Oxygen by Difference and Complete Analysis

Transcript

0:00Introduction

Hello everyone! Before we start, one quick question. In this chapter we found carbon and hydrogen by burning the compound, nitrogen by Dumas and by Kjeldahl, halogens and sulphur by Carius, and even phosphorus. But tell me, which common element did we never measure? Oxygen! You see, there is no simple direct method for oxygen. So how do we know how much oxygen a compound has? The answer is beautifully lazy: we don't measure it at all. And look, this is where the whole chapter comes together. In JEE, the full question, from combustion data to the molecular formula, is built on exactly this. So, let's begin.

0:42Oxygen: found by difference

So, first the bad news. There is no simple direct method for oxygen, okay? Liebig gives us carbon and hydrogen. Dumas or Kjeldahl gives nitrogen. Carius gives the halogens and sulphur. But for oxygen, nothing like that. Now think. The whole compound is one hundred percent. And we already know the percentage of every other element in it. So, what is left over? Yes, exactly! Whatever is left over must be oxygen. Percent oxygen equals one hundred, minus the sum of all the other percentages. That is why we call it oxygen by difference. Clear? And if the question gives masses instead of percentages, same idea. The mass of oxygen, m O, equals the sample mass w, minus the masses of carbon, hydrogen, nitrogen, and the little dots on screen mean every other element you found, okay? All the others, not just carbon and hydrogen. Remember this; it will save you in the second example.

1:44Complete analysis in four steps

Alright, now let's put the whole chapter together. This is called complete analysis. Step one: find every other element by its own method. Carbon and hydrogen from the carbon dioxide and water of combustion, nitrogen from Dumas or Kjeldahl, halogens and sulphur from Carius, okay? Step two: oxygen by difference. Now tell me, why must oxygen always come last? Because it needs all the others first, right? You cannot subtract a number you don't have yet. Step three: convert each percentage to moles per one hundred grams, and find the simplest ratio, exactly like we did in Chapter three. The arrow on screen just means: then. Percentages become grams, grams become moles, and then we divide by the smallest. And step four: use the molar mass. Molar mass divided by the empirical formula mass gives n, and n times the empirical formula is the molecular formula. Four steps. That's it, that is the complete analysis. Got it?

2:47Topper's trick: go straight to moles

Now a trick that toppers use. When the question gives you masses, don't go round through percentages. Go straight to moles of atoms. One mole of C O two contains one mole of carbon atoms; here the arrow on screen just means: gives. So moles of carbon are simply the mass of C O two, divided by forty four. Now water. One mole of water gives how many moles of hydrogen atoms? Two, that's right! H two O has two hydrogens. So divide the mass of water by eighteen, and then multiply by two. And nitrogen comes off as N two gas. At STP, twenty two thousand four hundred millilitres is one mole of N two. But one mole of N two means two moles of nitrogen atoms, okay? Students forget this factor of two again and again. Don't! And one bonus check. Suppose the masses of all the elements you found already add up to the full sample mass w, okay? Then what is left for oxygen? Nothing! There is no oxygen in that compound at all. So always check this before you put oxygen into the formula.

4:00Example 1: percent oxygen and empirical formula CH3O

Now, remember our question from the start? Here it is, with numbers. An organic compound contains thirty eight point seven percent carbon, nine point seven percent hydrogen, and no other element except oxygen. Find the percentage of oxygen, and the empirical formula. Pause the video, and try it first. Done? Okay, let's check. Oxygen by difference. Carbon plus hydrogen is forty eight point four. So oxygen is one hundred minus forty eight point four: fifty one point six percent. Nobody measured the oxygen, and still we know it. Opening question, solved! Now the formula. Carbon divided by twelve is three point two three, and hydrogen divided by one is nine point seven. That is one is to three. So the formula is C H three, right? No, no, careful! Fifty one point six percent oxygen just went missing. More than half the compound! Oxygen belongs in the formula too. So, moles per one hundred grams. Carbon: thirty eight point seven by twelve, three point two three. Hydrogen: nine point seven by one, nine point seven. Oxygen: fifty one point six by sixteen, three point two three again. Divide everything by the smallest, three point two three. One, three, one. So the empirical formula is C H three O, with the oxygen in it this time. Really simple, right?

5:33Example 2: molecular formula of glycine from analysis data

Now the full JEE style question. Read it with me. Zero point one five zero gram of a compound gives zero point one seven six gram of C O two, and zero point zero nine zero gram of water. Another zero point one five zero gram gives, at STP, twenty two point four millilitres of N two. Its molar mass is seventy five. Find the molecular formula. Pause the video and try it. Hint: go straight to moles. Done? Okay, let's check. Carbon first. Zero point one seven six by forty four is zero point zero zero four mole of carbon. Times twelve, that is zero point zero four eight gram. Hydrogen. Zero point zero nine zero by eighteen is zero point zero zero five mole of water. Two hydrogens in each, so zero point zero one zero mole of hydrogen atoms. Times one, zero point zero one zero gram. Nitrogen. Twenty two point four by twenty two thousand four hundred is zero point zero zero one mole of N two. Times two, remember the trick: zero point zero zero two mole of nitrogen atoms. Times fourteen, zero point zero two eight gram. Alright, now oxygen by difference. Sample minus carbon minus hydrogen: zero point one five zero minus zero point zero five eight, so zero point zero nine two gram, right? No, no! Who did we forget? Nitrogen! Subtract every other element, not just carbon and hydrogen. This is the JEE trap. Carbon, hydrogen and nitrogen together come to zero point zero eight six gram. Both samples weigh the same, so we can subtract directly. Oxygen is zero point one five zero minus zero point zero eight six: zero point zero six four gram. Divided by sixteen, zero point zero zero four mole. Now the mole ratio of carbon, hydrogen, nitrogen and oxygen, in thousandths of a mole: four is to ten is to two is to four. Divide by two: two is to five is to one is to two. So the empirical formula is C two H five N O two. Last step, the molar mass. Empirical formula mass: twenty four plus five plus fourteen plus thirty two, that is seventy five. The molar mass is also seventy five, so n is one. The molecular formula is C two H five N O two, and yes, this compound is glycine! Four elements, one never measured, and still the complete formula.

8:11Quick revision: oxygen by difference to molecular formula

So, quick revision. Oxygen has no simple direct method, so it is one hundred percent, minus the sum of all the other percentages. Subtract every other element, nitrogen and halogens too. Then moles, simplest ratio, molar mass, and the molecular formula. Now go and try the two practice problems on this page. You will find them easy, I promise. All the best!