In most of its ionic compounds, cobalt is either Co(II) or...

Key Concepts

Decomposition Reactions and Thermal Analysis

Decomposition reactions involve the breakdown of a compound into simpler substances when subjected to heat or other reagents. In this context, heating a hydroxide to yield cobalt(III) oxide provides a method to indirectly measure the cobalt content in the compound, serving as a basis for further stoichiometric analysis.

Ionic Compounds

Ionic compounds are chemical compounds composed of positively charged ions (cations) and negatively charged ions (anions) held together by electrostatic forces. Their formulas are determined by the need to balance the overall charge, and many such compounds often incorporate water molecules as waters of hydration, influencing their final composition and formula.

Balancing Chemical Equations

Balancing chemical equations is the process of ensuring that the number of atoms of each element is conserved in a chemical reaction. This is achieved by adjusting the stoichiometric coefficients, and it is essential for accurately determining the relationships among reactants and products when quantifying chemical reactions.

Empirical and Molecular Formula Determination

Empirical formula determination involves finding the simplest whole-number ratio of the elements in a compound based on the percent composition or mass measurements. When additional information, such as the molar mass, is available, the empirical formula can be scaled up to deduce the molecular formula of the compound.

Percent Composition

Percent composition refers to the mass percentage of each element within a compound. It is calculated by dividing the mass of an individual element by the total mass of the compound and multiplying by 100. This concept is crucial for understanding the elemental makeup of a compound and is used to deduce the empirical formula.

Stoichiometry

Stoichiometry involves calculating the relationships between the quantities of reactants and products in chemical reactions using their molar masses and balanced equations. It is fundamental for converting measured masses into moles, which then helps to determine the composition and ratios of the various elements in a compound.

Gravimetric Analysis

Gravimetric analysis is a quantitative method in analytical chemistry where the amount of an analyte is determined by converting it into an insoluble form, isolating it, and then weighing it. This technique was used when the chloride ion was precipitated as silver chloride, and the mass of the precipitate was used to infer the amount of chloride in the original compound.

Transcript

00:02 Okay, so in this problem we are asked to find the formula of a mystery cobalt chloride salt which also has waters of hydration.

00:11 So let's write a generic formula first.

00:14 We're going to write cobalt x, chlorine y, and we know that we will have z waters of hydration for my molecule.

00:26 So this is again a generic formula.

00:28 We do not know the values of x, y, and c at the moment.

00:32 But we are told that we take the compound and we subject the compound to a series of reactions and from those reactions we get the masses of cobalt and chloride.

00:45 So let's see what we have.

00:47 The first thing i'm going to do here is write the molar masses of the compounds that are relevant for the problem.

00:52 So the molar mass of silver chloride is 143 .32 grams per mole.

01:06 The molar mass of cobalt 3 oxide, and this is the formula for that, is 165 .86 grams per mole.

01:23 The molar mass of water is 18 grams per mole.

01:30 So we're going to be using these values later in the problem, so i just wanted to have them here.

01:36 So let's first look at silver chloride.

01:39 The problem tells us that we collect silver chloride from the first experiment and that we collect a mass of silver chloride that's equal to 0 .308 grams.

01:57 Now this is very important because in the first experiment we basically know that all of the chloride, all of the chloride in my original compound is going to end up in the silver chloride.

02:14 So this is a clue to actually find the total amount of chloride in my original compound.

02:20 So how do we do that? we know that in one mole of silver chloride, which is equal to 143 .32 grams, and this is from the molar mass of silver chloride, in one mole of silver chloride we have one mole of chlorine.

02:44 And one mole of chlorine equals 35 .5 grams of chlorine, and this is silver chloride.

02:51 This is the molar mass of chlorine basically.

02:54 One mole of chlorine is 35 .5 chlorine.

02:57 So using this proportionality we're going to say that if in 143 .32 grams of silver chloride we have 35 .5 grams of chlorine, then in 0 .308 grams of silver chloride, by proportionality we can get how much chlorine is present there, and we're going to get, if we do the proportionality calculation, 0 .076 grams of chlorine.

