Give the percent composition of each of the following substances: (a) Glucose, C6 H12 O6 (b) Sul

Give the percent composition of each of the following...

Key Concepts

Atomic Weights

Atomic weights, typically expressed in atomic mass units, are essential for converting between the atomic scale and the macroscopic scale. They serve as the conversion factors when calculating the mass contribution of each element in a compound and are pivotal in determining both molar mass and percent composition.

Chemical Formula Interpretation

Interpreting chemical formulas involves identifying the number of atoms of each element present in the compound. This skill is vital for calculating molar masses and allows for the accurate determination of how much mass each element contributes to the overall composition. It forms the basis for many quantitative chemical calculations.

Percent Composition

Percent composition is the measurement of the percentage by mass that each element contributes to a compound. It is calculated by dividing the total mass of an element in one mole of the compound by the compound’s molar mass and then multiplying by 100. This concept is crucial in analytical chemistry for understanding the elemental make-up of substances.

Molar Mass Calculation

Molar mass is the sum of the atomic masses of all atoms present in a compound's formula. By using the atomic weights from the periodic table, one can calculate the mass of one mole of the compound. This calculation is a fundamental step in determining percent composition and other stoichiometric computations in chemistry.

Transcript

00:01 In this problem, we will need to find the percent composition of each of the following substances, which includes glucose, sulfuric acid, potassium permanganate, and saccharine.

00:17 We will first start with glucose.

00:21 Glucose is given by the chemical formula c6, h12, and o6.

00:28 So to approach this kind of problem, we will first need to know the equation that the mass percent of each element is equal to the mass of that element over the total mass of the whole substance times 100 % to find the mass percent composition.

00:51 So before we begin, we will need to know the ratio between each of the element, in the following substance.

01:01 And in this case, we will need to know the ratio of each element, or each atom, and glucose.

01:08 So we know that there are six carbons, 12 hydrogens, and six oxygens.

01:14 We will first need to try to solve how many grams of each atom is within glucose.

01:21 We know that per one mole of glucose, we have six moles of carbon, and one mole of carbon has a mass of 12 grams.

01:32 We can find this information on our periodic table, which is our molar mass of that atom.

01:39 This will give us 72 grams of carbon and one mole of glucose.

01:45 We will do the same for hydrogen and oxygen.

01:48 So for one mole of glucose, we have 12 moles of hydrogen.

01:54 And per one mole of hydrogen, we have one point, 01 grams of hydrogen.

02:01 We do the multiplication and we will find that there are 12 .12 grams of hydrogen per 1 mole of glucose.

02:09 Lastly for oxygen, per 1 mole of glucose we have 6 moles of oxygen and per 1 mole of oxygen we have 16 grams of oxygen.

02:21 That's our molar mass.

02:23 And so in 1 mole of glucose we will have ourselves with 96 grams of oxygen.

02:28 Of oxygen.

02:29 Now going back to the equation of trying to find the percent composition of an element, we know that the mass, which we have just solved, is over the total mass to find the mass percent.

02:45 So we can add all of these masses together to find the total mass of glucose, which is 72 plus 12 .12 plus 96 grams, which gives us a total mass of 180 .12 grams.

03:02 So we have all the information to solve for the mass percent composition.

03:07 In the equation, we have the mass and the total mass.

03:10 All we need to do is separately calculate the mass percent or the percent composition of each of the atoms.

03:17 So for the percent composition of carbon, we know that the mass of carbon is 72 grams over the total mass times 100 % gives us 40 % composition of glucose.

03:34 So there's a 40 % composition of carbon in glucose.

03:39 And for hydrogen, we know that there are 12 .12 grams of hydrogen.

03:47 Over the total mass times 100%, which will give us 6 .7 % composition of hydrogen and glucose.

03:57 And lastly, for oxygen, we know that there are 180 .12 grams of oxygen, oh sorry, 96 grams of oxygen over our total mass, which is 180 .12 grams, times 100%.

04:12 And this will give us 53 .3 % composition of oxygen and glucose.

04:21 Next, we will solve for sulfuric acid.

04:27 Sulfuric acid is given by the chemical or empirical formula of h2s -o -4.

04:35 We can list the ratio of each atom, which in this case there are only three atoms, sulfur, and oxygen.

04:46 And from the empirical formula, we know that the ratio between the three atoms is 2 to 1 to 4.

04:52 There are 2 hydrogens, 1 sulfur, and 4 oxygen, and 1 in sulfuric acid.

05:00 Like before, we will start by solving for the mass of each element in 1 mole of sulfuric acid.

05:06 We know that in 1 mole of sulfuric acid, we have 2 moles of hydrogen, and 4.

05:13 For 1 mole of hydrogen, it is 1 .01 grams.

05:18 So, multiplying everything together, we will find ourselves with 2 .02 grams of hydrogen per 1 mole of sulfur.

05:27 Next, we will solve for the mass of sulfur.

05:31 So for 1 mole of sulfur acid, we'll have 1 mole of sulfur, and per 1 mole of sulfur, it will weigh, the molar mass is 32 .06 grams.

05:43 Of sulfur, which will give us 32 .06 grams of sulfur per one mole of sulfur acid.

05:52 Lastly, we can solve for the mass of oxygen.

05:56 We know that one mole of sulfuric acid, we have four moles of oxygen, and per one mole of oxygen, the molar mass of oxygen, is 16 grams.

06:08 We can multiply everything together, and we will find ourselves with 64 grams.

06:13 Of oxygen per 1 mole's sulfuric acid.

06:17 Like before, we will need to find a total mass, which is the sum of all of these individual masses of atoms in the substance.

06:30 So that will be the addition of all of these masses, which will give us 98 .08 grams.

06:39 Next, we can solve for the percent composition of the, following of this substance by using the formula given above in green like before...

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