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HELP with chemistry; number of particles/molecules/atoms

I have a rough understanding of working this out: Work out number of moles by doing mass/Mr, then multiplying the number of moles with Na(6.02*10^23).

"Calculate number of moles for phosphorus(III) chloride: PCl3" I got 1.204*10^23

"....atoms in 3.55g of chlorine:Cl2" How would I do this one?
buuuuuuuuuuump
Reply 2
Avatar for ranz
ranz
5
Original post by Perfection Ace
buuuuuuuuuuump


If i remember correctly, you first find the moles so

3.55/Mr Mr of Cl2= (35.5 x2) =71 so
3.55/71 = 0.04733 mole
then times moles by avagadros constant (6.023x10^23)

so 0.04733 x 6.023x10^23 = whatever.

this gives you the number of cl2 in one mole.

its been a while >.< gl :biggrin:
(edited 7 years ago)
Original post by ranz
If i remember correctly, you first find the moles so

3.55/Mr Mr of Cl2= (35.5 x2) =71 so
3.55/71 = 0.04733 mole
then times moles by avagadros constant (6.023x10^23)

so 0.04733 x 6.023x10^23 = whatever.

this gives you the number of cl2 in one mole.

its been a while >.< gl :biggrin:


That's what I thought too! But it's Cl2? So don't I multiply or divide by 2 or something...... :frown:
Reply 4
Avatar for ranz
ranz
5
Original post by Perfection Ace
That's what I thought too! But it's Cl2? So don't I multiply or divide by 2 or something...... :frown:


well, u would divide the moles by 2 only if they ask for number of atoms of Cl in Cl2 ,
however if they ask number of molecules of cl2 in x grams, then u dont divide.


u DONT divide grams and then find moles noo! u find moles then divide by 2.

example q:number of atoms of S in S8 as 2grams,
you find mole of S8 so: g/mr > 2/(8x32.1) = 0.25 mol
then u divide 0.25/8 =0.0311 mol [for one S]

then 0.0311 mol x avagadros constant for number of atoms of S in 2g of S8


but i advice u also ask ur teacher, its been a while since i did chem but from what i can remember this is how its done.


if they ask for ex. for number of molecules of S8 in x grams, then u do the method i mentioned before.



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(edited 7 years ago)

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