Percent of element in the original impure sample?

Question

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?

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Answer 1 · 2018-01-19T03:50:54

$0.11%$

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,

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

$0.11%$