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Q26. The Elemental analysis of a compound sample informed that it contains ? 83 &g of oxygen 435 g of CI . and 2.82 g of Na What is the empirical formula of the...

Question

Q26. The Elemental analysis of a compound sample informed that it contains ? 83 &g of oxygen 435 g of CI . and 2.82 g of Na What is the empirical formula of the compound?

Q26. The Elemental analysis of a compound sample informed that it contains ? 83 &g of oxygen 435 g of CI . and 2.82 g of Na What is the empirical formula of the compound?



Answers

A 175.0 g sample of a compound contains 56.15 g C, 9.43 g H, 74.81 g O, 13.11 g N, and 21.49 g Na. What is the compound’s empirical formula?

Change Tombo over Question seven which took some trickling in park or Melis. When you have the compliment decomposes into night so her she have to cover each each element in two moles on dividing by in marvels. So you have one pointing severally, Gramps. So for everyone more you have 14 rooms. You mean 0.12 moves for oxygen. You have 4.5 grand to lock team. So for everyone. So you have 16 rooms. You need 25 moles of oxygen. So you have been pretty and 0.12 ho went too far divided by the meets tomorrow this to yourself you think oh to

I even saw in discussion, they asked a sample of three owners want to contain .4-3 gram of carbon. So thank you buddy. Yes. Yeah. What rate of carbon? .4- three g? Yeah. Then went out chlorine, correct, flowed in. That is equal to 2.50 g. Come on. Mm went up flooring right now. Lauren confined 34 grand. Yeah man, the empirical formula of the compound. Yeah. So yeah, here in bags element then Graham and then again say and name simplest here man, whole number. Mhm. The seventies got a button something Point All two threes. The next 11 chlorine who claimed 40 Next Lemon Florentine, one point 34 So .4-3 upon 12 point zero and 1 30 sequel to 0.03 fire too. Then 2.50 upon 35 point for you. So that is the question, you know point 0705 1.34 a bomb 19 and that is equal to European 0705. So some blessed young. Yeah, yeah, yeah. Some flashed and 0.0352 Upon European 0352. That is he called one then 0 yeah, 07 You know for you. 1 0.0352. That is going to two Montecito Finance 070 Fighting a bond 0.0352. Again, that is the 1- two. So all the mouths are one two two. Since the ratio of carbon is to chlorine is to floor Een is one he stood to use to and that fall I mean they call formula the confound and a miracle formula of compound is equal to see see how you do F two. Okay.

In this problem, we have to determine the empirical formula off a compound for which were given masses off the elements which make up this compound. These are 10.52 g off and I 4.38 g off C and 5.10 g off in to find the empirical formula. First, we will convert these masses to moves. To do that, we'll have to divide each one of these by their Mueller masses. Which means that 10.5 g off and I is equal to 0.1792 months off and I 4.38 g off C is equal in tow. 0.3470 most off carbon 5.10 g off nitrogen is equal in tow. 0.3640 moles off nitrogen here. The smallest number off moles are that off and I, which is 0.1792 So we will divide each of these moles by 0.1792 which means that this compound has which means that this compound has these elements in the ratio off. One more off nickel. The two moves off carbon to two moles off nitrogen, which means that the empirical formula will be and I see two and two or which can be written as and I c n two.

Here we are continuing to take a look at the molecular formula calculations, empirical formula calculations and the masses of these. So firstly, what we need to do is calculate the number of moles. So we've got the moles of carbon, Which is the mass divided by the Molar Mass, gives us 1.215 mi. Next we have the moles of oxygen, so we've got 39, divided by 16 g per mole to give us 2.4-5 oxygen. And lastly we have flooring, so that gives us 2.437 moles. And so we divided up by the smallest number of moles, 1.2 on five. And what we get is a simple ratio of 1-2-2. And so our empirical formula is CEO two F 2 and the empirical formula max is 82g kamal, Which is also the same value of the molecular formula master we take 82, divided by 82. We get a conversion factor of one, so we multiply c 02, F two R empirical formula by one. To get the molecular formula which is equal to c 02, F two. So it's the exact same as the empirical formula


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