# Mass of C-12

1. Homework Statement

Find the 1/12 of the mass of the atom of $$\stackrel{12}{6}C$$ in kilograms.

2. Homework Equations

$$n=\frac{m}{M}$$

3. The Attempt at a Solution

How I will do this? I have the answer $$\frac{1}{1000N_A}kg$$, but don't know where it come from...

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I might be looking at this problem the wrong way, or you might want a highly specific answer, forgive me if so.

What is the mass of one mole of Carbon 12 atoms?
How many particles are there in one mole?
If X particles have a mass of Y, what is the mass of one particle?

In one mole have N (Avogadro's number) particles...

The mass of one mole C-12 atoms is:

probably 1mol=$$\frac{m}{M}$$

I don't know exactly what is M ($$\stackrel{12}{6}C$$)

$$\frac{X}{Y}=\frac{1}{Z}$$

$$Y=XZ$$

I have no clue what any of those equations mean. The mass of a mole of C-12 is 12 grams, which is equivalent to Avogadro's constant. Find the mass of one atom from this, and then 1/12 of that atom.

Can you please tell me what do you mean by the mass of a mole of C-12 is 12 grams, which is equivalent to Avogadro's constant? Can you start so I can continue?

I can't get any simpler by saying that the mass of one mole of C-12 is 12 grams, and that one mole of a substance has Avogadro's number of atoms in it. If you're not unsure of this, I advise you to read your notes or textbook.

$$1mol* 6.022 x 10^2^3mol^-^1=6.022 x 10^2^3=N$$

$$1mol=\frac{m}{12g/mol}$$

$$m=12g$$

$$\frac{1}{12}*6.022 x 10^2^3$$ =number of atoms in 1g of C-12

What to do nexT?

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If you have a number of particles equal to Avogadro's Constant, the mass of said quantity is equal to the relative molecular/formula/atomic mass in grams. 6.023*10^23 Carbon 12 atoms have a mass of 12 grams. 6.023*10^23 Hydrogen atoms have a mass of 1 gram.

So if you know the mass of Avogadro's Constant carbon-12 atoms, what is the mass of one?

If 6,022x10^23=number of atoms in 12g of C-12, than in one atom will be

$$\frac{6,022*10^23}{12g}=\frac{1}{x}$$
the mass of 1 atom will be $$\frac{12g}{6,022*10^2^3}=1,9926935*10^-^2^3$$
right? So to get the unified atomic mass unit, I should divide it by 12, so I get
$$1u=\frac{1,9926935*10^-^2^3}{12}=1,6605779*10^-^2^4g=1,6605779*10^-^2^7kg$$

and why they use 1/12 of the mass of one atom of C-12? Why not of N-12 or some other element?

right?

N-12 doesn't exist. C-12 is common.

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And how many protons and neutrons, the atom of carbon have? The isotope C-12 have 6 protons and 6 neutrons.

Right.

Carbon 14 also exists, which has 2 more neutrons. The identity of each element is defined by the amount of protons in it, so a carbon atom will always have 6 protons.

Carbon-12 is used because Carbon 12 is more abundant compared to its isotopes than other elements.

but why it is called isotope? Istope is called when there is same number of protons and different number of atomic mass number. Isn't there any "standard" carbon atom?

Carbon 14 also exists, which has 2 more neutrons. The identity of each element is defined by the amount of protons in it, so a carbon atom will always have 6 protons.

Carbon-12 is used because Carbon 12 is more abundant compared to its isotopes than other elements.
I remember in the pase , both H-1 and O-16 were used to define the "relative" mass of other natural isotopes/atoms

but why it is called isotope? Istope is called when there is same number of protons and different number of atomic mass number. Isn't there any "standard" carbon atom?
No, all of the carbon found in nature are isotopes of each other. You can't really call any one of them 'standard' because they all exist. You can call one of them the most common.