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molar mass of (Z)-4,4,4-trifluoro-3-methyl-2-butenoic acid

Input interpretation

(Z)-4, 4, 4-trifluoro-3-methyl-2-butenoic acid | molar mass
(Z)-4, 4, 4-trifluoro-3-methyl-2-butenoic acid | molar mass

Result

Find the molar mass, M, for (Z)-4, 4, 4-trifluoro-3-methyl-2-butenoic acid: 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: CF_3C(CH_3)=CHCO_2H Use the chemical formula to count the number of atoms, N_i, for each element:  | N_i  C (carbon) | 5  F (fluorine) | 3  H (hydrogen) | 5  O (oxygen) | 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) | 5 | 12.011  F (fluorine) | 3 | 18.998403163  H (hydrogen) | 5 | 1.008  O (oxygen) | 2 | 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)  C (carbon) | 5 | 12.011 | 5 × 12.011 = 60.055  F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489  H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040  O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998  M = 60.055 g/mol + 56.995209489 g/mol + 5.040 g/mol + 31.998 g/mol = 154.088 g/mol
Find the molar mass, M, for (Z)-4, 4, 4-trifluoro-3-methyl-2-butenoic acid: 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: CF_3C(CH_3)=CHCO_2H Use the chemical formula to count the number of atoms, N_i, for each element: | N_i C (carbon) | 5 F (fluorine) | 3 H (hydrogen) | 5 O (oxygen) | 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) | 5 | 12.011 F (fluorine) | 3 | 18.998403163 H (hydrogen) | 5 | 1.008 O (oxygen) | 2 | 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) C (carbon) | 5 | 12.011 | 5 × 12.011 = 60.055 F (fluorine) | 3 | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 5 | 1.008 | 5 × 1.008 = 5.040 O (oxygen) | 2 | 15.999 | 2 × 15.999 = 31.998 M = 60.055 g/mol + 56.995209489 g/mol + 5.040 g/mol + 31.998 g/mol = 154.088 g/mol

Unit conversion

0.15409 kg/mol (kilograms per mole)
0.15409 kg/mol (kilograms per mole)

Comparisons

 ≈ ( 0.21 ≈ 1/5 ) × molar mass of fullerene ( ≈ 721 g/mol )
≈ ( 0.21 ≈ 1/5 ) × molar mass of fullerene ( ≈ 721 g/mol )
 ≈ 0.79 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 0.79 × molar mass of caffeine ( ≈ 194 g/mol )
 ≈ 2.6 × molar mass of sodium chloride ( ≈ 58 g/mol )
≈ 2.6 × molar mass of sodium chloride ( ≈ 58 g/mol )

Corresponding quantities

Mass of a molecule m from m = M/N_A:  | 2.6×10^-22 grams  | 2.6×10^-25 kg (kilograms)  | 154 u (unified atomic mass units)  | 154 Da (daltons)
Mass of a molecule m from m = M/N_A: | 2.6×10^-22 grams | 2.6×10^-25 kg (kilograms) | 154 u (unified atomic mass units) | 154 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M:  | 154
Relative molecular mass M_r from M_r = M_u/M: | 154