Input interpretation
![3, 5-diiodo-L-tyrosine | molar mass](../image_source/3561d8fc75f713ae49df155eee8a14c8.png)
3, 5-diiodo-L-tyrosine | molar mass
Result
![Find the molar mass, M, for 3, 5-diiodo-L-tyrosine: 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_9H_9I_2NO_3 Use the chemical formula, C_9H_9I_2NO_3, to count the number of atoms, N_i, for each element: | N_i C (carbon) | 9 H (hydrogen) | 9 I (iodine) | 2 N (nitrogen) | 1 O (oxygen) | 3 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) | 9 | 12.011 H (hydrogen) | 9 | 1.008 I (iodine) | 2 | 126.90447 N (nitrogen) | 1 | 14.007 O (oxygen) | 3 | 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) | 9 | 12.011 | 9 × 12.011 = 108.099 H (hydrogen) | 9 | 1.008 | 9 × 1.008 = 9.072 I (iodine) | 2 | 126.90447 | 2 × 126.90447 = 253.80894 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 M = 108.099 g/mol + 9.072 g/mol + 253.80894 g/mol + 14.007 g/mol + 47.997 g/mol = 432.984 g/mol](../image_source/fd833445b8bc57200d5b1d17f404d8d5.png)
Find the molar mass, M, for 3, 5-diiodo-L-tyrosine: 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_9H_9I_2NO_3 Use the chemical formula, C_9H_9I_2NO_3, to count the number of atoms, N_i, for each element: | N_i C (carbon) | 9 H (hydrogen) | 9 I (iodine) | 2 N (nitrogen) | 1 O (oxygen) | 3 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) | 9 | 12.011 H (hydrogen) | 9 | 1.008 I (iodine) | 2 | 126.90447 N (nitrogen) | 1 | 14.007 O (oxygen) | 3 | 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) | 9 | 12.011 | 9 × 12.011 = 108.099 H (hydrogen) | 9 | 1.008 | 9 × 1.008 = 9.072 I (iodine) | 2 | 126.90447 | 2 × 126.90447 = 253.80894 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 M = 108.099 g/mol + 9.072 g/mol + 253.80894 g/mol + 14.007 g/mol + 47.997 g/mol = 432.984 g/mol
Unit conversion
![0.43298 kg/mol (kilograms per mole)](../image_source/2ec477ef47e72175fef16c81ae491506.png)
0.43298 kg/mol (kilograms per mole)
Comparisons
![≈ 0.6 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/95435278355d6c05927397d42fb4f50d.png)
≈ 0.6 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ 2.2 × molar mass of caffeine ( ≈ 194 g/mol )](../image_source/e9034255340032fbda6b067006971a1d.png)
≈ 2.2 × molar mass of caffeine ( ≈ 194 g/mol )
![≈ 7.4 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/02b3cc7780a351e951aa18a6e4f02ac9.png)
≈ 7.4 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 7.2×10^-22 grams | 7.2×10^-25 kg (kilograms) | 433 u (unified atomic mass units) | 433 Da (daltons)](../image_source/2ebe49895b5a36aac2cc30212a92d667.png)
Mass of a molecule m from m = M/N_A: | 7.2×10^-22 grams | 7.2×10^-25 kg (kilograms) | 433 u (unified atomic mass units) | 433 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 433](../image_source/99acd997ce7855db61289cbf5b27186e.png)
Relative molecular mass M_r from M_r = M_u/M: | 433