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What is the empirical formula of a compound if it has 80g Ca and 70g Cl?

1 Answer

Aug 12, 2017

$C a C l$ $CaCl$ ....

Explanation:

We determine the number of moles of calcium and chlorine respectively....

$Moles of calcium = \frac{80 \cdot g}{40.1 \cdot g \cdot m o l^{- 1}} = 2.0 \cdot m o l$ $"Moles of calcium"=(80*g)/(40.1*g*mol^-1)=2.0*mol$

$Moles of chlorine = \frac{70 \cdot g}{35.45 \cdot g \cdot m o l^{- 1}} = 2.0 \cdot m o l$ $"Moles of chlorine"=(70*g)/(35.45*g*mol^-1)=2.0*mol$

We divide thru by the smallest molar quantity to get $C a C l$ $CaCl$ ; a $C a (+ I)$ $Ca(+I)$ salt that I don't think is known.

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