Question #05d76

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Question #05d76

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Answer 1 · 2016-12-16T03:24:47

The $empirical formula$ $"empirical formula"$ is $A s_{2} O_{5}$ $As_2O_5$ .

Explanation:

As with all these problems, we ASSUME that there are $100 \cdot g$ $100*g$ of unknown compound.

Given the elemental percentages, thus there are:

$Moles of oxygen$ $"Moles of oxygen"$ $=$ $=$ $\frac{34.8 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}} = 2.18 \cdot m o l$ $(34.8*g)/(15.999*g*mol^-1)=2.18*mol$ .

$Moles of arsenick$ $"Moles of arsenick"$ $=$ $=$ $\frac{65.2 \cdot g}{74.9 \cdot g \cdot m o l^{- 1}} = 0.871 \cdot m o l$ $(65.2*g)/(74.9*g*mol^-1)=0.871*mol$ .

We divide thru by the smallest molar quantity to get, $O : A s$ $O:As$ , $2.5 : 1.0$ $2.5:1.0$ .

But by definition, the $empirical formula$ $"empirical formula"$ is the simplest WHOLE number ratio that defines constituent atoms in a species, so we take this preliminary determination, and DOUBLE it to give WHOLE numbers:

$A s_{2} O_{5}$ $As_2O_5$ . Capisce?

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Question #05d76

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