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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Einstein on February 25, 2014, 09:12:19 pm

Title: The Mole
Post by: Einstein on February 25, 2014, 09:12:19 pm

can someone help me out with these questions, not sure where to start.

Thanks,

Title: Re: The Mole
Post by: swagsxcboi on February 25, 2014, 09:51:50 pm

one mole = 6.023 x 10²³ particles
particles of CH4 = 5.2 x 10²³

n(CH4) = 5.2 x 10²³ / 6.023 x 10²³
a) = 0.86335 mols

n(CH4) x 1 = n(C)
b) n(C) = 0.86335 mols

c) n(CH4) x 4 = n(H)
n(H) = 3.453428 mols

Title: Re: The Mole
Post by: Lizzy7 on February 25, 2014, 10:04:22 pm

Basically what swagsxcboi said but instead of 5.2*10^23 I saw 5.2*10^24 from your screen shot :)


You have to use n=N/Na ; where n=number of moles, N= number of particles and Na= Avogadro's number

a) n (CH4) =N/Na
=5.2*10^24/6.02*10^23
=8.63787 (to 2 significant figs) = 8.6 mol of methane molecules

b) Here you have to use mole ratios. So you have 1 mole of CH4 and in that 1 mole of CH4 there is 1 mole of Carbon
n(C)= 8.63787 * 1 = 8.6 mol of Carbon atoms

c) In 1 mole of CH4 there is 4 mole  of Hydrogen atoms
n(H) =8.673787*4= 34.5515 (to 2 sig figs) = 35 mol of Hydrogen atoms

Title: Re: The Mole
Post by: Einstein on February 25, 2014, 10:41:13 pm

Thank you to both of you :) +1, thanks heaps :)))