Question

Percent of element in the original impure sample?

So the balance chemical equation is `Mn_2(CO_3)_5 -> Mn2O5 + 5CO2`. The mass of the impure sample of `Mn_2(CO_3)_5` is `28.222g` and we know that the reaction actually produced `3.787*10^-4 mol CO2`. How do I find the percent of `Mn_2(CO3)_5` that was in the original sample?

Answer 1

`0.11%`

Explanation:

From the reaction, we know that 5 moles of `CO_2` reacts with 1 mole of `Mn_2(CO_3)_5`.

No of moles of `CO_2` produced,

`= 3.787*10^(-4)`

So, from the above relation between `CO_2` and `Mn_2(CO_3)_5` we know that the no of moles of `Mn_2(CO_3)_5` are.

`=(3.787*10^(-4))/5`

`=0.7574*10^(-4)`

This is our pure sample present.
For finding the no of grams of this pure sample,

`0.7574*10^(-4) = (mass)/(molar* mass)`

`0.7574*10^(-4) = (mass)/409.9`

`mass = 301.4*10^(-4)gm`

For percentage purity,

`(301.4*10^(-4))/(28.222)*100`

`0.11%`