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molar mass of 5-fluoro-2-nitrobenzoic acid

Input interpretation

5-fluoro-2-nitrobenzoic acid | molar mass
5-fluoro-2-nitrobenzoic acid | molar mass

Result

Find the molar mass, M, for 5-fluoro-2-nitrobenzoic 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: FC_6H_3(NO_2)CO_2H Use the chemical formula, FC_6H_3(NO_2)CO_2H, to count the number of atoms, N_i, for each element:  | N_i  C (carbon) | 7  F (fluorine) | 1  H (hydrogen) | 4  N (nitrogen) | 1  O (oxygen) | 4 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)  C (carbon) | 7 | 12.011  F (fluorine) | 1 | 18.998403163  H (hydrogen) | 4 | 1.008  N (nitrogen) | 1 | 14.007  O (oxygen) | 4 | 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)  C (carbon) | 7 | 12.011 | 7 × 12.011 = 84.077  F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163  H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032  N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007  O (oxygen) | 4 | 15.999 | 4 × 15.999 = 63.996  M = 84.077 g/mol + 18.998403163 g/mol + 4.032 g/mol + 14.007 g/mol + 63.996 g/mol = 185.110 g/mol
Find the molar mass, M, for 5-fluoro-2-nitrobenzoic 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: FC_6H_3(NO_2)CO_2H Use the chemical formula, FC_6H_3(NO_2)CO_2H, to count the number of atoms, N_i, for each element: | N_i C (carbon) | 7 F (fluorine) | 1 H (hydrogen) | 4 N (nitrogen) | 1 O (oxygen) | 4 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) C (carbon) | 7 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 4 | 1.008 N (nitrogen) | 1 | 14.007 O (oxygen) | 4 | 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) C (carbon) | 7 | 12.011 | 7 × 12.011 = 84.077 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 4 | 15.999 | 4 × 15.999 = 63.996 M = 84.077 g/mol + 18.998403163 g/mol + 4.032 g/mol + 14.007 g/mol + 63.996 g/mol = 185.110 g/mol

Unit conversion

0.18511 kg/mol (kilograms per mole)
0.18511 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.26 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.26 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 0.95 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 0.95 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 3.2 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 3.2 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 3.1×10^-22 grams  | 3.1×10^-25 kg (kilograms)  | 185 u (unified atomic mass units)  | 185 Da (daltons)
Mass of a molecule m from m = M/N_A: | 3.1×10^-22 grams | 3.1×10^-25 kg (kilograms) | 185 u (unified atomic mass units) | 185 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 185
Relative molecular mass M_r from M_r = M_u/M: | 185