- Empirical, molecular, and structural formulas
- Worked example: Calculating mass percent
- Worked example: Determining an empirical formula from percent composition data
- Worked example: Determining an empirical formula from combustion data
- Elemental composition of pure substances
Worked example: Calculating mass percent
To calculate the mass percent of an element in a compound, we divide the mass of the element in 1 mole of the compound by the compound's molar mass and multiply the result by 100. Created by Sal Khan.
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- why does sal wait until the end of his calculations to worry about sig figs? shouldn't you have to account for sig figs at each step since we are multiplying, dividing, and adding measured numbers?(2 votes)
- You apply sig figs to the end of a calculation not any intermediate steps.(24 votes)
- Why did Sal use 1 mole of glucose rather than 1 molecule(if it's right way to say) of glucose for his calculation? Is it just for a practice of dimensional analysis or any other reason?
I did it following way using 1 molecule of glucose:
The ratio of mass of Carbon to 1 (molecule of) glucose
= total mass of Carbon in 1 glucose / total mass of 1 glucose
= 6*C / 6*C+12*H+6*O
= 6(12.01u) / 6(12.01u)+12(1.008u)+6(16.00u)
= 72.06u / 180.156u
= about 0.4(5 votes)
- You could really use any amount of glucose for calculating this percentage because you'd get the same result anyway. He probably used a mole because then the atomic mass units go to g/mol easily. Also, mass for a mole of glucose is way easier to measure or imagine than with just one molecule.(5 votes)
- I still don't understand where Sal got the molar mass from and why you multiply it by the number of moles of a element; is it to convert the moles to grams?(2 votes)
- moles = mass / molar mass
-> mass = moles x molar mass
You can get the molar mass of an element from the periodic table. For a molecule or compound you just add up the molar masses of all the elements that make it up.(8 votes)
- I have a question. How does significant figures work, like how do I know to round to 2 or 3? Thanks!(3 votes)
- You generally round your final answer to the significant figures of your given with the fewest sig figs. If your givens are 9.5, 5.07, 7.993, and 3.32, you'd round to 2 significant figures because the fewest in your givens is two in 9.5.
- Why is 12.01 u = 12.01 g/mole of c?(2 votes)
- Chemists defined the molar masses of elements such that a mole of atoms of a single element would be equal to its atomic mass, but in grams instead of u. So carbon's relative atomic mass is 12.01 u, but a mole of carbon atoms is 12.01 gram simply because we defined it to be so.
Hope that helps.(4 votes)
- Couldn't you just say
6(12.01u)/6(12.01u) + 12 (1.008u) + 6 (16.00u) = mass %
72.06u/72.06u + 12.096u + 96.00u = mass %
72.06u/180.156u = mass %
72.06/180.156 = mass %
0.4000 = mass & (Rounded to 4 significant figures)
Why does Sal convert unified atomic mass (u) into g/mol instead of solving for the mass % directly using unified atomic mass?(2 votes)
- "72.06u/72.06u + 12.096u + 96.00u = mass %
72.06u/180.156u = mass %"
Not sure how you arrived at that.
Sal begins with the element's relative atomic masses (in u though) and converts them into molar masses. He can do this because the magnitude of an element's relative atomic mass on the periodic table is defined to be the same as the magnitude of the mass of a mole of the element, but in grams. Then he assumes he has a mole of glucose and calculates the mass percent of carbon in grams technically. If you keep the mass in unified atomic mass units and assume you're dealing with a single molecule of glucose then the math is identical.
Hope that helps.(2 votes)
- dont u have to have to account for the sig figs during every step(0 votes)
- You should be accounting for sig figs at the end of the calculation otherwise you’re introducing rounding errors.(7 votes)
- still don't get the concept of sig figs, are we just checking for sig figs at the end calculation? why don't we account for constants when we check for it?(1 vote)
- When we are performing calculations it's important for us to take note of the actual significant figures of intermediate calculations, but there isn't a need to round off digits in the middle. We finally round off digits to have the correct sig figs at the last calculation and report our answer as such.
We do take constant's sig figs into account when we perform a calculation. It's just often measured values in a lab will have less sig figs than constants so answers are more often determined by the sig figs of measured values.
Hope that helps.(2 votes)
- Why do we have a scientific and a commonly used name. The reason I asked
is he called c6 h12 o6 glucose, but it is also called sugar(1 vote)
- Most chemicals will have both a modern scientific and older common name which are used. The reason multiple names are employed for the same chemical is because each has its own advantages and disadvantages which matter differently to different groups of people.
Scientific, or more properly systematic names, are newer names given by the IUPAC. They have the advantage of showing the structure of a chemical just from the name, but a disadvantage of usually being cumbersome and difficult to write or speak. Common names, or sometimes referred to as trivial names, are older names usually with a historical basis about its discovery or intended use. They have the advantage of being easier to communicate, but do little to tell someone about the structure of the chemical. Using water as an example, water would be the common name which everyone easily recognizes, but its systematic name is dihydrogen monoxide. The systematic name is less familiar to most people and is longer to write and speak compared to just "water", but it tell us the chemical formula of water if we break down the name. "di-" = 2, "hydrogen" = hydrogen (obi), "mon-"= 1, and "oxide" = oxygen. Putting that all together and we get H2O, water's chemical formula.
