Question 1

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

```

Accepted Answer

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.

Question 2

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?

Accepted Answer

You apply sig figs to the end of a calculation not any intermediate steps.

Question 3

I have a question. How does significant figures work, like how do I know to round to 2 or 3? Thanks!

Accepted Answer

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.

https://www.khanacademy.org/math/arithmetic-home/arith-review-decimals#arithmetic-significant-figures-tutorial

https://www.khanacademy.org/math/arithmetic-home/arith-review-decimals#arithmetic-significant-figures-tutorial

Question 4

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?

Accepted Answer

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.

Question 5

Why is 12.01 u = 12.01 g/mole of c?

Accepted Answer

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.

Hope that helps.

Question 6

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?

Accepted Answer

"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.

Question 7

Is the mass percent really independent of given mass? I mean what if I take 2 moles of glucose? Would it affect the mass percent?

Accepted Answer

If we had two moles of glucose, then the moles of carbon, hydrogen, and oxygen would be twice that for the one mole of glucose calculation (since we have twice as many glucose molecules). So yeah, the mass percent would be consistent regardless of how many glucose molecules we had.

Hope that helps.

Hope that helps.

Question 8

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?

Accepted Answer

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.

Question 9

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

Accepted Answer

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.

Question 10

dont u have to have to account for the sig figs during every step

Accepted Answer

You should be accounting for sig figs at the end of the calculation otherwise you’re introducing rounding errors.

Chemistry library

Course: Chemistry library > Unit 1

Worked example: Calculating mass percent

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