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molar mass of 1,1,3-trichlorotrifluoroacetone

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1, 1, 3-trichlorotrifluoroacetone | molar mass
1, 1, 3-trichlorotrifluoroacetone | molar mass

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Find the molar mass, M, for 1, 1, 3-trichlorotrifluoroacetone: 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_3Cl_3F_3O Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  C (carbon) | 3  Cl (chlorine) | 3  F (fluorine) | 3  O (oxygen) | 1 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) | 3 | 12.011  Cl (chlorine) | 3 | 35.45  F (fluorine) | 3 | 18.998403163  O (oxygen) | 1 | 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) | 3 | 12.011 | 3 × 12.011 = 36.033  Cl (chlorine) | 3 | 35.45 | 3 × 35.45 = 106.35  F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489  O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999  M = 36.033 g/mol + 106.35 g/mol + 56.995209489 g/mol + 15.999 g/mol = 215.38 g/mol
Find the molar mass, M, for 1, 1, 3-trichlorotrifluoroacetone: 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_3Cl_3F_3O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 3 Cl (chlorine) | 3 F (fluorine) | 3 O (oxygen) | 1 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) | 3 | 12.011 Cl (chlorine) | 3 | 35.45 F (fluorine) | 3 | 18.998403163 O (oxygen) | 1 | 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) | 3 | 12.011 | 3 × 12.011 = 36.033 Cl (chlorine) | 3 | 35.45 | 3 × 35.45 = 106.35 F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 36.033 g/mol + 106.35 g/mol + 56.995209489 g/mol + 15.999 g/mol = 215.38 g/mol