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Carleton College

Numerade Educator

Drexel University

University of Maryland - University College

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So in this problem we have 5 g of lithium sulfate reacting with 5 g of Baron. I did so for this problem. We will determine the empirical formula of the reactant and sure work using the following present compositions. So for this problem, we want to use the percent compositions and convert this into grams, which we can then use to figure out the moles of each component and then from there, determine a ratio between the atoms of each molecule and then figure out the empirical formula. And so first, let's work with lithium sulfate. So for lithium sulfate, we have lithium, which has a percent composition of 12.63%. Yeah, for sulfur, we have 29.17%. Yeah, For oxygen, we have 58.21%. So the first thing we want to do is to convert this into grams. So if we have 100 g of material, this is pretty easy to do on DSO. In this problem, we can use the fact that you're using 5 g. But that's not necessary for the problem, because three percentages 5 g of lithium sulfate and one grandma looking sulfate is the same. And so now these will all be in grants. And so from here we can use the mess of each atom to determine the number of moles for each component. And so to do this, um, we can multiply this by one more over 6.94 1 g of lithium, which is from the PR table. So we get 1.8200 moles, and we're going to do the same thing for sulfur. And so we're going to divide this by the mass of sulfur, which is 32.7 g to obtain 0.9 whoa 96 moles. And lastly, we're going to do the same thing for oxygen. And so we're going to divide this by 16.0 g. So we end up with 3.638 moles, and from there we have obtained the moles, lithium, sulfur and oxygen. So now for the next step, we're going divide all of these moles by the smallest mole number that we've obtained. And so in this problem, the smallest number is 0.996 Then we're going to divide all the values by this number. And so from here we get the following ratios to one and four. So that means the miracle formula for lithium sulfate is lithium to sulfate. Um, which makes sense because, according to our period of table lithium is in the group one group. And so this has a charge of plus one. Sophie, um is a poly atomic and ion with a charge of negative too. And so we have to lithium each carrying you, plus one charge with a total charge l plus do and sulfate, which has a total charge of negative. And so this molecule is neutral and our answer makes sense. And so now that's the first part of the problem. And now we need Thio. Figure out the empirical formula of the other reactant, which is barium idea. And to do this, we're going to do the same thing as before. So we have the percent composition of each Adam so far, Bear, um, it is 35.11%. I died is 64.89% and so we can convert this into grams if we love 100 g of the material. Andi, we're going to do the same thing for iodide. And then from there we divide this by the atomic mouse of each bottom. And so for Berry, um, this would be 137 0.3 g. We have to a bowl of 0.2557 and for I died, you divide this by its atomic mass, which is 126 29 4. And so we end up with 0.5113 And from here we again divide all of the rules by the smallest small number that we've obtained. And in this problem, this would be 0.2557 And so when we do this, we end up with a ratio of one to do. And so our empirical formula is very, um I died. So be a two. Which makes sense because from our periodic table, barium is a group to element and so has a charge of plus two. And I died is a hey lied with the charge of minus one. And so we have a total of two. I'd IEDs with a total charge of minus two and barren with which has a charge plus two. And so we know this molecule is neutral. And so we have confirmed that this is indeed the empirical formula as well as the empirical formula for lithium sulfate.

Brown University

Aqueous Solutions

Thermochemistry

Electronic Structure

Periodic Table properties

Chemical Bonding