An unknown compound was found to have a percent composition as follows: 47.0% potassium, 14.5% carbon, and 38.5% oxygen. What is empirical formula? If the true molar mass of the compound is 166.22 g/mol, what is its molecular formula?

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Answer 1 · 2017-03-31T15:56:51

The $"empirical formula"$ is $KCO_2$

As with all these problems, we assume a $100*g$ mass of unknown compound, and then we work out the molar quantity:

$"Moles of potassium"=(47.0*g)/(39.10*g*mol^-1)=1.20*mol$

$"Moles of carbon"=(14.5*g)/(12.011*g*mol^-1)=1.21*mol$

$"Moles of oxygen"=(38.5*g)/(16.0*g*mol^-1)=2.41*mol$

We divide thru by the smallest molar quantity to give the empirical formula:

$KCO_2$ .

Now the molecular formula is always a whole number of the empirical formula:

i.e. $"molecular formula"=nxx"empirical formula"$

And thus with the molecular mass, we can solve for $n$ .

$166.2*g*mol^-1=nxx(39.1+12.011+2xx16.00)*g*mol^-1$

$166.2*g*mol^-1=nxx(83.1)*g*mol^-1$

Clearly, $n=2$ , and the $"molecular formula"=K_2C_2O_4$

The compound is LIKELY the potassium salt of oxalic acid, $K^(+)""^(-)O(O=)C-C(=O)O^(-)K^+$ , i.e. $"potassium oxalate."$