Input interpretation
![iron(II) tartrate | molar mass](../image_source/3874813701fabb6d4cf87890c8d874e6.png)
iron(II) tartrate | molar mass
Result
![Find the molar mass, M, for iron(II) tartrate: 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: FeC_4H_4O_6 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 4 Fe (iron) | 1 H (hydrogen) | 4 O (oxygen) | 6 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) | 4 | 12.011 Fe (iron) | 1 | 55.845 H (hydrogen) | 4 | 1.008 O (oxygen) | 6 | 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) | 4 | 12.011 | 4 × 12.011 = 48.044 Fe (iron) | 1 | 55.845 | 1 × 55.845 = 55.845 H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032 O (oxygen) | 6 | 15.999 | 6 × 15.999 = 95.994 M = 48.044 g/mol + 55.845 g/mol + 4.032 g/mol + 95.994 g/mol = 203.915 g/mol](../image_source/92dc4637a689d34a790aee9c6093117d.png)
Find the molar mass, M, for iron(II) tartrate: 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: FeC_4H_4O_6 Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 4 Fe (iron) | 1 H (hydrogen) | 4 O (oxygen) | 6 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) | 4 | 12.011 Fe (iron) | 1 | 55.845 H (hydrogen) | 4 | 1.008 O (oxygen) | 6 | 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) | 4 | 12.011 | 4 × 12.011 = 48.044 Fe (iron) | 1 | 55.845 | 1 × 55.845 = 55.845 H (hydrogen) | 4 | 1.008 | 4 × 1.008 = 4.032 O (oxygen) | 6 | 15.999 | 6 × 15.999 = 95.994 M = 48.044 g/mol + 55.845 g/mol + 4.032 g/mol + 95.994 g/mol = 203.915 g/mol
Unit conversion
![0.20392 kg/mol (kilograms per mole)](../image_source/b7cf9a0bfc27f65842c6c14908625709.png)
0.20392 kg/mol (kilograms per mole)
Comparisons
![≈ 0.28 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/5317631b8725a36341132e2aac9a3a8d.png)
≈ 0.28 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ 1.1 × molar mass of caffeine ( ≈ 194 g/mol )](../image_source/750a8f37564cf4c2c72be65821ae92e6.png)
≈ 1.1 × molar mass of caffeine ( ≈ 194 g/mol )
![≈ 3.5 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/5a49613adcb2d83e9aa3585b9b6a99bf.png)
≈ 3.5 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 3.4×10^-22 grams | 3.4×10^-25 kg (kilograms) | 204 u (unified atomic mass units) | 204 Da (daltons)](../image_source/67cff4e61f4a6039580cdc6972fa0744.png)
Mass of a molecule m from m = M/N_A: | 3.4×10^-22 grams | 3.4×10^-25 kg (kilograms) | 204 u (unified atomic mass units) | 204 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 204](../image_source/4f0591df58c8e02191088f5678580b58.png)
Relative molecular mass M_r from M_r = M_u/M: | 204