Finding Chemical Formulas
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- Опубліковано 8 вер 2024
- In this video, we discuss chemical formulas. Specifically, we define the terms "empirical formula" and "molecular formula". We also go over how to calculate the empirical formula of a compound given the masses of the elements in that compound. In addition, we go over how to calculate the molecular formula of a compound given its empirical formula and its molar mass.
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passing chem, cause of you. thank you!
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Hey Ben. How's everything going? I just wanted to know why no reaction occurs when NaF and AgNO3 are mixed together. Thanks.
Ronaldo Eid Everything is going well.
Ok, so I'm assuming that you're talking about AQUEOUS NaF and AQUEOUS AgNO_3.
The reason why this is important is because an aqueous solution of an ionic compound is simply a collection of the cations and anions floating around in water. Aqueous NaF is simply Na^+ ions and F^- ions floating around in solution. Aqueous AgNO_3 is simply Ag^+ ions and NO_3^- ions floating around in solution.
So this means that the ionic bonds are BROKEN, and you just have ions floating around. In a chemical reaction, new chemical bonds must form.
If you mix these two solutions together, those ions aren't going to form any new ionic bonds. In other words, Na^+ will not bond with NO_3^- and Ag^+ will not bond with F^-. This is because NaNO_3 and AgF, the two possible products of the reaction, are both SOLUBLE ionic compounds.
If either of those two compounds were INSOLUBLE, then we would see a reaction, because the ions would rather form ionic bonds with one another than float freely around in the solution.
I would recommend two videos on this subject.
First is my video on solubility rules. This will help you determine whether an ionic compound is soluble or insoluble. If you're interested, follow the URL below.
ua-cam.com/video/E7lXJ5F8zIw/v-deo.html
The next is my video on precipitation reactions, which is what happens when you get an insoluble product from two soluble reactants. If you're interested, follow the URL below.
ua-cam.com/video/E7lXJ5F8zIw/v-deo.html
Hope this helps!
-Ben
Hi teacher I hope you are good. I have a question. Here it is: When 0.273 g Mg is heated in a nitrogen atmosphere, a chemical reaction occurs. The mass of the product formed is 0.378 g. Find the empirical formula of the compound containing magnesium and nitrogen. Actually I solved it and found the right answer but there is something I don't understand in my solution. I subtracted 0.237 from 0.378 to find the mass of the nitrogen gas. Then I found the mol of magnesium and nitrogen gas. Then divided them by the smallest mole. Finally, I found 1 mol N2 and 3 mol Mg so the answer is Mg3N2. But what I did not understand is that I found the mole of Nitrogen gas not Nitrogen. Is it wrong? which way is accurate? I hope I could explain the point that I am so confused 😕
I'm confused. Is the original mass of Mg 0.273 g or 0.237 g?
@@BensChemVideos I am soo sorry I wrote it wrong it is 0.273 g Mg
Gotcha...so you were definitely on the right track.
If you start with 0.273 g of Mg, and you end up with 0.378 g Mg + N, then the mass of N is 0.105g (0.378 g - 0.273 g)
So we have 0.273 g Mg and 0.105 g N. It's actually not necessary to convert to moles at this point. We know that the chemical formula must contain whole numbers of Mg and N--for instance, it could be MgN_2, but not MgN_2.5.
So all we have to do is to divide both mass values (0.273 g and 0.105 g), by the smaller value of the two (in this case 0.105 g) and see if both values are whole-number multiples of the smallest value. If that doesn't work, then we try one half of the smallest value. If that doesn't work, then we try one third of the smallest value, and so on and so fourth until we find a number of which 0.273 g and 0.105 g are whole-number multiples.
If we divide both values by 0.105, then we get Mg_2.6 N_1...doesn't work, because 2.6 isn't a whole number.
If we divide both values by HALF of 0.105, then we get Mg_5.2 N_2...doesn't work, because 5.2 isn't a whole number.
If we divide both values by a THIRD of 0.105, then we get Mg_7.8 N_3...doesn't work, because 7.8 isn't a whole number.
If we divide both values by a QUARTER of 0.105, then we get Mg_10.4 N_4...doesn't work, because 10.4 isn't a whole number.
If we divide both values by a FIFTH of 0.105, then we get Mg_13 N_5...this works, because 13 and 5 are both whole numbers.
So the formula would be Mg_13_N_5
Does that make sense?
@@BensChemVideos you are amazing 💜 I was very touched when I saw your answer to my question. Thank you very much for having your time 😸
Actually I did not understand why you did not convert grams into moles. And when I write this question on internet I found the answer. It says the answer is Mg3N2.
The point that I don't understand is whether I should find the mole of nitrogen or the mole of nitrogen gas. I solved it finding the mole of nitrogen gas, since Mg is heated in the nitrogen gas. But the solution of this problem on the internet is not like that. They found the mole of nitrogen not nitrogen gas. Actually I found the correct answer already. But I just wonder if the way I solved this problem is true or not.
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