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molar mass of 3,5-diiodoanthranilic acid

Input interpretation

3, 5-diiodoanthranilic acid | molar mass
3, 5-diiodoanthranilic acid | molar mass

Result

Find the molar mass, M, for 3, 5-diiodoanthranilic 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: I_2C_6H_2(NH_2)CO_2H Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  C (carbon) | 7  H (hydrogen) | 5  I (iodine) | 2  N (nitrogen) | 1  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)  C (carbon) | 7 | 12.011  H (hydrogen) | 5 | 1.008  I (iodine) | 2 | 126.90447  N (nitrogen) | 1 | 14.007  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)  C (carbon) | 7 | 12.011 | 7 × 12.011 = 84.077  H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040  I (iodine) | 2 | 126.90447 | 2 × 126.90447 = 253.80894  N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007  O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998  M = 84.077 g/mol + 5.040 g/mol + 253.80894 g/mol + 14.007 g/mol + 31.998 g/mol = 388.931 g/mol
Find the molar mass, M, for 3, 5-diiodoanthranilic 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: I_2C_6H_2(NH_2)CO_2H Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 7 H (hydrogen) | 5 I (iodine) | 2 N (nitrogen) | 1 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) C (carbon) | 7 | 12.011 H (hydrogen) | 5 | 1.008 I (iodine) | 2 | 126.90447 N (nitrogen) | 1 | 14.007 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) C (carbon) | 7 | 12.011 | 7 × 12.011 = 84.077 H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040 I (iodine) | 2 | 126.90447 | 2 × 126.90447 = 253.80894 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 M = 84.077 g/mol + 5.040 g/mol + 253.80894 g/mol + 14.007 g/mol + 31.998 g/mol = 388.931 g/mol

Unit conversion

0.38893 kg/mol (kilograms per mole)
0.38893 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.54 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.54 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 2 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 2 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 6.7 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 6.7 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 6.5×10^-22 grams  | 6.5×10^-25 kg (kilograms)  | 389 u (unified atomic mass units)  | 389 Da (daltons)
Mass of a molecule m from m = M/N_A: | 6.5×10^-22 grams | 6.5×10^-25 kg (kilograms) | 389 u (unified atomic mass units) | 389 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 389
Relative molecular mass M_r from M_r = M_u/M: | 389