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molar mass of 6-(morpholin-4-yl)pyridine-3-boronic acid pinacol ester

Input interpretation

6-(morpholin-4-yl)pyridine-3-boronic acid pinacol ester | molar mass
6-(morpholin-4-yl)pyridine-3-boronic acid pinacol ester | molar mass

Result

Find the molar mass, M, for 6-(morpholin-4-yl)pyridine-3-boronic acid pinacol ester: 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_15H_23BN_2O_3 Use the chemical formula, C_15H_23BN_2O_3, to count the number of atoms, N_i, for each element:  | N_i  N (nitrogen) | 2  C (carbon) | 15  O (oxygen) | 3  B (boron) | 1  H (hydrogen) | 23 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)  N (nitrogen) | 2 | 14.007  C (carbon) | 15 | 12.011  O (oxygen) | 3 | 15.999  B (boron) | 1 | 10.81  H (hydrogen) | 23 | 1.008 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)  N (nitrogen) | 2 | 14.007 | 2 × 14.007 = 28.014  C (carbon) | 15 | 12.011 | 15 × 12.011 = 180.165  O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997  B (boron) | 1 | 10.81 | 1 × 10.81 = 10.81  H (hydrogen) | 23 | 1.008 | 23 × 1.008 = 23.184  M = 28.014 g/mol + 180.165 g/mol + 47.997 g/mol + 10.81 g/mol + 23.184 g/mol = 290.17 g/mol
Find the molar mass, M, for 6-(morpholin-4-yl)pyridine-3-boronic acid pinacol ester: 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_15H_23BN_2O_3 Use the chemical formula, C_15H_23BN_2O_3, to count the number of atoms, N_i, for each element: | N_i N (nitrogen) | 2 C (carbon) | 15 O (oxygen) | 3 B (boron) | 1 H (hydrogen) | 23 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) N (nitrogen) | 2 | 14.007 C (carbon) | 15 | 12.011 O (oxygen) | 3 | 15.999 B (boron) | 1 | 10.81 H (hydrogen) | 23 | 1.008 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) N (nitrogen) | 2 | 14.007 | 2 × 14.007 = 28.014 C (carbon) | 15 | 12.011 | 15 × 12.011 = 180.165 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 B (boron) | 1 | 10.81 | 1 × 10.81 = 10.81 H (hydrogen) | 23 | 1.008 | 23 × 1.008 = 23.184 M = 28.014 g/mol + 180.165 g/mol + 47.997 g/mol + 10.81 g/mol + 23.184 g/mol = 290.17 g/mol

Unit conversion

0.2902 kg/mol (kilograms per mole)
0.2902 kg/mol (kilograms per mole)

Comparisons

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≈ 0.4 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 1.5 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 1.5 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 5 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 5 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 4.8×10^-22 grams  | 4.8×10^-25 kg (kilograms)  | 290 u (unified atomic mass units)  | 290 Da (daltons)
Mass of a molecule m from m = M/N_A: | 4.8×10^-22 grams | 4.8×10^-25 kg (kilograms) | 290 u (unified atomic mass units) | 290 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 290
Relative molecular mass M_r from M_r = M_u/M: | 290