Ethylene glycol, commonly used as automobile antifreeze,...

Ethylene glycol, commonly used as automobile antifreeze, contains only carbon, hydrogen, and oxygen. Combustion analysis of a 23.46 $\mathrm{mg}$ sample yields 20.42 $\mathrm{mg}$ of $\mathrm{H}_{2} \mathrm{O}$ and 33.27 $\mathrm{mg}$ of $\mathrm{CO}_{2}$ . What is the empirical formula of ethylene glycol? What is its molecular formula if it has a molecular mass of 62.0 amu?

Key Concepts

Molecular Formula Determination

The molecular formula indicates the actual number of atoms of each element in a molecule and is a multiple of the empirical formula. Once the empirical formula and its corresponding molar mass are known, the molecular formula is found by comparing the empirical formula mass to the compound's given molecular mass. The ratio of these masses determines the multiplier that transforms the empirical formula to the molecular formula.

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of the elements in a compound. By using the results from combustion analysis, one converts the masses of CO2 and H2O into moles of carbon and hydrogen. Oxygen, if present, is determined either by difference or directly if its mass is measured. The mole ratios are then simplified to yield the empirical formula of the compound.

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of a compound by burning a known sample and measuring the amounts of products, typically carbon dioxide and water, formed. This method relies on the fact that all the carbon in the compound is converted to CO2 and all the hydrogen is converted to H2O, allowing for the calculation of the number of moles of carbon and hydrogen present in the original compound.

Transcript

00:01 This is a carbon -hydrogen, oxygen containing samples.

00:07 So all of the hydrogen is contained in the water.

00:12 All of the carbon is contained in the carbon dioxide.

00:16 And there's a confounding variable in the oxygen because the combustion process introduces oxygen.

00:23 So i'm going to use...

00:24 I don't know how much oxygen there is either.

00:27 But i can find the amount of...

00:32 Hydrogen by converting milligrams of water into moles of hydrogen.

00:42 So how i'll do that is first i'll work with grams.

00:50 I divide it by thousands to get this value.

00:53 I'm going to divide by the molar mass of water 18 .2 grams of water in every one mole of water.

01:02 I know from the formula that there are moles of hydrogen with every one mole of water and so what i find is that this original sample contains 0 .002266 moles of hydrogen.

01:30 Okay, i'm going to repeat the same process, carbon, give the arrows, and it's the same.

01:47 Here's carbon dioxide here's the molar mass of carbon dioxide and i'll include this for completeness and that my units cancel accordingly.

02:12 This sample contains 0 .17, there's a top there, 560 moles of carbon.

02:32 Now this compound only contained carbon and hydrogen.

02:35 It could be done.

02:35 But i don't know how much oxygen there is left.

02:40 So i'm going to take an intermediate step and convert moles of hydrogen into milligrams of hydrogen.

02:52 Actually, this is the grams of hydrogen first.

02:58 I'm multiplying by the molary mass of just hydrogen.

03:04 We'll find 0 .00 -2284 grams of hydrogen, which is all.

03:19 Also 2 .284 milligrams of hydrogen and multiplied by 1 ,000 to get that value.

03:28 We'll do the same for carbons to get 9080 grams of carbon for 9 .080 grams of carbon for 9 .080 grams, not milligrams, not grams, milligrams of carbon.

03:53 The reason i wanted in milligrams was so that i could attract from the sample...

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