Answer:
Empirical formula = C5H4
Molecular formula = C10H8
Explanation:
When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:
CxHy + O2 = CO2 + H2O
We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:
CO2 = 12,011 + 2 (15,999) = 44,009 g
Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C
Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C
Mass H = 3 mg - 2.8105 mg = 0.1895 mg H
Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H
To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:
C = 0.2340 mmol C / 0.1895 mol H = 1.25
H = 0.1895 mmol H / 0.1895 mmol H = 1
We multiply the coefficients by 4, and we have the empirical formula:
C1.25 * 4H1 * 4 = C5H4
The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:
Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g
m = 130 g / 64.09 g = 2.02 = 2
Therefore we have the molecular formula:
(C5H4)2 = C10H8
