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molar mass of 4-bromo-3-fluorophenylzinc iodide

Input interpretation

4-bromo-3-fluorophenylzinc iodide | molar mass
4-bromo-3-fluorophenylzinc iodide | molar mass

Result

Find the molar mass, M, for 4-bromo-3-fluorophenylzinc iodide: 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: BrC_6H_3(F)ZnI Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  Br (bromine) | 1  C (carbon) | 6  F (fluorine) | 1  H (hydrogen) | 3  I (iodine) | 1  Zn (zinc) | 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)  Br (bromine) | 1 | 79.904  C (carbon) | 6 | 12.011  F (fluorine) | 1 | 18.998403163  H (hydrogen) | 3 | 1.008  I (iodine) | 1 | 126.90447  Zn (zinc) | 1 | 65.38 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)  Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904  C (carbon) | 6 | 12.011 | 6 × 12.011 = 72.066  F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163  H (hydrogen) | 3 | 1.008 | 3 × 1.008 = 3.024  I (iodine) | 1 | 126.90447 | 1 × 126.90447 = 126.90447  Zn (zinc) | 1 | 65.38 | 1 × 65.38 = 65.38  M = 79.904 g/mol + 72.066 g/mol + 18.998403163 g/mol + 3.024 g/mol + 126.90447 g/mol + 65.38 g/mol = 366.28 g/mol
Find the molar mass, M, for 4-bromo-3-fluorophenylzinc iodide: 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: BrC_6H_3(F)ZnI Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 6 F (fluorine) | 1 H (hydrogen) | 3 I (iodine) | 1 Zn (zinc) | 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) Br (bromine) | 1 | 79.904 C (carbon) | 6 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 3 | 1.008 I (iodine) | 1 | 126.90447 Zn (zinc) | 1 | 65.38 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) Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 6 | 12.011 | 6 × 12.011 = 72.066 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 3 | 1.008 | 3 × 1.008 = 3.024 I (iodine) | 1 | 126.90447 | 1 × 126.90447 = 126.90447 Zn (zinc) | 1 | 65.38 | 1 × 65.38 = 65.38 M = 79.904 g/mol + 72.066 g/mol + 18.998403163 g/mol + 3.024 g/mol + 126.90447 g/mol + 65.38 g/mol = 366.28 g/mol

Unit conversion

0.36628 kg/mol (kilograms per mole)
0.36628 kg/mol (kilograms per mole)

Comparisons

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 ≈ 1.9 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 1.9 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 6.3 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 6.3 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 6.1×10^-22 grams  | 6.1×10^-25 kg (kilograms)  | 366 u (unified atomic mass units)  | 366 Da (daltons)
Mass of a molecule m from m = M/N_A: | 6.1×10^-22 grams | 6.1×10^-25 kg (kilograms) | 366 u (unified atomic mass units) | 366 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 366
Relative molecular mass M_r from M_r = M_u/M: | 366