With glucose, the common name would simply be glucose (or dextrose), but the systematic name is: (2R,3S,4R,5R)-2,3,4,5,6-Pentahydroxyhexanal. The systematic name is quite long because it shows glucose's structure which is more complex than something like water. Now almost everyone, including scientists, will rather just use glucose when referring to it because that systematic name is just so inconvenient to use.
As for the whole sugar thing, that just a bunch of nomenclature. Glucose is part of a wide range of biomolecules (molecules which living things use routinely) called carbohydrates, or simply 'carbs'. A carbohydrate is composed of only carbon, hydrogen, and oxygen in a ratio of 1:2:1. The name comes the chemical formulae of these molecules looking like they were carbon atoms with water molecules bonded to them. Glucose for example has the formula C6H12O6 and looks like six carbon atoms bonded to six water molecules if you don't know what its structure is. Carbohydrates are further divided into monosaccharaides, disaccharides, and polysaccharides. Monosaccharaides and disaccharides are the smaller carbohydrates and are collectively referred to as sugars. Glucose is a monosaccharide and would therefore be considered a sugar as well. What we usually thing of as 'sugar' in everyday life is known as sucrose which is a disaccharide composed of two bonded monosaccharaides; glucose and fructose. But in chemistry sucrose isn't THE sugar, just one of many sugars which includes glucose.
So glucose would the common name of itself, and is in the broad group of carbohydrates and within that carbohydrate group it is a monosaccharide which is a synonym of sugar.
Hope that helps.(2 votes)
- If the mass of glucose is 40% then what would it be in grams?
And if mL and Molar concentration were in the equation would that change anything?
And he said that glucose is 40%...40% of what exactly?(1 vote)
- Sal was trying to calculate how much of glucose's mass is composed of carbon atoms. The 40% is saying that glucose's mass is 40% carbon.(2 votes)
- [Instructor] So right over here, I have the molecular formula for glucose. And so let's just say that I had a sample of pure glucose right over here, this is my little pile of glucose. I'm not even gonna tell you its mass, but based on the molecular formula, can you figure out the percentage of carbon by mass of my sample? Pause this video and think about it. And as a hint, I've given you the average atomic masses of carbon, hydrogen, and oxygen. All right, now let's work through this together. Now, the reason why the amount of glucose doesn't matter is because the percent carbon by mass should be the same regardless of the amount. But to help us think this through, we can imagine amount. Let's just assume that this is a mole, this is a mole of glucose. So one way we could think about it is, we say okay, for every mole of glucose, we have six moles of carbon. Because every glucose molecule has six carbon atoms. So we could say, what is going to be the mass of six moles of carbon divided by the mass of one mole of glucose? And once again, the reason why it's six moles of carbon divided by one mole of glucose is because this, if we assume this is a mole of glucose, every molecule of glucose has six carbons. So it's going to be six times as many carbon atoms or six moles of carbon. Now, what is this going to be? Well, this is going to be equal to, it's going to be in our numerator, we're going to have six moles of carbon times the molar mass of carbon. Well, what's that going to be? Well, we can get that from the average atomic mass of carbon. If the average atomic mass is 12.01 universal atomic mass units, the molar mass is going to be 12.01 grams per mole of carbon. So times 12.01 grams per mole of carbon. And notice the numerator will be just left with grams. And then in the denominator, what are we going to have? Well, the mass of one mole of glucose, for every glucose molecule, we have six carbons, 12 hydrogens, and six oxygens. So it's going to be the mass of six moles of carbon, 12 moles of hydrogen, and six moles of oxygen. So it's going to be what we just had up here, it's going to be six moles of carbon times the molar mass of carbon, 12.01 grams per mole of carbon. To that, we are going to add the mass of 12 moles of hydrogen. So 12 moles of hydrogen times the molar mass of hydrogen, which is going to be 1.008 grams per mole of hydrogen. Plus six moles of oxygen, times the molar mass of oxygen, which is going to be 16.00 grams per mole of oxygen. And the good thing is, down here, the units cancel out, so we're left with just grams in the denominator. And that makes sense. We're gonna end up with grams in the numerator, grams in the denominator, the units will cancel out, and we'll get a pure percentage at the end. So let's see, in the numerator, six times 12.01 is 72.06. And then in the denominator, I'm just going to do the pure calculation first, and then I'm gonna worry about significant figures. So in the denominator, we have 72.06 plus, let's see, 12 times 1.008 is 12.096, and then we have plus six times 16 is 96.00, and this will be equal to 72, if we're just thinking about the pure calculation, before we think about significant figures, 72.06 divided by, let's see, if I add 72 to 12, I get 84, plus 96, I get 180.156. Did I do that right? If I were just to add up everything, not even think about significant figures. So we can type this into a calculator but we should remind ourselves that our final answer should have no more than four significant figures. Because even down here, if we were just doing this blue calculation here, that should only have four significant figures, it would have gotten us to the hundredths place, and so when we add things together, we should get no more than the hundredths place, but even if we rounded over there for significant figures purposes, we would still have at least four, we'd actually have five significant figures. So this four significant figures is our significant figures limiting factor. So we just have to calculate this and round to four significant figures. 72.06 divided by 180.156 is equal to, and if we round to four significant figures, this will be .4000. So this will be, I'll say, approximately equal to 0.4000. Or we could say 40% or 40.00% carbon by mass when we round to four significant figures. And we are done.