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molar mass of 3-bromo-4'-fluorobenzophenone

Input interpretation

3-bromo-4'-fluorobenzophenone | molar mass
3-bromo-4'-fluorobenzophenone | molar mass

Result

Find the molar mass, M, for 3-bromo-4'-fluorobenzophenone: 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: BrC_6H_4COC_6H_4F Use the chemical formula, BrC_6H_4COC_6H_4F, to count the number of atoms, N_i, for each element:  | N_i  Br (bromine) | 1  C (carbon) | 13  F (fluorine) | 1  H (hydrogen) | 8  O (oxygen) | 1 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) | 13 | 12.011  F (fluorine) | 1 | 18.998403163  H (hydrogen) | 8 | 1.008  O (oxygen) | 1 | 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) | 13 | 12.011 | 13 × 12.011 = 156.143  F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163  H (hydrogen) | 8 | 1.008 | 8 × 1.008 = 8.064  O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999  M = 79.904 g/mol + 156.143 g/mol + 18.998403163 g/mol + 8.064 g/mol + 15.999 g/mol = 279.108 g/mol
Find the molar mass, M, for 3-bromo-4'-fluorobenzophenone: 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: BrC_6H_4COC_6H_4F Use the chemical formula, BrC_6H_4COC_6H_4F, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 13 F (fluorine) | 1 H (hydrogen) | 8 O (oxygen) | 1 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) | 13 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 8 | 1.008 O (oxygen) | 1 | 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) | 13 | 12.011 | 13 × 12.011 = 156.143 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 8 | 1.008 | 8 × 1.008 = 8.064 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 79.904 g/mol + 156.143 g/mol + 18.998403163 g/mol + 8.064 g/mol + 15.999 g/mol = 279.108 g/mol

Unit conversion

0.27911 kg/mol (kilograms per mole)
0.27911 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.39 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.39 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 1.4 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 1.4 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 4.8 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 4.8 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 4.6×10^-22 grams  | 4.6×10^-25 kg (kilograms)  | 279 u (unified atomic mass units)  | 279 Da (daltons)
Mass of a molecule m from m = M/N_A: | 4.6×10^-22 grams | 4.6×10^-25 kg (kilograms) | 279 u (unified atomic mass units) | 279 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 279
Relative molecular mass M_r from M_r = M_u/M: | 279