Because we have, for the fixed of an atom of carbon 12, call it $m(\ce^12C)$, that

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

and furthermore

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


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

So ultimately we acquire that $\pu1 g/mol = \pu1 amu$ .

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However, my teacher is informing me the those room two fully different things and that i am confused in between the mass per atom and the mass every $6.022\cdot10^23$ atoms. I can"t understand how, and this is really bugging me, so help is really appreciated.

Note that this requires the mole to it is in a number (or a "constant"), which might be wherein I"m wrong.

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edited Mar 14 "18 in ~ 13:33

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You room correct, however to do it a little much more clear you can include the presume "atom" in the denominator of 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 various other words, the ratio the amu/atom is the very same as the proportion of g/mol. The meanings of amu and also moles to be intentionally liked to make that take place (I"m surprised your teacher didn"t explain this, actually). This permits us to quickly relate masses in ~ the atomic scale to masses at the macroscopic scale.

To check this, look at the fixed of an amu when converted to grams:

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

Now divide one gram by one mole:

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

It"s the exact same number! Therefore:

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

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edited Oct 3 "18 at 11:27
answer Jun 20 "14 in ~ 16:26

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You must be more careful v your units. The erroneous result is the you room equating a value in amu (a measure of mass, choose grams) with a worth in grams every mole (an invariant residential property of an facet or compound, regardless of the amount girlfriend have).

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reply Jun 20 "14 at 14:00

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There room two points that routinely mystify scientific research students:

anything to do with lot of substance (now to be dubbed "chemical amount"), the mole, and the Avogadro constant (or the Avogadro number), and also

anything to do with the now-you-see-me-now-you-don"t radian. Let me deal with the first.

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If we have actually a general variety of entities of type X (e.g. X is the chemistry symbol) represented by N(X), the equivalent chemical amount of X is denoted by n(X), which is an accumulation of N(X) entities. In symbols:n(X) = N(X) ent, wherein ent to represent an amount of one reality (atom, molecule, ion, sub-atomic particle, . . .), i.e. The reality itself.

The Avogadro number is the (dimensionless) ratio of one gram come one "atomic massive unit" (now called dalton, Da): g/Da. One mole is one an Avogadro variety of entities: mol = (g/Da) ent. For this reason we have the essential relationship: Da/ent = g/mol = kg/kmol, exactly. In various other words, in ~ the atom level, the suitable unit for amount-specific massive ("molar" mass) is dalton every entity--and, since of the mole meaning as an Avogadro number of entities, dalton per entity is specifically equal to the macroscopic systems gram every mole or kilogram per kilomole.

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