Carbon and Hydrogen: Liebig's Method

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I take a tiny pinch of an organic compound, just point two grams, and I burn it completely. Its carbon and hydrogen fly away as gases. So how can a chemist still tell you exactly how much carbon was inside? But you cannot weigh something that has flown away, right? Or can you? Today's answer is three steps: burn it, trap it, weigh it. And look, JEE loves this experiment: the percentage formulas, the order of the tubes, the empirical formula. So, let's begin.

0:37Burn it: carbon and hydrogen become gases

So, this is Liebig's combustion method. First, weigh the sample. We call its mass small w, in grams, okay? Then heat it with excess copper two oxide, the C u O on screen, in a stream of oxygen gas. Excess means more than enough, got it? Now tell me. When the sample burns, where does its carbon go? And where does its hydrogen go? Yes, exactly! Carbon becomes carbon dioxide, and hydrogen becomes water. Both leave as gases, so we must catch them, right? But wait. Why copper oxide at all? Why not burn it in oxygen alone? See, copper oxide gives oxygen mixed right into the sample, and it turns any carbon monoxide, the C O you see on screen, into carbon dioxide. So the combustion is complete, and all the carbon ends up as carbon dioxide. Clear?

1:28Trap it: two weighed tubes

Now, how do we catch them? With two tubes, both weighed before we start. The water goes into a U-tube of anhydrous calcium chloride, the C a C l two on screen. It absorbs the water. And the carbon dioxide goes into a tube of concentrated potassium hydroxide, KOH, okay? It absorbs the carbon dioxide. Afterwards, weigh both tubes again. The calcium chloride tube is heavier: that gain is the mass of water. The KOH tube is heavier: that gain is the mass of carbon dioxide. One small care point. The oxygen and the apparatus must be dry, and free of carbon dioxide. Why? Because any water or carbon dioxide coming in from outside also gets absorbed, and counted as if it came from the sample. Your results come out too high. Got it?

2:22JEE trap: which tube comes first?

Alright, examiners love this one. Does the order of the two tubes matter? Each tube catches its own gas, so any order is fine, right? No, no, careful! This is the JEE trap. The calcium chloride tube must come first, and then the KOH tube. Always. But why? Because concentrated KOH also absorbs water! It does not take only carbon dioxide. So in the reverse order, the KOH tube would gain the mass of both gases, water and carbon dioxide. Water gets counted as carbon dioxide, and your carbon comes out too high. So remember: water first, then carbon dioxide, clear?

3:05Weigh it: from gas mass to percentage

Now, weigh it. The tube gives the mass of carbon dioxide, but we want carbon. So tell me: the molar mass of carbon dioxide is forty four. How much of that forty four is carbon? Twelve, yes! Exactly. One carbon atom, twelve, and two oxygen atoms, thirty two: total forty four. So the mass of carbon is twelve by forty four, times the mass of carbon dioxide. Divide by w, times hundred: percent carbon equals twelve by forty four, times m C O two by w, times hundred. Here m C O two is the gain of the KOH tube, the mass of carbon dioxide, okay? Now hydrogen, from water. The molar mass of water is eighteen, and the atomic mass of hydrogen is one. So one by eighteen, right? No, no! Count the atoms first. Water is H two O: two hydrogen atoms in every molecule. So in every eighteen grams of water, two grams are hydrogen. Percent hydrogen equals two by eighteen, times m H two O by w, times hundred, where m H two O is the gain of the calcium chloride tube, the mass of water. Forget this two, and your hydrogen comes out half. Classic mistake, okay? And oxygen? Our tubes catch only water and carbon dioxide, so oxygen is never weighed, right? If the compound has only carbon, hydrogen and oxygen, find oxygen by difference: hundred, minus percent carbon, minus percent hydrogen. Nice and simple.

4:38Topper's trick: for a formula, go straight to moles

Now a topper's trick. When the question asks for the empirical formula, don't stop at percentages. Go straight to moles. Each carbon dioxide molecule has one carbon atom, so moles of carbon equals the mass of carbon dioxide by forty four. That's it! Each water molecule has two hydrogen atoms, so moles of hydrogen equals two times the mass of water by eighteen. There is our two again, see? And oxygen, by difference, but in grams: w, minus the mass of carbon, minus the mass of hydrogen. Then divide by sixteen to get its moles. Never subtract moles from grams. Clear?

5:18Example 1: percentage of C, H and O in a sample

Now, example one. Remember our pinch of point two grams from the start? Here it is. Point two zero zero grams of an organic compound gives point two nine three grams of carbon dioxide, and point one two zero grams of water. Find the percentages of carbon and hydrogen, and of oxygen by difference. Pause the video, and try it first. Done? Okay, let's check. Carbon first. Twelve by forty four, times point two nine three, gives point zero seven nine nine grams of carbon. Divide by point two zero zero, times hundred: forty point zero percent. So the carbon that flew away? We caught it, and we weighed it! Hydrogen: two by eighteen, times point one two zero by point two zero zero, times hundred. Six point six seven percent. And see, two by eighteen, not one by eighteen, okay? Oxygen, by difference. Forty plus six point six seven is about forty six point seven. Hundred minus forty six point seven: fifty three point three percent oxygen. And we're done!

6:24Example 2: empirical formula C2H6O from tube masses

Now a harder one, JEE level. In a Liebig estimation on point two three zero grams of a compound containing carbon, hydrogen and oxygen, the calcium chloride tube gains point two seven zero grams, and the KOH tube gains point four four zero grams. Find the empirical formula. Pause and try. Hint: go straight to moles. Done? Okay, let's check. First, which gain is which gas? Calcium chloride catches water, so the water is point two seven zero grams. KOH catches carbon dioxide, so that is point four four zero grams. Don't swap them, okay? Carbon: point four four zero by forty four, that is point zero one zero mole. Times twelve: point one two zero grams of carbon. Now hydrogen. Point two seven zero by eighteen is point zero one five. So, point zero one five mole of hydrogen, right? No, no, careful! That is moles of water, not of hydrogen. Each water has two hydrogen atoms, so times two: point zero three zero mole of hydrogen, which is point zero three zero grams. Oxygen by difference, in grams. Carbon plus hydrogen is point one five zero grams. Point two three zero minus point one five zero: point zero eight zero grams of oxygen. Divide by sixteen: point zero zero five mole. Now the ratio: point zero one zero, to point zero three zero, to point zero zero five. Divide each by the smallest, point zero zero five. Two, six, one! So the empirical formula is C two H six O, okay? See? Burn it, trap it, weigh it, and out comes the formula.

8:18Quick revision: tube order and the 12/44, 2/18 formulas

So, quick revision. Burn with excess copper oxide. Water goes into the calcium chloride tube first, then carbon dioxide into the KOH tube. Percent carbon: twelve by forty four, percent hydrogen: two by eighteen, each times the gas mass by w, times hundred. Oxygen comes by difference, and for a formula, go straight to moles. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!