Search

molar mass of barium tetracyanoplatinate

Input interpretation

barium tetracyanoplatinate | molar mass
barium tetracyanoplatinate | molar mass

Result

Find the molar mass, M, for barium tetracyanoplatinate: 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: C_4BaN_4Pt Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  Ba (barium) | 1  C (carbon) | 4  N (nitrogen) | 4  Pt (platinum) | 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)  Ba (barium) | 1 | 137.327  C (carbon) | 4 | 12.011  N (nitrogen) | 4 | 14.007  Pt (platinum) | 1 | 195.084 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)  Ba (barium) | 1 | 137.327 | 1 × 137.327 = 137.327  C (carbon) | 4 | 12.011 | 4 × 12.011 = 48.044  N (nitrogen) | 4 | 14.007 | 4 × 14.007 = 56.028  Pt (platinum) | 1 | 195.084 | 1 × 195.084 = 195.084  M = 137.327 g/mol + 48.044 g/mol + 56.028 g/mol + 195.084 g/mol = 436.483 g/mol
Find the molar mass, M, for barium tetracyanoplatinate: 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: C_4BaN_4Pt Use the chemical formula to count the number of atoms, N_i, for each element: | N_i Ba (barium) | 1 C (carbon) | 4 N (nitrogen) | 4 Pt (platinum) | 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) Ba (barium) | 1 | 137.327 C (carbon) | 4 | 12.011 N (nitrogen) | 4 | 14.007 Pt (platinum) | 1 | 195.084 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) Ba (barium) | 1 | 137.327 | 1 × 137.327 = 137.327 C (carbon) | 4 | 12.011 | 4 × 12.011 = 48.044 N (nitrogen) | 4 | 14.007 | 4 × 14.007 = 56.028 Pt (platinum) | 1 | 195.084 | 1 × 195.084 = 195.084 M = 137.327 g/mol + 48.044 g/mol + 56.028 g/mol + 195.084 g/mol = 436.483 g/mol

Unit conversion

0.43648 kg/mol (kilograms per mole)
0.43648 kg/mol (kilograms per mole)

Comparisons

 ≈ 0.61 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 0.61 × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 2.2 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 2.2 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 7.5 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 7.5 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 7.2×10^-22 grams  | 7.2×10^-25 kg (kilograms)  | 436 u (unified atomic mass units)  | 436 Da (daltons)
Mass of a molecule m from m = M/N_A: | 7.2×10^-22 grams | 7.2×10^-25 kg (kilograms) | 436 u (unified atomic mass units) | 436 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 436
Relative molecular mass M_r from M_r = M_u/M: | 436