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molecular mass of 1,3-didecyl-2-methylimidazolium chloride

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1, 3-didecyl-2-methylimidazolium chloride | molecular mass
1, 3-didecyl-2-methylimidazolium chloride | molecular mass

Result

Find the molecular mass, m, for 1, 3-didecyl-2-methylimidazolium chloride: 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_24H_47ClN_2 Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  Cl (chlorine) | 1  C (carbon) | 24  N (nitrogen) | 2  H (hydrogen) | 47 Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table:  | N_i | m_i/u  Cl (chlorine) | 1 | 35.45  C (carbon) | 24 | 12.011  N (nitrogen) | 2 | 14.007  H (hydrogen) | 47 | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: Answer: |   | | N_i | m_i/u | mass/u  Cl (chlorine) | 1 | 35.45 | 1 × 35.45 = 35.45  C (carbon) | 24 | 12.011 | 24 × 12.011 = 288.264  N (nitrogen) | 2 | 14.007 | 2 × 14.007 = 28.014  H (hydrogen) | 47 | 1.008 | 47 × 1.008 = 47.376  m = 35.45 u + 288.264 u + 28.014 u + 47.376 u = 399.10 u
Find the molecular mass, m, for 1, 3-didecyl-2-methylimidazolium chloride: 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_24H_47ClN_2 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Cl (chlorine) | 1 C (carbon) | 24 N (nitrogen) | 2 H (hydrogen) | 47 Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | N_i | m_i/u Cl (chlorine) | 1 | 35.45 C (carbon) | 24 | 12.011 N (nitrogen) | 2 | 14.007 H (hydrogen) | 47 | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: Answer: | | | N_i | m_i/u | mass/u Cl (chlorine) | 1 | 35.45 | 1 × 35.45 = 35.45 C (carbon) | 24 | 12.011 | 24 × 12.011 = 288.264 N (nitrogen) | 2 | 14.007 | 2 × 14.007 = 28.014 H (hydrogen) | 47 | 1.008 | 47 × 1.008 = 47.376 m = 35.45 u + 288.264 u + 28.014 u + 47.376 u = 399.10 u