The combustion of 40.5 mg of a compound extracted from the bark of the sassafras tree and known

The combustion of 40.5 mg of a compound extracted from the...

Key Concepts

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of elements in a compound. After calculating the number of moles of each element from experimental data, these values are divided by the smallest number of moles to obtain the empirical formula.

Molecular Formula Determination

While the empirical formula gives the simplest ratio of elements, the molecular formula indicates the actual number of each type of atom in a molecule. The molecular formula is determined by using the compound's molar mass to scale up the empirical formula, ensuring the molecular formula's total mass matches the experimental molar mass.

Combustion Analysis

Combustion analysis is an experimental technique used to determine the elemental composition of a compound by burning it in oxygen. The amounts of carbon dioxide (CO2) and water (H2O) produced are measured, and these values are then used to calculate the amounts of carbon, hydrogen, and sometimes oxygen present in the original compound.

Mass-to-Mole Conversion

This concept involves converting the measured masses of reaction products to moles using their respective molar masses. In combustion analysis, the mass of CO2 is converted to moles of carbon and the mass of H2O is converted to moles of hydrogen, which is essential for determining the composition of the compound.

Transcript

00:01 Okay, so for this question, they want us to calculate the empirical in the molecular formula of the unknown compound.

00:07 And the information they've given us so far is there is 40 .5 milligrams of the unknown compound x.

00:18 There's 110 milligrams of co2.

00:23 There is 22 .5 milligrams of h2l.

00:27 So we can set up the equation.

00:28 They said that the starting compound x contains carbon.

00:34 We don't know how many.

00:34 So i put an x here.

00:36 It contains hydrogen.

00:37 We don't know how many.

00:38 So it's y.

00:39 And oxygen.

00:40 And since we don't know how many, i'm going to label it as z.

00:43 So a combustion reaction is always a hydrocarbon reacting with oxygen to form carbon dioxide and water.

00:53 Okay, so let's calculate how much oxygen we started with, right? so the law of conservation of mass, meaning that the mass of the reactant, so if we add this one with this one, it should equal adding this one with this one.

01:14 Right.

01:14 So the product, we have 110 milligram plus 22 .5 milligram.

01:23 This will equal 110 plus 22 .5 is equal to 132.

01:31 So we do 132 .5 subtracted from 40 .5 milligrams, that's compound x.

01:41 He gets 92 milligrams.

01:45 So we started off with 92 milligrams of o2.

01:54 Now the next thing we do is we look on the right side, we look at co2.

01:58 We see that there is only one source of carbon.

02:01 So let's calculate how many milligrams out of the 110 belongs to carbon.

02:08 So what we do now is we find the amululatory weight of carbon dioxide, which is c, which is 12, oxygen is 16.

02:16 But since we have two of them, we multiply it by two, we add these two together.

02:21 We get 44 grams.