CHEM!! HELP (1 Viewer)

dumNerd

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Just started this stuff, need help with this question: Calculate the number of hydrogen atoms in 0.35mol of methane (CH4)
How do I find the mol(H4)
 

username_2

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Mol of hydrogen atoms = n(CH4)*4 as there are 4 hydrogens for every molecule of CH4. Therefore, n(H) = 0.35*4 = 1.4 mol
 

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Mol of hydrogen atoms = n(CH4)*4 as there are 4 hydrogens for every molecule of CH4. Therefore, n(H) = 0.35*4 = 1.4 mol
Assuming by "number" it actually means the number of H atoms, you'd then multiply it by avogadro's number (6.022*10^23)
 

dumNerd

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Mol of hydrogen atoms = n(CH4)*4 as there are 4 hydrogens for every molecule of CH4. Therefore, n(H) = 0.35*4 = 1.4 mol
wait what how is four hydrogen a greater value than four hydrogen + carbon. Also isnt there 4 hydrogen for every carbon
 

idkkdi

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wait what how is four hydrogen a greater value than four hydrogen + carbon. Also isnt there 4 hydrogen for every carbon
0.35 mol of CH4 = 0.35 mol x (CH4) = 0.35 mol x C + 0.35 mol x H4

0.35 mol x H4 = 0.35 mol x 4 x H
= 1.4 mol H
 

Minari243

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wait what how is four hydrogen a greater value than four hydrogen + carbon. Also isnt there 4 hydrogen for every carbon
I guess just try to see moles as a count of how many particles there are in a substance. So for every particle of CH4, there are 4 particles of hydrogen. Thus, if you have 0.35 mols of CH4, you get 0.35x4, which is 1.4 mols of Hydrogen
 

dumNerd

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0.35 mol of CH4 = 0.35 mol x (CH4) = 0.35 mol x C + 0.35 mol x H4

0.35 mol x H4 = 0.35 mol x 4 x H
= 1.4 mol H
Oh I see I thought it meant that there was 0.35 moles of the compound as a wholw
 

idkkdi

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Oops u get what I meant
The thing with moles is that you don't interpret it like how you were doing so before. Rather when you talk about moles, you assign it to a particular substance indicating that there is 6.022 x 10^23 particles of what you specify per mole. If you specify H4, then 6.022 x 10^23 of H4 which is obviously equivalent to that value x 4 for h. If you specify CH4, you have 6.022 x 10^23 of CH4 which you can further express in its components if need be.
 
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