Carbon and Hydrogen

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two students each burn an organic compound. Both collect exactly zero point six six grams of carbon dioxide, and zero point two seven grams of water. Same products, same masses. So, are the two compounds the same? Hold on to your answer; by the end of this lesson, you will check it yourself. Today we learn how chemists find the percentage of carbon and hydrogen in a compound, by Liebig's combustion method. Questions on this do appear in JEE Main, so let's begin.

0:38Liebig's method: burn the sample completely

So, here is Liebig's combustion method. First, we weigh the sample. Then we burn it in a stream of dry oxygen, the O two on screen. The small two written low is a subscript: two oxygen atoms in each molecule. Okay? And it burns over heated copper two oxide, the C u O on screen. The Roman two in brackets tells you that copper is in its plus two state. Right? Now let me ask you this. When the sample burns completely, what do its carbon and its hydrogen become? Yes, exactly! Carbon becomes C O two, carbon dioxide, and hydrogen becomes H two O, which is water. The arrow means becomes. Both leave as gases, so we must catch them and weigh them. Clear? But wait. We already have a stream of oxygen. So why do we need the copper oxide at all? Because copper oxide supplies oxygen, and makes the combustion complete. All the carbon goes fully to carbon dioxide, and not to carbon monoxide, the C O at the end of the line. With me?

1:47Catch the gases in two weighed tubes

Now, how do we catch these gases? With two tubes, both weighed before we start. The gases pass first through a U-tube of anhydrous calcium chloride, the C a C l two on screen. Anhydrous means without water, and it absorbs the water. Magnesium perchlorate can be used instead. Got it? Then the gases pass through a tube of concentrated potassium hydroxide, KOH, which chemists call potash. It absorbs the carbon dioxide. Okay? After the combustion, we weigh both tubes again. The gain in mass of each tube is exactly the mass of the gas it absorbed. Right? Now a question that comes up often. Each tube catches its own gas, so the order does not matter, right? Oh no, careful, that is exactly the trap! The water tube must come first, and the potash tube after it, always. So tell me, why? Because KOH solution absorbs water as well as carbon dioxide. Clear? So imagine the wrong order. The potash tube catches water too, so carbon comes out too high. And the calcium chloride tube gets too little water, so hydrogen comes out too low. With me?

3:04From the mass of each gas to percentages

Now let's turn these masses into percentages. One mole of carbon dioxide weighs forty four grams. So tell me, how many grams of carbon are inside it? Twelve grams, yes, exactly! One carbon atom in each molecule. So forty four grams of carbon dioxide contains twelve grams of carbon. Okay? So here is the formula. Per cent carbon equals twelve over forty four, times the mass of carbon dioxide over w, times a hundred. On screen, that mass is the letter m, with C O two as a subscript. And w, the small letter below it, is the mass of the sample we burnt. Right? Now let's do the same for hydrogen. Water is eighteen grams per mole, and hydrogen is one gram per mole. So the fraction is one over eighteen, right? Oh no, careful, that is a very common slip! Each water molecule has two hydrogen atoms. So eighteen grams of water contains two grams of hydrogen, not one. Clear? So, per cent hydrogen equals two over eighteen, times the mass of water over w, times a hundred. Got it? And what about oxygen? This method gives us only carbon and hydrogen. So if the compound has only carbon, hydrogen and oxygen, per cent oxygen is a hundred, minus per cent carbon, minus per cent hydrogen. This is oxygen by difference.

4:32Example 1: percent C, H and O from CO2 and water masses

Okay? Now, Example one, an easy start. Zero point two nine grams of an organic compound gives zero point six six grams of carbon dioxide, and zero point two seven grams of water. Find the percentages of carbon and hydrogen, and of oxygen if no other element is present. Pause the video and try it first. Done? Okay, let's check. Let's do carbon first. Twelve over forty four, times zero point six six, gives zero point one eight grams of carbon. Then zero point one eight over zero point two nine, times a hundred, gives sixty two point one per cent. Right? Now let's do hydrogen. Remember, two over eighteen, times zero point two seven, gives zero point zero three grams of hydrogen. Then zero point zero three over zero point two nine, times a hundred, gives ten point three per cent. Clear? Now look. Sixty two point one plus ten point three is only seventy two point four per cent. So what makes up the rest of the compound? Oxygen, yes, exactly! No other element is present. So per cent oxygen is a hundred, minus sixty two point one, minus ten point three, which gives twenty seven point six per cent. And notice, this compound is sixty two point one per cent carbon. Keep that in mind. With me?

5:51Example 2: empirical formula CH2O, oxygen by difference

Now a harder one, Example two. Zero point four five grams of a compound of carbon, hydrogen and oxygen gives zero point six six grams of carbon dioxide, and zero point two seven grams of water. Find its empirical formula. And look, these are the same gas masses as in our puzzle! Pause the video and try it first. Done? Okay, let's check. Same carbon dioxide, so the same zero point one eight grams of carbon. For a formula we need moles, so divide by twelve, the molar mass of carbon: that gives zero point zero one five mol, short for moles. Okay? Hydrogen is again zero point zero three grams. Divide by one, the molar mass of hydrogen, and we get zero point zero three moles of hydrogen atoms. Right? Now we need the oxygen. The tubes give us only carbon and hydrogen, and the question gives no mass of oxygen. So how do we find it? By difference, yes! Zero point four five, minus the sum of zero point one eight and zero point zero three in the brackets, gives zero point two four grams of oxygen. Divide by sixteen, the molar mass of oxygen, to get zero point zero one five moles. Clear? Now the mole ratio; each colon on screen is read as to. Zero point zero one five, to zero point zero three, to zero point zero one five. Divide each by the smallest, and we get one to two to one. Got it? So the empirical formula is C H two O: one carbon, two hydrogens and one oxygen. With me? And now, our puzzle. Here per cent carbon is zero point one eight over zero point four five, times a hundred, which is forty per cent. In Example one, it was sixty two point one per cent. So the answer is no! Same gas masses, but different sample masses, so different compositions, and different compounds! Always divide by the mass of the sample you burnt, okay?

7:57Quick revision: tube order, percent C and H formulas

So, quick revision. Burn over copper oxide; the water goes into calcium chloride first, then the carbon dioxide into potash. Per cent carbon is twelve over forty four, times the mass of carbon dioxide over w, times a hundred. Per cent hydrogen is two over eighteen, times the mass of water over w, times a hundred. Oxygen comes by difference, and for a formula, change masses to moles. Now go and try the three practice problems on this page. All the best!