03:30 This is the grams of chlorine in the silver chloride that we obtained, but it's also the same as the grams of chlorine present in the original compound.

03:40 Now we know that for the first experiment we started with 0 .256 grams of the original compound.

03:51 So we know that in 0 .256 grams of the compound we have 0 .076 grams of chlorine, and from that we can get the percentage mass of chlorine in the compound.

04:07 So for that we're going to do 0 .076 over 0 .256 grams, this is chlorine and this is compound, and we are going to get, and we're going to multiply it by 100 to get a percentage, and that percentage is going to be 29 .7%, and this is for chlorine, so we already have the first component, chlorine, that's my percentage.

04:34 We will do now the same, but for cobalt.

04:37 So we are also told that we do a second experiment, we take 0 .416 grams of the compound, we do a set of reactions, and at the end we obtain a mass of cobalt oxide of 0 .145.

04:55 So let's write it here, the mass of cobalt oxide for my second experiment is 0 .145 grams.

05:11 Now from that we can also assume that all of the cobalt in the cobalt oxide forms is coming from the original compound as well.

05:21 So all of that mass of cobalt in the cobalt oxide is going to belong to the original compound.

05:31 So we know that in one mole of cobalt oxide, the mass of that is 165 .86 grams, and this is from the molar mass of cobalt oxide.

05:51 One mole of cobalt oxide is 165 .86 grams.

05:55 Now in one mole of cobalt oxide, we actually have two moles of cobalt.

06:03 Now two moles of cobalt equals 117 .86 grams of cobalt.

06:11 This is simply using the molar mass of cobalt.

06:15 So two times the molar mass of cobalt is 117 .86 grams.

06:21 So now i can use that for proportionality.

06:24 In 165 .86 grams of cobalt oxide, i have 117 .86 grams of cobalt.

06:38 So in 0 .145 grams of cobalt oxide by proportionality, sorry this is three, then i'm going to get 0 .103 grams of cobalt.

06:57 Now i know that that's the amount of cobalt that i have in 0 .416 grams of my compound, starting compound.

07:15 So remember that i started with 0 .416 grams of compound for this experiment.

07:22 Now in that experiment, i know that i have 0 .103 grams of cobalt.

07:31 Now we can get the percentage by doing 0 .103 grams of cobalt over the total grams of compound for this experiment times 100, and that's going to be 24 .8 % of cobalt.

07:53 That's the second component of my salt.

07:57 So now i know the percentage of chlorine and the percentage of cobalt.

08:00 Now let's continue to the water.

08:04 So the remaining percentage, the remaining percentage is going to belong to the water because we only have chlorine, cobalt, and water.

08:20 So we can say that 100 minus 29 .7 minus 24 .8, that's going to be equal to 45 .5%, and that is going to be the percentage of water in my compound, simply by difference.

08:46 Now let's go a little further and obtain the separate percentages of hydrogen and oxygen because the problem is asking us for the mass percentages of all the components of my compound.

09:00 So we know that in 18 grams of water, which is one mole, one mole of water, we have two grams of hydrogen, two moles of hydrogen.

09:15 So in one mole of water, we have two moles of hydrogen, and that is equivalent to two grams of hydrogen.

09:23 So we can say also that from the percentage of water, let's take the first experiment.

09:41 Now in the first experiment, the original mass for that experiment was 0 .256 grams.

09:52 Now i know that the percentage of water is 45 .5.

09:55 So if i multiply by 0 .455, i'm going to get 0 .116 grams, and that's going to be the mass of water in my compound, okay? so this is the original mass of compound for the first experiment.

10:14 That's the percentage of water.

10:16 Therefore i have 0 .116 grams of water for my first experiment.

10:22 So in 0 .116 grams of water, i can get the amount of oxygen.

10:31 By proportionality, that's going to be 0 .013 grams of hydrogen.

10:38 Again, this refers to the first experiment, so i know that i have 0 .013 grams of hydrogen in the compound amount, which was 0 .4256 grams, times 100, and that is going to be 5 .1%.

10:56 This is the percentage of hydrogen.

11:00 That's the percentage of hydrogen in my compound...

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