What is the empirical formula of a species that is $7.19 %$ $7.19%$ by mass with respect to phosphorus, and $92.81 %$ $92.81%$ by mass with respect to bromine? What is the molecular formula is the molecular mass is $431 \cdot g \cdot m o l^{- 1}$ $431gmol^-1$ ?

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What is the empirical formula of a species that is $7.19 %$ $7.19%$ by mass with respect to phosphorus, and $92.81 %$ $92.81%$ by mass with respect to bromine? What is the molecular formula is the molecular mass is $431 \cdot g \cdot m o l^{- 1}$ $431gmol^-1$ ?

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Answer 1 · 2017-07-23T06:51:58

I think you have quoted a question from this [site.](http://bergen.edu/Portals/0/Docs/Student%20Services/Tutor%20Resources/Chemistry/CHM%20100%20%20140%20-%20Empirical%20and%20Molecular%20Formulas%20%282%29.pdf)

Explanation:

And this gives (i) microanalysis of $P (7.19 %)$ $P(7.19%)$ , and $B r (92.81 %)$ $Br(92.81%)$ . You are lucky that I bothered to search for the question, as I am no computer buff, and these data should have been included with the question.......

For these data we determine the empirical formula......by assuming an $100 \cdot g$ $100*g$ mass of compound.

$Moles of phosphorus$ $"Moles of phosphorus"$ $=$ $=$ $\frac{7.19 \cdot g}{31 \cdot g \cdot m o l^{- 1}} = 0.232 \cdot m o l$ $(7.19*g)/(31*g*mol^-1)=0.232*mol$ .

$Moles of bromine$ $"Moles of bromine"$ $=$ $=$ $\frac{92.81 \cdot g}{79.9 \cdot g \cdot m o l^{- 1}} = 1.161 \cdot m o l$ $(92.81*g)/(79.9*g*mol^-1)=1.161*mol$ .

In each case I divided the elemental mass thru by the molar mass of the ELEMENT.

We divide thru by the lowest molar quantity (that of $P$ $P$ ), to get an empirical formula of $P : \frac{0.232}{0.232} = 1$ $P:(0.232)/(0.232)=1$ and $B r : \frac{1.161}{0.232} = 5$ $Br:(1.161)/(0.232)=5$ .. to give an empirical formula, the simplest whole number ratio of constituent elements in a species, of $P B r_{5}$ $PBr_5$ .

Now, we know that.........

${empirical formula} \times n = molecular formula$ ${"empirical formula"}xxn="molecular formula"$ , and from (ii) the molecular mass, we determine that.....

${31.00 + 5 \times 79.9} \cdot g \cdot m o l^{- 1} \times n = 431 \cdot g \cdot m o l^{- 1}$ ${31.00+5xx79.9}*g*mol^-1xxn=431*g*mol^-1$

Clearly, $n = 1$ $n=1$ , and so.........

$empirical formula = molecular formula = P B r_{5}$ $"empirical formula" = "molecular formula" = PBr_5$

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What is the empirical formula of a species that is $7.19 %$ $7.19%$ by mass with respect to phosphorus, and $92.81 %$ $92.81%$ by mass with respect to bromine? What is the molecular formula is the molecular mass is $431 \cdot g \cdot m o l^{- 1}$ $431gmol^-1$ ?

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What is the empirical formula of a species that is 7.19%7.19%7.19% by mass with respect to phosphorus, and 92.81%92.81%92.81% by mass with respect to bromine? What is the molecular formula is the molecular mass is 431*g*mol^-1431⋅g⋅mol−1431*g*mol^-1?

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What is the empirical formula of a species that is $7.19 %$ $7.19%$ by mass with respect to phosphorus, and $92.81 %$ $92.81%$ by mass with respect to bromine? What is the molecular formula is the molecular mass is $431 \cdot g \cdot m o l^{- 1}$ $431gmol^-1$ ?