mass/g·mol^(-1) Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 8 | 12.011 | 8 × 12.011 = 96.088 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 6 | 1.008 | 6 × 1.008 = 6.048 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 M = 79.904 g/mol + 96.088 g/mol + 18.998403163 g/mol + 6.048 g/mol + 31.998 g/mol = 233.036 g/mol

Unit conversion

0.23304 kg/mol (kilograms per mole)
0.23304 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.32 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.32 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 1.2 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 1.2 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 4 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 4 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 3.9×10^-22 grams  | 3.9×10^-25 kg (kilograms)  | 233 u (unified atomic mass units)  | 233 Da (daltons)
Mass of a molecule m from m = M/N_A: | 3.9×10^-22 grams | 3.9×10^-25 kg (kilograms) | 233 u (unified atomic mass units) | 233 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 233
Relative molecular mass M_r from M_r = M_u/M: | 233