Input interpretation
![3, 5-bis(trifluoromethyl)benzenamine | molar mass](../image_source/495bfcc4ce819b3435d3b9029ed7e50c.png)
3, 5-bis(trifluoromethyl)benzenamine | molar mass
Result
![Find the molar mass, M, for 3, 5-bis(trifluoromethyl)benzenamine: 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: (CF_3)_2C_6H_3NH_2 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 8 F (fluorine) | 6 H (hydrogen) | 5 N (nitrogen) | 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) | 8 | 12.011 F (fluorine) | 6 | 18.998403163 H (hydrogen) | 5 | 1.008 N (nitrogen) | 1 | 14.007 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) | 8 | 12.011 | 8 × 12.011 = 96.088 F (fluorine) | 6 | 18.998403163 | 6 × 18.998403163 = 113.990418978 H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 M = 96.088 g/mol + 113.990418978 g/mol + 5.040 g/mol + 14.007 g/mol = 229.125 g/mol](../image_source/44b425f468b672272f63f0de4e90bc54.png)
Find the molar mass, M, for 3, 5-bis(trifluoromethyl)benzenamine: 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: (CF_3)_2C_6H_3NH_2 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 8 F (fluorine) | 6 H (hydrogen) | 5 N (nitrogen) | 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) | 8 | 12.011 F (fluorine) | 6 | 18.998403163 H (hydrogen) | 5 | 1.008 N (nitrogen) | 1 | 14.007 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) | 8 | 12.011 | 8 × 12.011 = 96.088 F (fluorine) | 6 | 18.998403163 | 6 × 18.998403163 = 113.990418978 H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 M = 96.088 g/mol + 113.990418978 g/mol + 5.040 g/mol + 14.007 g/mol = 229.125 g/mol
Unit conversion
![0.22913 kg/mol (kilograms per mole)](../image_source/432d403b19cacfcea1ff08c9cda6bf74.png)
0.22913 kg/mol (kilograms per mole)
Comparisons
![≈ 0.32 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/01254248a6b9ac9b002e48ea6d27c8fa.png)
≈ 0.32 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ 1.2 × molar mass of caffeine ( ≈ 194 g/mol )](../image_source/66ad59191f9291e2a9502f1020864c99.png)
≈ 1.2 × molar mass of caffeine ( ≈ 194 g/mol )
![≈ 3.9 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/88322d3a150e5a3f576243d8b42bccbf.png)
≈ 3.9 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 3.8×10^-22 grams | 3.8×10^-25 kg (kilograms) | 229 u (unified atomic mass units) | 229 Da (daltons)](../image_source/7f4593f16fbfb45a43e8831657451f30.png)
Mass of a molecule m from m = M/N_A: | 3.8×10^-22 grams | 3.8×10^-25 kg (kilograms) | 229 u (unified atomic mass units) | 229 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 229](../image_source/d506f46ebfe8c52c160a98de42dfb40f.png)
Relative molecular mass M_r from M_r = M_u/M: | 229