Combustion of 135.0 mg of a hydrocarbon produces 440.0 mg CO2 and 135.0 mg H2 O . The molar mass

Combustion of 135.0 mg of a hydrocarbon produces 440.0 mg...

Combustion of 135.0 mg of a hydrocarbon produces $440.0 \mathrm{mg} \mathrm{CO}_{2}$ and $135.0 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .$ The molar mass of the hydrocarbon is $270 \mathrm{g} / \mathrm{mol}$. What are the empirical and molecular formulas of this compound?

Key Concepts

Combustion Analysis

Combustion analysis is a method used to determine the elemental composition of a compound, particularly for organic compounds containing carbon and hydrogen. By burning the compound and measuring the amounts of CO2 and H2O produced, one can calculate the moles of carbon and hydrogen present in the original sample.

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of elements in a compound. It is determined by converting the masses of each element obtained from analysis (such as combustion analysis) into moles, then normalizing these amounts by dividing by the smallest mole value.

Molecular Formula Determination

The molecular formula provides the actual number of atoms of each element in a molecule. It is found by comparing the molar mass of the compound with the molar mass of its empirical formula and multiplying the subscripts in the empirical formula by the appropriate factor.

Stoichiometric Conversions

Stoichiometric conversions involve using molar masses to convert between mass and moles of substances. This process is essential in quantitative chemical analyses, such as combustion analysis, to accurately determine the composition of a compound from the measured masses of reaction products.

Transcript

00:01 Let's do some practice on empirical formulas in combustion reactions.

00:07 So we have our combustion reaction here.

00:10 And let's say we're given some hydrocarbon.

00:15 So a hydrocarbon is going to have the formula, x, h, y.

00:21 And this just means there's some number of carbon and some number of hydrogen.

00:26 And it's going to react with oxygen and create carbon dioxide and water vapor.

00:32 So if we're given, let's say we're given 0 .135 grams of this hydrocarbon.

00:42 And we have 0 .4 grams of carbon dioxide created and 0 .135 grams of water vapor created.

00:58 And we need to find the empirical formula.

01:04 Our only source of carbon is going to be this carbon dioxide and our only source of hydrogen is going to be this h2o.

01:12 So let's go ahead and find out how much of each are in this hydrocarbon here.

01:19 So we have this 0 .44 grams of carbon dioxide and we're going to divide that by the weight of carbon dioxide.

01:36 Now, we're worried about the carbon only.

01:41 So we're just going to take the part that's carbon.

01:43 So we take the atomic mass of carbon right there, and we get 0 .44 divided by 44, which is going to be 0 .01.

01:54 And then we multiply that by 12, and we get 0 .12.

02:00 So the amount of carbon here equals 0 .1.

02:05 One two grams alright so we have our amount of carbon and we're gonna find the amount of hydrogen and it should be the difference between this and this but we're just gonna double check and make sure we're doing it right so we take the point one three five and you're gonna divide that over the weight h2 o and then we're gonna multiply that by the parts of hydrogen so hydrogen is one atomic mass, and then there's two molecules.

02:45 So 0 .135 divided by 18, and then we're going to multiply that by 2...

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