Question 4.17: Obtaining an Empirical Formula from Combustion Analysis Upo...

1. Write down as given the masses of each combustion product and the mass of the sample (if given).

GIVEN:  1.83 \ g \ CO_2, 0.901 \ g \ H_2O
FIND: empirical formula

2. Convert the masses of CO_2 andH_2Ofrom Step 1 to moles by using the appropriate molar mass for each compound as a conversion factor.

1.83 \ \cancel{g \ CO_2} \times \frac{1 \ mol \ CO_2}{44.01 \ \cancel{g \ CO_2}}
= 0.0416 \ mol \ CO_2

0.901 \ \cancel{g \ H_2O} \times \frac{1 \ mol \ H_2O}{18.02 \ \cancel{g \ H_2O}}
= 0.0500 \ mol \ H_2O

3. Convert the moles of CO_2 and moles of H_2O from Step 2 to moles of C and moles of H using the conversion factors inherent in the chemical formulas of CO_2 and H_2O

0.0416 \ \cancel{mol \ CO_2} \times \frac{1 \ mol \ C}{1 \ \cancel{\ mol \ CO_2}}
= 0.0416 \ mol \ C

0.0500 \ \cancel{mol \ H_2O} \times \frac{2 \ mol \ H}{\cancel{1 \ mol \ H_2O}}
= 0.100 \ mol \ H

4. If the compound contains an element other than C and H, find the mass of the other element by subtracting the sum of the masses of C and H (obtained in Step 3) from the mass of the sample. Finally, convert the mass of the other element to moles.

The sample contains no elements other than C and H, so proceed to next step.

5. Write down a pseudoformula for the compound using the number of moles of each element (from Steps 3 and 4) as subscripts.

C_{0.0416}H_{0.100}

6. Divide all the subscripts in the formula by the smallest subscript. (Round all subscripts that are within 0.1 of a whole number.)

C\frac{0.0416}{0.0416} H\frac{0.100}{0.0416} \longrightarrow C_1H_{2.4}