Combustion analysis of a hydrocarbon produces 33.01 g CO 2 and 13.51 g H 2 O. Calculate the empirical formula of the hydrocarbon.

All right. Hello everyone. So this question says that combustion analysis of a hydrocarbon produces 33.01 g of carbon dioxide and 13.51 g of h2o or water. Calculate the empirical formula of the hydrocarbon option AC C two, option BC C two H four, option three says C three H five and option D says C two H six. Now first let's recall a few things, right. A hydrocarbon refers to a compound composed only of carbon and hydrogen and an empirical formula is a molecular formula that shows the simplest hole number ratio between all of the elements in a given chemical compound. In other words, our task is to find the simplest ratio between carbon and hydrogen in this hydrocarbon. So to find the simplest ratio of carbon and hydrogen, we're going to go ahead and make use of the data provided in the combustion analysis. So first lets go ahead and recall the molar masses of both carbon dioxide and water. The molar mass of carbon dioxide is 44.01 g per more. Whereas the molar mass or water is 18.02 g per month Now, the reason why this is relevant is because we're given these data of the combustion analysis in units of grams. However, we're interested in in understanding the moles produced of each compound as a means of finding the molar ratios between carbon and hydrogen. So first, if I go ahead and I take the mass of carbon dioxide produced, right, that's 33.01 g. I'm going to go ahead and divide this by the molar mass of carbon dioxide to find how many moles were produced. So 33.01 g of carbon dioxide divided by 44.01 g per mole equals 0.7501 moles of carbon dioxide. And I'm rounding this number to four significant figures. So now lets repeat this with water. I'm going to take the mass of water produced and divide that by the molder mass of water. So that's 13.51 grams of H2O divided by 18.02 grams per mole of H two up, this equals 0.7497 moles of H two. And it's worth mentioning that units of grams cancel out in both cases. All right. So now at this point, we have the moles of both carbon dioxide and water. However, this is not going to be sufficient to produce the ratios that we need, right? Because carbon dioxide contains carbon and water contains hydrogen. However, they both contain oxygen as Well, right. So in this case, we have to consider how many moles of carbon are in carbon dioxide and how many moles of hydrogen are in water. And to do that, we're going to need their mole ratio. So let me go ahead and make some space on the side here next to my moles of carbon dioxide. Right, in this case, notice how one more of carbon dioxide is composed of one mole of carbon elemental carpet. Meaning that the moles of carbon dioxide are equivalent to the moles of carbon by itself, meaning that this is equal to 0.7501 moles of carbon. And now we can do the same thing with water by considering the subscripts in its molecular formula, right? Because one mole of water or H2O is made up of two moles of hydrogen, meaning that the number of moles of water must be multiplied by two to find the moles of hydrogen. So 0.7497 multiplied by two, gives us 1.499 moles of hydrogen. And again, this was rounded to four significant figures. So at this point, I do need to scroll down to open up some more space. But now we're going to go ahead and divide the number of moles of both carbon and hydrogen by the smaller of the two. So this means that the number of moles of carbon and hydrogen would be divided by 0.7501 considering that is the smaller number. And this gives us the subscripts necessary in the empirical formula. In other words, the smallest ratio, right. So here or carpet specifically, we would take the number of moles of carbon that's 0.7501 and divide that 50.7501 to give us one for its coefficient. Excuse me, it's subscript in the empirical formula. Whereas hydrogen, we would take the number of moles of hydrogen that's 1.499 and divide this by 0.7501 to give us approximately two. In other words, the empirical formula of this hydrocarbon is C one H two or simply CH two. So our answer is CH two and that corresponds with option A in the multiple choice and there you have it. So if you stuck around until the end of this video, thank you so very much for watching. And I hope you found this helpful.

Ch.3 - Molecules, Compounds & Chemical Equations

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Ch.3 - Molecules, Compounds & Chemical Equations

Problem 97

Chapter 3, Problem 97

Combustion analysis of a hydrocarbon produces 33.01 g CO₂ and 13.51 g H₂O. Calculate the empirical formula of the hydrocarbon.

Verified step by step guidance

Determine the moles of carbon in CO2 by using the molar mass of CO2 (44.01 g/mol) and the fact that each mole of CO2 contains one mole of carbon.

Calculate the moles of hydrogen in H2O by using the molar mass of H2O (18.02 g/mol) and the fact that each mole of H2O contains two moles of hydrogen.

Convert the moles of carbon and hydrogen to a mole ratio by dividing each by the smallest number of moles calculated in the previous steps.

If necessary, multiply the mole ratio by a whole number to obtain the smallest whole number ratio of carbon to hydrogen.

Write the empirical formula using the whole number ratio of carbon to hydrogen.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Combustion Analysis

Combustion analysis is a technique used to determine the composition of a hydrocarbon by burning it in excess oxygen. The products of this reaction are typically carbon dioxide (CO2) and water (H2O). By measuring the masses of these products, one can infer the amounts of carbon and hydrogen in the original hydrocarbon, which is essential for calculating its empirical formula.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements present in that compound. It is derived from the moles of each element calculated from the mass of the products obtained in combustion analysis. This formula is crucial for understanding the basic composition of the hydrocarbon before determining its molecular structure.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of particles it contains. One mole of any substance contains Avogadro's number (approximately 6.022 x 10^23) of entities, whether atoms, molecules, or ions. In combustion analysis, converting the mass of CO2 and H2O to moles allows for the determination of the number of carbon and hydrogen atoms in the hydrocarbon, which is necessary for calculating the empirical formula.