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molar mass of ruthenium(III) nitrosyl chloride hydrate

Input interpretation

ruthenium(III) nitrosyl chloride hydrate | molar mass
ruthenium(III) nitrosyl chloride hydrate | molar mass

Result

Find the molar mass, M, for ruthenium(III) nitrosyl chloride hydrate: 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: Ru(NO)Cl_3·xH_2O Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  Cl (chlorine) | 3  H (hydrogen) | 2  N (nitrogen) | 1  O (oxygen) | 2  Ru (ruthenium) | 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)  Cl (chlorine) | 3 | 35.45  H (hydrogen) | 2 | 1.008  N (nitrogen) | 1 | 14.007  O (oxygen) | 2 | 15.999  Ru (ruthenium) | 1 | 101.07 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)  Cl (chlorine) | 3 | 35.45 | 3 × 35.45 = 106.35  H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016  N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007  O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998  Ru (ruthenium) | 1 | 101.07 | 1 × 101.07 = 101.07  M = 106.35 g/mol + 2.016 g/mol + 14.007 g/mol + 31.998 g/mol + 101.07 g/mol = 255.44 g/mol
Find the molar mass, M, for ruthenium(III) nitrosyl chloride hydrate: 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: Ru(NO)Cl_3·xH_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Cl (chlorine) | 3 H (hydrogen) | 2 N (nitrogen) | 1 O (oxygen) | 2 Ru (ruthenium) | 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) Cl (chlorine) | 3 | 35.45 H (hydrogen) | 2 | 1.008 N (nitrogen) | 1 | 14.007 O (oxygen) | 2 | 15.999 Ru (ruthenium) | 1 | 101.07 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) Cl (chlorine) | 3 | 35.45 | 3 × 35.45 = 106.35 H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016 N (nitrogen) | 1 | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 Ru (ruthenium) | 1 | 101.07 | 1 × 101.07 = 101.07 M = 106.35 g/mol + 2.016 g/mol + 14.007 g/mol + 31.998 g/mol + 101.07 g/mol = 255.44 g/mol

Unit conversion

0.2554 kg/mol (kilograms per mole)
0.2554 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.35 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.35 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 1.3 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 1.3 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 4.4 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 4.4 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 4.2×10^-22 grams  | 4.2×10^-25 kg (kilograms)  | 255 u (unified atomic mass units)  | 255 Da (daltons)
Mass of a molecule m from m = M/N_A: | 4.2×10^-22 grams | 4.2×10^-25 kg (kilograms) | 255 u (unified atomic mass units) | 255 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 255
Relative molecular mass M_r from M_r = M_u/M: | 255