Menthol (ℳ=156.3 g / mol) , the strong-smelling substance in many cough drops, is a compound of

Menthol (ℳ=156.3 g / mol) , the strong-smelling substance in...

Key Concepts

Atomic Mass and Molar Mass Concepts

Understanding atomic masses and molar masses is crucial in translating between mass and number of moles. Atomic mass units (amu) represent the mass of individual atoms, and when combined according to a compound's formula, they yield the molar mass, which is used as a conversion factor in both analyzing combustion products and identifying the molecular formula of the compound.

Molecular Formula Determination

The molecular formula is the actual number of atoms of each element present in a molecule and is usually a whole-number multiple of the empirical formula. In this process, the empirical formula weight is compared to the experimentally determined molar mass of the compound; the ratio of these values indicates how many times the empirical formula must be multiplied to yield the molecular formula.

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of atoms in a compound. After calculating the moles of the individual elements (typically carbon and hydrogen from CO2 and H2O, with any oxygen being determined by difference), these values are normalized to the smallest number of moles to provide the simplest ratio that represents the compound's composition.

Stoichiometry and Mole Conversions

Stoichiometry involves using the mole relationships between reactants and products in a chemical reaction to calculate quantities like mass, moles, or concentration. In combustion analysis, the known masses of CO2 and H2O are converted to moles using their molar masses, which then allows for the determination of the moles of carbon and hydrogen in the original compound.

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of organic compounds by burning the sample and measuring the amounts of the combustion products, typically CO2 and H2O. This method relies on the fact that all the carbon in the compound converts to CO2 and all the hydrogen converts to H2O, which provides a basis for calculating the original amounts of these elements in the compound.

Transcript

00:01 We have a chemical compound compound with the molymouth 1 .56 .3 gram per mole.

00:05 So what would be the molecule formula? when we have our compound being undergo a combustion reaction, we're acting with oxygen to produce carbon dioxide and water only.

00:15 Okay, so first of all, we have to find out the number of most of carbon, and also number of more hydrogen, and also the number of more oxygen.

00:24 And then it will help us to help determine the molecular formula.

00:28 Okay, so from here, how can we find out the number of most of carbon, also carbon.

00:31 If we look at the combustion reaction over here, but keep in mind that this is not balanced.

00:36 This is not balanced.

00:38 So the carbon will convert to carbon dioxide and also the hydrogen will convert to hydrogen, i mean water.

00:47 And the oxygen, we have to add external source of oxygen, so it may not be helpful for us to find out the actual amount of oxygen by simply from co2 and also water.

01:00 But co2 and water, again, have us to find out the number of coal of carbon and also hydrogen.

01:05 Okay, so let's find out the number of mode of co2 first.

01:10 So, number of co2 would equal to our 0 .449 gram per mole, divided by our motor mass of co2.

01:21 So it would be equals to 44 gram per mole.

01:25 Let's say 44 gram per mole.

01:27 So 0 .449 divided by 44.

01:29 So we have 0 .0102 moles of co2.

01:36 Because for each co2 we have one carbon.

01:40 So the number of mobile carbon will be also equals to 0 .0102 mole.

01:47 Okay, so same thing we're going to look for the number of more hydrogen, but we start with water.

01:52 So we are 0 .184 gram, and then for water 80 gram per mole.

01:57 And then we are 0 .184 diroids.

02:00 18, so we have 0 .0 .02 mode of water, of water.

02:08 However, we have to file the actual number of more for individual hydrogen.

02:15 So we are going to multiply this number by 2, the water by 2.

02:20 Again, because for each water, we have 2, sorry, we have 2 hydrogen.

02:27 So for hydrogen, we are 0 .02 4 modes and then for our carbon we are 0 .02 volts.

02:36 Alright, so how about oxygen? how about oxygen? how can we find our oxygen? so from here, actually i missed one of the information given.

02:50 From here we have 0 .15 -95 gram of metal being burned.

02:57 So yeah, i left that information.

03:00 So it's corresponding to the mass of the mantle of this being burned.

03:04 Okay, so we have the lump of mole of carbon, and then we have the number of modes of our hydrogen, right? so we just have to convert that to mass, and then we use our total mass of our mantle to subtract the hydrogen and carbon, and then we have the remaining mass to be oxygen, and then from there we can file the number of moles.

03:27 Okay, so left, let's convert to the mass of carbon very quickly.

03:36 So we have 0 .0 .0 .0 .02 mules times 12...

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