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molar mass of copper(II) tetrafluoroborate hydrate

Input interpretation

copper(II) tetrafluoroborate hydrate | molar mass
copper(II) tetrafluoroborate hydrate | molar mass

Result

Find the molar mass, M, for copper(II) tetrafluoroborate 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: Cu(BF_4)_2·xH_2O Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  B (boron) | 2  Cu (copper) | 1  F (fluorine) | 8  H (hydrogen) | 2  O (oxygen) | 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)  B (boron) | 2 | 10.81  Cu (copper) | 1 | 63.546  F (fluorine) | 8 | 18.998403163  H (hydrogen) | 2 | 1.008  O (oxygen) | 1 | 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)  B (boron) | 2 | 10.81 | 2 × 10.81 = 21.62  Cu (copper) | 1 | 63.546 | 1 × 63.546 = 63.546  F (fluorine) | 8 | 18.998403163 | 8 × 18.998403163 = 151.987225304  H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016  O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999  M = 21.62 g/mol + 63.546 g/mol + 151.987225304 g/mol + 2.016 g/mol + 15.999 g/mol = 255.17 g/mol
Find the molar mass, M, for copper(II) tetrafluoroborate 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: Cu(BF_4)_2·xH_2O Use the chemical formula to count the number of atoms, N_i, for each element: | N_i B (boron) | 2 Cu (copper) | 1 F (fluorine) | 8 H (hydrogen) | 2 O (oxygen) | 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) B (boron) | 2 | 10.81 Cu (copper) | 1 | 63.546 F (fluorine) | 8 | 18.998403163 H (hydrogen) | 2 | 1.008 O (oxygen) | 1 | 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) B (boron) | 2 | 10.81 | 2 × 10.81 = 21.62 Cu (copper) | 1 | 63.546 | 1 × 63.546 = 63.546 F (fluorine) | 8 | 18.998403163 | 8 × 18.998403163 = 151.987225304 H (hydrogen) | 2 | 1.008 | 2 × 1.008 = 2.016 O (oxygen) | 1 | 15.999 | 1 × 15.999 = 15.999 M = 21.62 g/mol + 63.546 g/mol + 151.987225304 g/mol + 2.016 g/mol + 15.999 g/mol = 255.17 g/mol

Unit conversion

0.25517 kg/mol (kilograms per mole)
0.25517 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