Because us have, because that the mass of one atom the carbon 12, call it $m(\ce^12C)$, that

$$m(\ce^12C) = \pu12 amu$$

and furthermore

$$\pu1 mol \cdot m(\ce^12C) = \pu12 g$$

therefore

$$m(\ce^12C) = \pu12 amu = \pu12 g/mol$$

So finally we obtain that $\pu1 g/mol = \pu1 amu$ .

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However, my cg-tower.com teacher is telling me that those are two completely different things and that ns am confused in between the mass per atom and also the mass every $6.022\cdot10^23$ atoms. I can"t know how, and also this is yes, really bugging me, so aid is really appreciated.

Note that this needs the mole to be a number (or a "constant"), which may be wherein I"m wrong.


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edited Mar 14 "18 in ~ 13:33
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mhchem
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You are correct, yet to do it a little an ext clear friend can encompass the suspect "atom" in the denominator that amu:

$$\beginalignm_\ceC^12 &= \pu12amu atom^-1 \\ \\m_\ceC^12 &= \pu12g mol^-1 \\ \\\pu12amu atom^-1 &= \pu12g mol^-1 \\ \\\pu1amu atom^-1 &= \pu1g mol^-1\endalign$$

In other words, the ratio the amu/atom is the same as the ratio of g/mol. The interpretations of amu and also moles were intentionally favored to do that take place (I"m surprised your teacher didn"t define this, actually). This enables us to easily relate masses at the atomic range to masses at the macroscopic scale.

To examine this, look at the massive of one amu when converted to grams:

$\pu1amu= \pu1.6605E-24 g$

Now division one gram by one mole:

$\pu1g mol^-1= \frac\pu1 g\pu6.022E23 atom = \pu1.6605E-24 g atom^-1$

It"s the very same number! Therefore:

$\pu1g mol^-1= \pu 1 amu atom^-1$


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edited Oct 3 "18 in ~ 11:27
answered Jun 20 "14 at 16:26
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You must be more careful v your units. The erroneous an outcome is that you room equating a worth in amu (a measure of mass, choose grams) through a value in grams every mole (an invariant home of an aspect or compound, regardless of the amount girlfriend have).


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answered Jun 20 "14 in ~ 14:00
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There room two points that on regular basis mystify science students:

anything to carry out with quantity of substance (now come be called "chemical amount"), the mole, and also the Avogadro consistent (or the Avogadro number), and also

anything to execute with the now-you-see-me-now-you-don"t radian. Allow me address the first.

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If we have actually a general number of entities of type X (e.g. X is the chemical symbol) represented by N(X), the equivalent chemical lot of X is denoted through n(X), i m sorry is an aggregate of N(X) entities. In symbols:n(X) = N(X) ent, wherein ent represents an amount of one entity (atom, molecule, ion, sub-atomic particle, . . .), i.e. The entity itself.

The Avogadro number is the (dimensionless) ratio of one gram come one "atomic fixed unit" (now referred to as dalton, Da): g/Da. One mole is one an Avogadro number of entities: mol = (g/Da) ent. Hence we have the crucial relationship: Da/ent = g/mol = kg/kmol, exactly. In various other words, at the atomic level, the suitable unit for amount-specific fixed ("molar" mass) is dalton every entity--and, due to the fact that of the mole definition as one Avogadro variety of entities, dalton per entity is precisely equal come the macroscopic devices gram per mole or kilogram per kilomole.

The vital problem is that IUPAC walk not have actually a recognised symbol for one entity. That is sometimes (incorrectly) thought of together the (dimensionless) number one. In which instance the "mole" is simply one more name because that the Avogadro number: "mol = g/Da". In this case we have actually the (incorrect) relationship: "Da = g/mol". Tables that "atomic weights" perform the numerical values of atomic-scale masses in daltons--e.g. Ar(O) = ma(O)/Da = 16. The equivalent amount-specific massive is M(O) = 16 Da/ent; and also this is (exactly) same to 16 g/mol or 16 kg/kmol.