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0.250 g of an element, M, reacts with excess fluorine to produce 0.547 g of the...

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lak-86 | Student, Undergraduate | Salutatorian

Posted August 2, 2013 at 10:57 AM via web

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0.250 g of an element, M, reacts with excess fluorine to produce 0.547 g of the hexafluoride, MF6. What is the element?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted August 2, 2013 at 11:59 AM (Answer #1)

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First we have to do the chemical reaction of the unknown element ‘M’ with fluorine.

`M + F_2 -> MF_6`

Balancing it will make it:

`M + 3 F_2 -> MF_6`

We assume that all of the mass of the element M is consumed.

1. Get the mass of F reacted

`Mass MF_6 = mass M + mass F `

`0.547 g = 0.250 + mass F `

`Mass F = 0.297 grams F `

2. get the amount of moles of M via stoichiometry

`0.297 grams F * (1 mol e F)/(19.0 grams F) * (1 mol e M)/(6 mol es F)`

`= 0.002605 mol es M`

3. Identify the unknown element

`At omic mass = (mass of M)/(mol es of M) `

`At omic mass = 0.250/ 0.002605 `

Atomic mass = 95.95 = 96.0 g/mol = Molybdenum (Mo) -> answer

MoF6 -> Molybdenum hexafluoride

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