Respuesta :
Empirical formula of compound is XF3
Compound consist of 65% F
In 100g of compound there is 65 g of F
= 65 / 19 moles of Fluorine = 3.421 moles
So moles of X = 3.421 / 3 = 1.140 moles
And in 100 g X consist of 35 g
So the molar mass of X = 35 / 1.140 = 30.71 g = 31 approximately
And it is the mass of phosphorus
So the empirical formula for the compound is PX3
The atomic mass of X is 30.69 u.
Further Explanation
Mass percent is the ratio of an element's mass to the mass of the compound. It is mathematically expressed as:
[tex]\% \ mass \ of \ A \ = \frac{mass \ of \ element \ A}{mass \ of \ compound} \times 100[/tex]
In the problem, the mass percent of F is given and can be used to determine the atomic mass of element X. To do this, just set up the mass percent equation for F as shown below:
[tex]mass \ \% \ of \ F = \frac{3 \times 18.998 \ u}{ X + 56.994 u}\\\\mass \% \ of \ F = \frac{56.994 \ u}{X + 56.994 \ u} = 65\%\\[/tex]
Solving for X,
[tex]0.65 (X + 56.994) = 56.994\\0.65X \ + \ 37.0461 = 56.994\\0.65X = 19.9479\\\\\boxed {\boxed {X = 30.69 \ u}}[/tex]
To check if the atomic mass of X is correct, use this value to calculate the %F. It is is equal to 65%, then the calculated value for X is correct.
[tex]\% F \ = \frac{3 \times 18.998 u}{30.69u \ + 56.994 u} \times 100\\\\\%F \ = \frac{56.994u}{87.684u} \times 100\\\\\\boxed {\%F = 64.999\%}[/tex]
Since substituting 30.69 as the atomic mass of X gives a mass % of 65 for F, then the calculated atomic mass is correct.
Learn More
- Learn more about empirical formula brainly.com/question/8516072
- Learn more about chemical formula brainly.com/question/4697698
- Learn more about percent composition https://brainly.com/question/11042307
Keywords: mass percent, empirical formula
