. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar mass of nicotine to the nearest gram

Nicotine has the formula [tex]C_xH_yN_z[/tex] . To determine its composition, a sample is burned in excess oxygen, producing the following results:

1.0 mol of CO₂

0.70 mol of H₂O

0.20 mol of NO₂

Assume that all the atoms in nicotine are present as products

Solution

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

1.0 mol of CO₂ has 1.0 mol of C

0.70 mol of H₂O has 1.4 mol of H

0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

C: 1.0

H: 1.4

N: 0.20

Divide all by the smallest number: 0.20

C: 1.0 / 0.20 = 5

H: 1.4 / 0.20 = 7

N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

C: 5mol × 12g/mol = 60g

H: 7mol × 1g/mol = 7 g

N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.