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Complete combustion (reaction with O₂) of a 0.8078-gram sample of an unknown that contains carbon, hydrogen, and oxygen yields 2.3036 grams of CO₂ and 0.5388 grams of H₂O. If the molar mass of this unknown compound is 216.2 g/mol, what is it's molecular formula?​

User Verma
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Final answer:

The combustion analysis involves determining the amounts of carbon and hydrogen in an unknown compound using the measured masses of CO₂ and H₂O produced, allowing us to calculate the molecular formula when combined with the molar mass of the compound.

Step-by-step explanation:

The question involves using combustion analysis to determine the molecular formula of an unknown compound containing carbon, hydrogen, and oxygen. The molar mass of the compounds involved in combustion such as CO₂ and H₂O, as well as the molar mass of the unknown compound, are used in calculations.

Firstly, from the mass of CO₂ produced, we can calculate the moles of CO₂, and consequently, the moles and mass of carbon in the original compound. We use the molar mass of CO₂ (44.01 g/mol) to find that:

2.3036 g CO₂ x (1 mol / 44.01 g) = 0.05234 mol of C

0.05234 mol C x (12.01 g/mol) = 0.6286 g of carbon

Then, we calculate the moles and mass of hydrogen using the mass of H₂O:

0.5388 g H₂O x (1 mol / 18.016 g) = 0.02991 mol of H₂O

0.02991 mol H₂O x (2 mol H / 1 mol H₂O) x (1.008 g/mol) = 0.0602 g of hydrogen

To find the mass of oxygen in the compound, we subtract the masses of carbon and hydrogen from the total mass of the compound:

0.8078 g sample - (0.6286 g C + 0.0602 g H) = 0.1190 g of oxygen

Knowing the masses of each element, we can now calculate the moles of each element in the unknown compound and therefore its empirical formula. From the empirical formula and the given molar mass of the unknown compound, we can then determine the molecular formula.

It's important to note that additional calculations are needed to obtain the final molecular formula, which includes finding the empirical formula first through the mole ratios of the elements and then using the molar mass of the unknown compound to find the actual molecular formula. Details of these calculations are not provided in this response.

User Betaorbust
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