Combustion analysis of a hydrocarbon produced 33.01 g g CO2 C O 2 and 10.15 g g H2O H 2 O . Part A Calculate the empirical formula of the hydrocarbon.

The empirical formula is C2H3

Explanation:

Step 1: Data given

Mass of CO2 = 33.01 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 10.15 grams

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 33.01 grams / 44.01 g/mol

Moles CO2 = 0.7501 moles

Step 3: Calculate moles C

In 1 mol CO2 we have 1 mol C

In 0.7501 moles CO2 we have 0.7501 moles C

Step 4: Calculate moles H2O

Moles H2O = 10.15 grams / 18.02 g/mol

Moles H2O = 0.5633 moles

Step 5 Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.5633 moles H2O we have 2*0.5633 moles = 1.1266 moles

Step 6: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.7501 moles / 0.7501 moles = 1

H: 1.1266 moles / 0.7501 moles = 1.5

For each C atom we have 1.5 H atoms   OR for each 2 C atoms we have 3 H atoms

The empirical formula is C2H3

im pretty sure its b