I know that relative atomic mass of $\ce{^{12}C}$ is $12~\mathrm{u}$.

Therefore mass of $1~\mathrm{mol}~\ce{C} = 12~\mathrm{g}$ \begin{align} \text{mass of }6.022\cdot 10^{23} \text{ C atoms} &= 12~\mathrm{g}\\ \text{mass of }1 \text{ C atom} &= \frac{12~\mathrm{g}}{6.022\cdot10^{23}}\\ &=\boxed{1.99\cdot10^{-23}~\mathrm{g}},\\ \end{align}


\begin{align} 1~\mathrm{u} &= 1.66\cdot10^{-24}~\mathrm{g}\\ \implies \text{Mass of }1~\ce{^{12}C}\text{ atom} &= 1.66\cdot 12 \cdot 10^{-24}~\mathrm{g}\\ &= 19.92\cdot 10^{-24}~\mathrm{g}\\ &= 1.992\cdot 10^{-23}~\mathrm{g}.\\ \end{align}

Which is correct?

Also what is dimension formula of relative atomic mass, molar mass?

Why is unit of molar mass $\dfrac{\text{gram}}{\text{mole}}$ and not just $\text{gram}$ Since when has this $\text{mol}$ become a unit? It's just a number.

  • 1
    $\begingroup$ The number in the rectangle was off by 46 orders of magnitude! (Sign error: should have been $10^{-23}$ not $10^{23}$ as it was originally.) I'm edited it to be correct, presuming it was just a typo, but let me know if you have any confusion about why. $\endgroup$
    – Curt F.
    Apr 23 '15 at 13:28
  • 3
    $\begingroup$ The mole became a SI unit in 1971. $\endgroup$ Apr 23 '15 at 13:29
  • $\begingroup$ Thank you both! @NicolauSakerNeto actually it was just a typo! the question was about which approach is correct? some of my friends were saying that second approach is dimensionally incorrect. $\endgroup$
    – Max Payne
    Apr 23 '15 at 13:30
  • $\begingroup$ Sorry if my previous comment seemed condescending, I should have realized it was a typo but recently I commented a question involving an incorrect exponent sign and that led me astray. $\endgroup$ Apr 23 '15 at 14:28

Both approaches are correct.

Avogadro's number is $6.02214129\times 10^{23}$ and represents the number of carbon-12 atoms in 12 grams of unbound carbon-12 in the ground electronic state.

$12$grams$/6.02214129\times 10^{23} = 1.9926467\times 10^{-23}$grams

The unified atomic mass unit (u) is $1.660538921 \times 10^{-24}$ grams

$12 \times 1.660538921 \times 10^{-24}$ grams $ = 1.9926467\times 10^{-23}$grams

  • 2
    $\begingroup$ The Avogadro constant has a unit: $L = 6.022 141 79 30 \cdot 10^{23}~\color{red}{\mathrm{mol}^{−1}}$. $\endgroup$ Apr 23 '15 at 14:15
  • 4
    $\begingroup$ The difference between the Avogadro constant (dimensional) and Avogadro's number (dimensionless) is quite subtle, and often overlooked. $\endgroup$ Apr 23 '15 at 14:22
  • $\begingroup$ Yea Niclau Saker and Martin enlightened me! actually we are dividing by Avogadaros number and not avogadaro constant, so the answer is dimensionally correct! $\endgroup$
    – Max Payne
    Apr 23 '15 at 14:33
  • $\begingroup$ @Martin I changed answer to say avogrado's number instead $\endgroup$
    – DavePhD
    Apr 23 '15 at 14:33
  • $\begingroup$ If you want to use units then actually L=$6.0221417930 \times 10^{23}$ atoms/mole $\endgroup$
    – MaxW
    Jan 21 '16 at 16:30

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