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molar mass of samarium(III) bromate nonahydrate

Input interpretation

samarium(III) bromate nonahydrate | molar mass
samarium(III) bromate nonahydrate | molar mass

Result

Find the molar mass, M, for samarium(III) bromate nonahydrate: 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: Sm(BrO_3)_3·9H_2O Use the chemical formula, Sm(BrO_3)_3·9H_2O, to count the number of atoms, N_i, for each element:  | N_i  Br (bromine) | 3  H (hydrogen) | 18  O (oxygen) | 18  Sm (samarium) | 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) | 3 | 79.904  H (hydrogen) | 18 | 1.008  O (oxygen) | 18 | 15.999  Sm (samarium) | 1 | 150.36 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) | 3 | 79.904 | 3 × 79.904 = 239.712  H (hydrogen) | 18 | 1.008 | 18 × 1.008 = 18.144  O (oxygen) | 18 | 15.999 | 18 × 15.999 = 287.982  Sm (samarium) | 1 | 150.36 | 1 × 150.36 = 150.36  M = 239.712 g/mol + 18.144 g/mol + 287.982 g/mol + 150.36 g/mol = 696.20 g/mol
Find the molar mass, M, for samarium(III) bromate nonahydrate: 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: Sm(BrO_3)_3·9H_2O Use the chemical formula, Sm(BrO_3)_3·9H_2O, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 3 H (hydrogen) | 18 O (oxygen) | 18 Sm (samarium) | 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) | 3 | 79.904 H (hydrogen) | 18 | 1.008 O (oxygen) | 18 | 15.999 Sm (samarium) | 1 | 150.36 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) | 3 | 79.904 | 3 × 79.904 = 239.712 H (hydrogen) | 18 | 1.008 | 18 × 1.008 = 18.144 O (oxygen) | 18 | 15.999 | 18 × 15.999 = 287.982 Sm (samarium) | 1 | 150.36 | 1 × 150.36 = 150.36 M = 239.712 g/mol + 18.144 g/mol + 287.982 g/mol + 150.36 g/mol = 696.20 g/mol

Unit conversion

0.6962 kg/mol (kilograms per mole)
0.6962 kg/mol (kilograms per mole)

Comparisons

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≈ 0.97 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 3.6 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 3.6 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 12 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 12 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 1.2×10^-21 grams  | 1.2×10^-24 kg (kilograms)  | 696 u (unified atomic mass units)  | 696 Da (daltons)
Mass of a molecule m from m = M/N_A: | 1.2×10^-21 grams | 1.2×10^-24 kg (kilograms) | 696 u (unified atomic mass units) | 696 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 696
Relative molecular mass M_r from M_r = M_u/M: | 696