A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yi

A compound contains only carbon, hydrogen, and oxygen....

Key Concepts

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of a compound by burning it in oxygen. In organic compounds composed of carbon, hydrogen, and oxygen, combustion produces CO? and H?O, from which the amounts of carbon and hydrogen in the original compound can be deduced.

Mass-to-Mole Conversion

Converting mass measurements into moles using the respective molar masses is essential to quantify the number of atoms present. This stoichiometric method allows one to move from experimental data (mass of products) to the moles of constituent elements.

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It is determined by calculating the moles of each element from the combustion products and, if necessary, inferring the amount of oxygen by difference, then simplifying these ratios to the smallest integers.

Molecular Formula Determination

The molecular formula gives the actual number of each type of atom in a molecule. It is obtained by comparing the mass of the empirical formula with the given molar mass of the compound and finding the appropriate multiplication factor that scales the empirical formula to the molecular formula.

Transcript

00:01 This is a question about the combustion of an organic compound containing carbon hydrogen and oxygen to produce co2 water and 10 .68 milligrams of it was combusted and it produced 16 .01 milligrams of co2 and 4 .37mogram of water and the molar mass of the compound is 1 .76 .1 gram per more.

00:29 So the question was us to determine the molar mass, wants us to determine the empirical formula and the molecular formula for the compound.

00:39 So we know that 16 .01 is the mass of a co2 that was produced and we want to determine the mass of the carbon in the co2.

00:53 So if we have 16 .01 milligram of co2, we have 16 .01 milligram of co2, we have to multiply that by one millimole we were trying to keep everything in melee we could as well convert the 16 .01 milligram to grams that since our massive sign milligrams so we have keeping everything in melee so that's why we have one million more of co2 so one million more of co2 divided by the molar mass of co2 which is 44 .0 so we should be in grams but we're keeping everything in a millie.

01:38 So we're divided by 44 .01 of co2 and we're going to multiply that by one more of co2.

01:44 And we're going to multiply that by one more of carbon and we'll divide it by one millimole of co2.

01:56 So when we do that we'll come up with 0 .36378 moles.

02:06 Minimal of carbon and we can convert that to mass by multiplying by the mass of carbon which is 12 .01 and that would give us 4 .369 milligrams so we have to do the same thing with water and we know that water is 4 .37 milligram of it that was produced so go to the next page so 4 .37 milligram of water multiplied by one millimeter of water divided by 18 .06 milligram of water multiply by one sorry much by two because we have two hydrogen atoms here the water molecule so multiplied by by two millimeter more of hydrogen divided by one minimum water and that will be equals to 0 .485 .2 million more of hydrogen so we can also convert that to mass by multiplying by the mass of one hydrogen atom which is 1 .008 and that divided by 1 mm of hydrogen and that would give us 0 .489 milligram of hydrogen and the next thing would be to determine the mass of the oxygen since we know the mass of hydrogen we know the mass of carbon we can find a mass of oxygen and we know that the cshy or z as a mass of 10 .6 so we can then determine the mass of oxygen by summing up the mass of carbon and nitrogen and subtracting it from the 10 .68 milligram and that would give us 5 .h2 now that we know the mass of that, we can then go ahead to determine the number of more of each of those masses.

04:34 But if you remember we've determined a number of more for carbon to be 0 .36378 more and we'll determine the number of more for the water to be 0 .48512 milamore...

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