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molar mass of 2-(trifluoromethoxy)iodobenzene

Input interpretation

2-(trifluoromethoxy)iodobenzene | molar mass
2-(trifluoromethoxy)iodobenzene | molar mass

Result

Find the molar mass, M, for 2-(trifluoromethoxy)iodobenzene: 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_6H_4IOCF_3 Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  C (carbon) | 7  F (fluorine) | 3  H (hydrogen) | 4  I (iodine) | 1  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) | 7 | 12.011  F (fluorine) | 3 | 18.998403163  H (hydrogen) | 4 | 1.008  I (iodine) | 1 | 126.90447  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) | 7 | 12.011 | 7 × 12.011 = 84.077  F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489  H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032  I (iodine) | 1 | 126.90447 | 1 × 126.90447 = 126.90447  O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999  M = 84.077 g/mol + 56.995209489 g/mol + 4.032 g/mol + 126.90447 g/mol + 15.999 g/mol = 288.008 g/mol
Find the molar mass, M, for 2-(trifluoromethoxy)iodobenzene: 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_6H_4IOCF_3 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 7 F (fluorine) | 3 H (hydrogen) | 4 I (iodine) | 1 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) | 7 | 12.011 F (fluorine) | 3 | 18.998403163 H (hydrogen) | 4 | 1.008 I (iodine) | 1 | 126.90447 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) | 7 | 12.011 | 7 × 12.011 = 84.077 F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032 I (iodine) | 1 | 126.90447 | 1 × 126.90447 = 126.90447 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 84.077 g/mol + 56.995209489 g/mol + 4.032 g/mol + 126.90447 g/mol + 15.999 g/mol = 288.008 g/mol

Unit conversion

0.28801 kg/mol (kilograms per mole)
0.28801 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.4 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 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 )
 ≈ 4.9 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 4.9 × 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)  | 288 u (unified atomic mass units)  | 288 Da (daltons)
Mass of a molecule m from m = M/N_A: | 4.8×10^-22 grams | 4.8×10^-25 kg (kilograms) | 288 u (unified atomic mass units) | 288 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 288
Relative molecular mass M_r from M_r = M_u/M: | 288