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molar mass of 3,6-dibromo-2-fluorobenzaldehyde

Input interpretation

3, 6-dibromo-2-fluorobenzaldehyde | molar mass
3, 6-dibromo-2-fluorobenzaldehyde | molar mass

Result

Find the molar mass, M, for 3, 6-dibromo-2-fluorobenzaldehyde: 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_7H_3Br_2FO Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  Br (bromine) | 2  C (carbon) | 7  F (fluorine) | 1  H (hydrogen) | 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)  Br (bromine) | 2 | 79.904  C (carbon) | 7 | 12.011  F (fluorine) | 1 | 18.998403163  H (hydrogen) | 3 | 1.008  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)  Br (bromine) | 2 | 79.904 | 2 × 79.904 = 159.808  C (carbon) | 7 | 12.011 | 7 × 12.011 = 84.077  F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163  H (hydrogen) | 3 | 1.008 | 3 × 1.008 = 3.024  O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999  M = 159.808 g/mol + 84.077 g/mol + 18.998403163 g/mol + 3.024 g/mol + 15.999 g/mol = 281.906 g/mol
Find the molar mass, M, for 3, 6-dibromo-2-fluorobenzaldehyde: 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_7H_3Br_2FO Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 2 C (carbon) | 7 F (fluorine) | 1 H (hydrogen) | 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) Br (bromine) | 2 | 79.904 C (carbon) | 7 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 3 | 1.008 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) Br (bromine) | 2 | 79.904 | 2 × 79.904 = 159.808 C (carbon) | 7 | 12.011 | 7 × 12.011 = 84.077 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 3 | 1.008 | 3 × 1.008 = 3.024 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 159.808 g/mol + 84.077 g/mol + 18.998403163 g/mol + 3.024 g/mol + 15.999 g/mol = 281.906 g/mol

Unit conversion

0.28191 kg/mol (kilograms per mole)
0.28191 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.39 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.39 × 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 )
 ≈ 4.8 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 4.8 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 4.7×10^-22 grams  | 4.7×10^-25 kg (kilograms)  | 282 u (unified atomic mass units)  | 282 Da (daltons)
Mass of a molecule m from m = M/N_A: | 4.7×10^-22 grams | 4.7×10^-25 kg (kilograms) | 282 u (unified atomic mass units) | 282 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 282
Relative molecular mass M_r from M_r = M_u/M: | 282