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What is the empirical formula of vanadiu...

What is the empirical formula of vanadium oxide if 2.74 g of metal oxide contains 1.53g of metal?

A

`V_(2)O_(3)`

B

VO

C

`V_(2)O_(5)`

D

`V_(2)O_(7)`

Text Solution

Verified by Experts

The correct Answer is:
C
% of `V=(1.53)/(2.74)xx100=55.83`
`therefore % "of" O=44.17`
`{:("Element",%,"Atomic ratio","Simplest ratio"),(V,55.83,(55.83)/(52)=1.1,(1.1)/(1.1)=1),(O,44.17,(44.17)/(16)=2.76,(2.76)/(1.1)=2.5):}`
`V:O=2:5`
Thus, empirical formula =`V_(2)O_(5)`.
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Knowledge Check

  • What is the emprical formula of vanadium oxide , if 2.74g of the metal oxide contains 1.53g of metal ?

    A
    `V_(2)O_(3)`
    B
    `VO`
    C
    `V_(2)O_(5)`
    D
    `V_(2)O_(7)`

  • The atomic mass of a metal M is 56, then the empirical formulae of its oxide containing 70% metal

    A
    `MO`
    B
    `M_(2)O_(3)`
    C
    `M_(3)O_(2)`
    D
    `M_(3)O_(5)`

  • Metal X forms wo oxides. Formula of the first oxide is XO_(2) . The first oxide contains 50% of oxygen. If the second oxide contains 60% of oxygen, the formula of the second oxide is

    A
    `X_(2)O`
    B
    `XO_(3)`
    C
    `X_(2)O_(3)`
    D
    `X_(3)O_(2)`

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