Input interpretation
![praseodymium(III) carbonate octahydrate | molar mass](../image_source/5220e323bb6e13893fed20e437d7a420.png)
praseodymium(III) carbonate octahydrate | molar mass
Result
![Find the molar mass, M, for praseodymium(III) carbonate octahydrate: 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: Pr_2(CO_3)_3·8H_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 3 H (hydrogen) | 16 O (oxygen) | 17 Pr (praseodymium) | 2 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) | 3 | 12.011 H (hydrogen) | 16 | 1.008 O (oxygen) | 17 | 15.999 Pr (praseodymium) | 2 | 140.90766 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) | 3 | 12.011 | 3 × 12.011 = 36.033 H (hydrogen) | 16 | 1.008 | 16 × 1.008 = 16.128 O (oxygen) | 17 | 15.999 | 17 × 15.999 = 271.983 Pr (praseodymium) | 2 | 140.90766 | 2 × 140.90766 = 281.81532 M = 36.033 g/mol + 16.128 g/mol + 271.983 g/mol + 281.81532 g/mol = 605.959 g/mol](../image_source/cb5debe2e8bc1928bbba9ee7f9e0ae84.png)
Find the molar mass, M, for praseodymium(III) carbonate octahydrate: 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: Pr_2(CO_3)_3·8H_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 3 H (hydrogen) | 16 O (oxygen) | 17 Pr (praseodymium) | 2 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) | 3 | 12.011 H (hydrogen) | 16 | 1.008 O (oxygen) | 17 | 15.999 Pr (praseodymium) | 2 | 140.90766 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) | 3 | 12.011 | 3 × 12.011 = 36.033 H (hydrogen) | 16 | 1.008 | 16 × 1.008 = 16.128 O (oxygen) | 17 | 15.999 | 17 × 15.999 = 271.983 Pr (praseodymium) | 2 | 140.90766 | 2 × 140.90766 = 281.81532 M = 36.033 g/mol + 16.128 g/mol + 271.983 g/mol + 281.81532 g/mol = 605.959 g/mol
Unit conversion
![0.60596 kg/mol (kilograms per mole)](../image_source/9069d09ae820fea34234f9014b21f6ef.png)
0.60596 kg/mol (kilograms per mole)
Comparisons
![≈ 0.84 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/2953979da460391834b3b71d13d98e42.png)
≈ 0.84 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ 3.1 × molar mass of caffeine ( ≈ 194 g/mol )](../image_source/3359f6604d6b244b4ae2b5877e34fdd6.png)
≈ 3.1 × molar mass of caffeine ( ≈ 194 g/mol )
![≈ 10 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/13fbf8bcaf6da64febebfdbd97a1922f.png)
≈ 10 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 1×10^-21 grams | 1×10^-24 kg (kilograms) | 606 u (unified atomic mass units) | 606 Da (daltons)](../image_source/eda07cc38b7c8005b4b179722096486a.png)
Mass of a molecule m from m = M/N_A: | 1×10^-21 grams | 1×10^-24 kg (kilograms) | 606 u (unified atomic mass units) | 606 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 606](../image_source/0da7fe2f99ce4af5e66fcc1bffec5c0d.png)
Relative molecular mass M_r from M_r = M_u/M: | 606