how do we culculate the number of calcium ions in 50g of calcium carbonate?
do i use this formulea n=m/M?
or Avogadras number?
i have tried using the avogadra's number but i don't know how to use on ions.
|
how do we culculate the number of calcium ions in 50g of calcium carbonate?
do i use this formulea n=m/M?
or Avogadras number?
i have tried using the avogadra's number but i don't know how to use on ions.
pretty much.
use the molar mass of CaCO3 which is about 100g/mol
moles CaCO3 will be mass/molar mass which you can work out. lets call it x
then if you look at the stoichiometry
CaC0<sub>3</sub> ----> Ca<sup>2+</sup> + CO<sub>3</sub> <sup>2-</sup>
so one mole of CaCO3 contains one mole of Ca<sup>2+</sup> ions meaning that x moles contains ..... of Ca<sup>2+</sup> ions
OH, I get it now ^^, tyvm for your help![]()
i am having a hard time trying to get the empirical formula of a hydrocarbon that contain 82.7% carbon and 17.3% hydroegn. the answer is C2H5.
but when i tried to do it is not the same.
82.7 divided by the molar mass of carbon which is 12, gives 6.89 mols.
and 17.3 divided by the molar mass of hydrogen which is 1, gives 17.3.
so the ratio is 6.89:17.3, is there a way so makes it easier to read like 2:5?
thanks in advance.
divide each number in the ratio,by the smallest value
so in this case divide each number by 6.89
gives you (6.89/6.89)17.3/6.89)
= (1:2.5)
then multiply each number by 2 to make it an interger value, as half an atom doesnt really make sense
(1*2)2.5*2) = 2:5
so do I multiply the lowest number in order for it to become a interger number?
cause if i multiply by 3 the answer would be wrong![]()
thanks in advance![]()
i'm not really sure of the question, you want an empirical formula which is the ratio of atoms in its simplest, interger form.
if you multiplied the ratio by 3 you would get 3:7.5 which is the correct ratio. but it isn't the simplest nor is it an interger form so it is not the empirical formula
if you multiplied by 4 you would get a ratio of 4:10 this is correct ratio and interger
but this is not the simplest form.
you multiply by the smallest number possible to achieve interger values which is 2 in this case
sorry if this doesnt answer the question but i'm not really sure what the the question was
C2H5 would get the proportions right (I rely on your calculations here), but I am not sure there is such a compound.
Butane C4H10 is probably what you are looking for.
Hope this helps.
C<sub>4</sub>H<sub>10</sub> is not the empirical formula which is what the question was asking.
opps, my mistake, sorry if i wasn't clear, but you answered my question there.Originally Posted by organic god
thanks :-D
A precipitate of 1.34 g of Fe2O3 was obtained by treating a vitamin supplement
dissolved in water with NaOH and then heating. What is the mass of iron in the
tablet?
the answer is:
(i) n(Fe2O3) = 1.34 g /159.8 g mol−1
= 8.39 x 10−3 mol
(ii) n(Fe) = 2 x 8.39 x 10−3 mol
= 1.68 x 10−2 mol
(iii) m(Fe) = 1.68 x 10−2 mol x 55.9 g mol−1
= 0.938 g
what i don't get is in (iii) why is the molar mass for Fe 55.9? isn't the molar mass of Fe2 111.8?
thanks in advance. :-D
You have already multiplied the number of moles by two in part ii) to find how many moles of iron are present. The Fe2 in this formula is only a ratio; the iron oxide does not actually form molecules, but exists in a lattice.
As a result, treat the 2 after Fe as if it was in front of the symbol;
i.e 2Fe
as this is the form it would take is seperated from the oxgen. Each Fe has a molar mass of 55.9g mol·¹, but there are twice as many moles. So you double the moles, instead of the molar mass.
Make sense?
Mathmatically, it makes little difference which way around you do it, since you are still doubling it. But this is symbolic of the way metal behaves as ions and as an element.
:? umm, intresting, what about when you are calculating the mass of Fe2O3 in a 1.39g?
in the case why don't you only use one molar mass of Fe instead of 2?
thanks in advance.
Because the total mass of the substance accounts for iron and oxygen in a ratio of 2:3.
So, in every mole of Fe2O3, there are 2 moles of Fe.
tyvm![]()
Fe2O3 + 3CO ---> 2Fe + 3CO2
The mole ratio for Fe2O3 is 1, 3CO is 3, 2Fe is 2 and 3CO2 is 2.
Why is the last one 2 and not 3?
thanks in advance.
I'm not sure what you mean. Unless you mean the ration of O:C - there are two O per single C.
Fe2O3 + 3CO ---> 2Fe + 3CO2
On my text book it says there is one mole of Fe2O3 for every 3 mole of CO and so on.... but says that there is 2 mol for CO2, i think it is a printing error, can anyone comfirm?
thanks in advance. :-D
It sounds like a printing error.
« Any tips please? | Humble Origins » |