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molar mass of iron(II) fluoride tetrahydrate

Input interpretation

iron(II) fluoride tetrahydrate | molar mass
iron(II) fluoride tetrahydrate | molar mass

Result

Find the molar mass, M, for iron(II) fluoride tetrahydrate: 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: FeF_2·4H_2O Use the chemical formula, FeF_2·4H_2O, to count the number of atoms, N_i, for each element:  | N_i  F (fluorine) | 2  Fe (iron) | 1  H (hydrogen) | 2  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)  F (fluorine) | 2 | 18.998403163  Fe (iron) | 1 | 55.845  H (hydrogen) | 2 | 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)  F (fluorine) | 2 | 18.998403163 | 2 × 18.998403163 = 37.996806326  Fe (iron) | 1 | 55.845 | 1 × 55.845 = 55.845  H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016  O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999  M = 37.996806326 g/mol + 55.845 g/mol + 2.016 g/mol + 15.999 g/mol = 111.857 g/mol
Find the molar mass, M, for iron(II) fluoride tetrahydrate: 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: FeF_2·4H_2O Use the chemical formula, FeF_2·4H_2O, to count the number of atoms, N_i, for each element: | N_i F (fluorine) | 2 Fe (iron) | 1 H (hydrogen) | 2 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) F (fluorine) | 2 | 18.998403163 Fe (iron) | 1 | 55.845 H (hydrogen) | 2 | 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) F (fluorine) | 2 | 18.998403163 | 2 × 18.998403163 = 37.996806326 Fe (iron) | 1 | 55.845 | 1 × 55.845 = 55.845 H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 37.996806326 g/mol + 55.845 g/mol + 2.016 g/mol + 15.999 g/mol = 111.857 g/mol

Unit conversion

0.11186 kg/mol (kilograms per mole)
0.11186 kg/mol (kilograms per mole)

Comparisons

 ≈ ( 0.16 ≈ 1/6 ) × molar mass of fullerene ( ≈ 721 g/mol )
≈ ( 0.16 ≈ 1/6 ) × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 0.58 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 0.58 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 1.9 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 1.9 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 1.9×10^-22 grams  | 1.9×10^-25 kg (kilograms)  | 112 u (unified atomic mass units)  | 112 Da (daltons)
Mass of a molecule m from m = M/N_A: | 1.9×10^-22 grams | 1.9×10^-25 kg (kilograms) | 112 u (unified atomic mass units) | 112 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 112
Relative molecular mass M_r from M_r = M_u/M: | 112