Question

Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670 g of H2O. The molar mass is 532.7 g/mol. What is the molecular formula of the compound?

Answer 1

Solution:

Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670 g of H2O. The molar mass is 532.7 g/mol. What is the molecular formula of the compound?

carbon: 1.043 g x (12.011 g / 44.0098 g) = 0.2846 g

hydrogen: 0.5670 g x (2.0158 g / 18.0152 g) = 0.06344 g

1b) Determine the grams of oxygen in the sample by subtraction.

0.600 - (0.285 g + 0.06344) = 0.2519 g

Notice that the subtraction is the mass of the sample minus the sum of the carbon and hydrogen in the sample. Also, it is quite typical of these problems to specify that only C, H and O are involved.

Remember, sometimes the problem will give the CO₂ and H₂O values, but FAIL to say that C, H, and O are involved. Make sure you add the C and H values (or sometimes the C, H, and N values) and check against the mass of the sample. Any difference would be an amount of oxygen present .

2) Convert grams of C, H and O to their respective amount of moles.

carbon: 0.2846 g / 12.011 g / mol = 0.02369 mol

hydrogen: 0.06344 g / 1.0079 g/mol = 0.06294 mol

oxygen: 0.2519 g / 15.9994 g/mol = 0.01574 mol

3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers.

carbon: 0.02369 mol / 0.01574 mol = 1.505, I am taking this fig as 2

hydrogen: 0.06294 mol / 0.01572 mol = 4

oxygen: 0.01572 mol / 0.01572 mol = 1,

so mole ratio is 2:4:1,

Now 5) Multiply by three to get the whole-number ratio:

2 : 4 : 1

empirical formula = C₂H₄O

6) The weight of the empirical formula is 44:

532.7 / 44 = 12

the molecular formula is C₂₄H₄₈O₁₂