Input interpretation
![2, 4-diamino-6-(2-fluorophenyl)-1, 3, 5-triazine | molar mass](../image_source/e7d363406df1ff3c41da2119e492e8eb.png)
2, 4-diamino-6-(2-fluorophenyl)-1, 3, 5-triazine | molar mass
Result
![Find the molar mass, M, for 2, 4-diamino-6-(2-fluorophenyl)-1, 3, 5-triazine: 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_9H_8FN_5 Use the chemical formula, C_9H_8FN_5, to count the number of atoms, N_i, for each element: | N_i C (carbon) | 9 F (fluorine) | 1 H (hydrogen) | 8 N (nitrogen) | 5 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) | 9 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 8 | 1.008 N (nitrogen) | 5 | 14.007 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) | 9 | 12.011 | 9 × 12.011 = 108.099 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 8 | 1.008 | 8 × 1.008 = 8.064 N (nitrogen) | 5 | 14.007 | 5 × 14.007 = 70.035 M = 108.099 g/mol + 18.998403163 g/mol + 8.064 g/mol + 70.035 g/mol = 205.196 g/mol](../image_source/52292816a2d90b3da9d628e8cf531835.png)
Find the molar mass, M, for 2, 4-diamino-6-(2-fluorophenyl)-1, 3, 5-triazine: 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_9H_8FN_5 Use the chemical formula, C_9H_8FN_5, to count the number of atoms, N_i, for each element: | N_i C (carbon) | 9 F (fluorine) | 1 H (hydrogen) | 8 N (nitrogen) | 5 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) | 9 | 12.011 F (fluorine) | 1 | 18.998403163 H (hydrogen) | 8 | 1.008 N (nitrogen) | 5 | 14.007 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) | 9 | 12.011 | 9 × 12.011 = 108.099 F (fluorine) | 1 | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 8 | 1.008 | 8 × 1.008 = 8.064 N (nitrogen) | 5 | 14.007 | 5 × 14.007 = 70.035 M = 108.099 g/mol + 18.998403163 g/mol + 8.064 g/mol + 70.035 g/mol = 205.196 g/mol
Unit conversion
![0.2052 kg/mol (kilograms per mole)](../image_source/6e265d343f23ba3b2423d061317a4e1a.png)
0.2052 kg/mol (kilograms per mole)
Comparisons
![≈ 0.28 × molar mass of fullerene ( ≈ 721 g/mol )](../image_source/39258102d7ce79b1bd62b72fafc36687.png)
≈ 0.28 × molar mass of fullerene ( ≈ 721 g/mol )
![≈ 1.1 × molar mass of caffeine ( ≈ 194 g/mol )](../image_source/6a73928dbeb36235cc710d75cf0868e0.png)
≈ 1.1 × molar mass of caffeine ( ≈ 194 g/mol )
![≈ 3.5 × molar mass of sodium chloride ( ≈ 58 g/mol )](../image_source/8da176d3f0179edabc942dffeadaa3ad.png)
≈ 3.5 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
![Mass of a molecule m from m = M/N_A: | 3.4×10^-22 grams | 3.4×10^-25 kg (kilograms) | 205 u (unified atomic mass units) | 205 Da (daltons)](../image_source/5d1c1ae5204efe250a2e030177208ec6.png)
Mass of a molecule m from m = M/N_A: | 3.4×10^-22 grams | 3.4×10^-25 kg (kilograms) | 205 u (unified atomic mass units) | 205 Da (daltons)
![Relative molecular mass M_r from M_r = M_u/M: | 205](../image_source/110fc3c75e60a68f2401001266b539af.png)
Relative molecular mass M_r from M_r = M_u/M: | 205