How many $"hydrogen ATOMS"$ are present in a $23.7*g$ mass of $"methane"$ ?

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Answer 1 · 2016-10-03T07:43:15

There are approx. $N_Axx4xx2$ hydrogen atoms in such a mass of methane.

In $16.04*g$ methane, there are $N_A$ methane molecules, where $N_A="Avogadro's number"=6.022xx10^23$ .

And so $"moles of methane"$ $=$ $(23.7*g)/(12.04*g*mol^-1)$ $=$ $1.97*mol$

And thus $"atoms of hydrogen"$ $=$ $(23.7*g)/(12.04*g*mol^-1)xx"4 H atoms"*mol^-1xxN_A" H atoms"$

$~=$ $4.80xx10^24" hydrogen atoms"$

How many "hydrogen ATOMS""hydrogen ATOMS" are present in a 23.7*g23.7*g mass of "methane""methane"?