Input interpretation
![3-fluorophenylmagnesium bromide | molar mass](../image_source/a64e2e5c9c5ccd83517c209f0a6be9ef.png)
3-fluorophenylmagnesium bromide | molar mass
Result
![Find the molar mass, M, for 3-fluorophenylmagnesium bromide: 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: FC_6H_4MgBr Use the chemical formula, FC_6H_4MgBr, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 6 F (fluorine) | 1 H (hydrogen) | 4 Mg (magnesium) | 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) | 4 | 1.008 Mg (magnesium) | 1 | 24.305 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) | 4 | 1.008 | 4 × 1.008 = 4.032 Mg (magnesium) | 1 | 24.305 | 1 × 24.305 = 24.305 M = 79.904 g/mol + 72.066 g/mol + 18.998403163 g/mol + 4.032 g/mol + 24.305 g/mol = 199.305 g/mol](../image_source/0431e616a294669bf8c4060de6d42f95.png)
Find the molar mass, M, for 3-fluorophenylmagnesium bromide: 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: FC_6H_4MgBr Use the chemical formula, FC_6H_4MgBr, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 6 F (fluorine) | 1 H (hydrogen) | 4 Mg (magnesium) | 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) | 4 | 1.008 Mg (magnesium) | 1 | 24.305 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) | 4 | 1.008 | 4 × 1.008 = 4.032 Mg (magnesium) | 1 | 24.305 | 1 × 24.305 = 24.305 M = 79.904 g/mol + 72.066 g/mol + 18.998403163 g/mol + 4.032 g/mol + 24.305 g/mol = 199.305 g/mol
Unit conversion
![0.19931 kg/mol (kilograms per mole)](../image_source/8669493da0faec6bd651c5e81c1aa416.png)
0.19931 kg/mol (kilograms per mole)
Comparisons
![≈ 0.28 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/8fd24927fbe89daeb7afa195ffeaf9cb.png)
≈ 0.28 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ molar mass of caffeine ( ≈ 194 g/mol )](../image_source/7ac0f5014ebda055b433db770e040ff5.png)
≈ molar mass of caffeine ( ≈ 194 g/mol )
![≈ 3.4 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/78cc02b177847e5a84f8649b971317f4.png)
≈ 3.4 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 3.3×10^-22 grams | 3.3×10^-25 kg (kilograms) | 199 u (unified atomic mass units) | 199 Da (daltons)](../image_source/3edb815fa3fef7096eb678f93491b8a9.png)
Mass of a molecule m from m = M/N_A: | 3.3×10^-22 grams | 3.3×10^-25 kg (kilograms) | 199 u (unified atomic mass units) | 199 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 199](../image_source/316ec2e157e6aa96eaf73a0172e45aea.png)
Relative molecular mass M_r from M_r = M_u/M: | 199