combustion chemistry question

Question

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine produces 18.13 g of CO₂, 4.639 g of H₂O, and 2.885 g of N₂. Determine the molar mass of the compound if it is between 150 and 210 g/mol.

Answer 1

8.13 g of CO₂

----------------- = 0.412 mol CO₂ = 0.412 mol C × 12 g/mol = 4.945 g C

44 g/mol

4.639 g of H₂O

----------------- = 0.2577 mol H₂O = 0.5154 mol H × 1 g/mol = 0.5154 g H

18 g/mol

2.885 g of N₂

----------------- = 0.1030 mol N₂ = 0.2061 mol N × 14 g/mol = 2.885 g N

28 g/mol

10 g of caffeine – (4.945 g C + 0.5154 g H + 2.885 g H) = 1.6545 g O

.6545 g of O

----------------- = 0.1034 mol O

16 g/mol

0.1034 mol O is the smallest molar quantity, so O is the atom that occurs the fewest times in the formula

We get the other numbers in our empirical formula by dividing the molar quantities by 0.1034

(0.412 mol C)/(0.1034 mol O) = C : O = 4 : 1

(0.5154 mol H)/(0.1034 mol O) = H : O = 5 : 1

(0.2061 mol N)/(0.1034 mol O) = N : O = 2 : 1

So our empirical formula is C₄H₅N₂O

If this were our molecular formula, the molar mass would be 4×12 + 5×1 + 2×14 + 1×16 = 97

If the molar mass of caffeine is between 150 and 210 g/mol, this is consistent with a molecular formula where all of the numbers in the empirical formula are doubled: C₈H₁₀N₄O₂ (molar mass = 2×97 = 194 g/mol.)

Answer

The molar mass of caffeine is 194 g/mol

Note

The odd number of hydrogens and even number of nitrogens in the empirical formula C₄H₅N₂O also indicate that it cannot be the molecular formula.