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

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.