2. A compound with an empirical formula of C2H3Br2 has a molar mass of 373.69 g/mol. What is
the molecular formula?
(The molar mass of C2H3Br2 = 186.85 g/mol)
E) C4H6Br4
А) C2H3Br2
B) CHBr
C) C6H9Br6 D) C4H6Br2
(0.5 Points)
A

Answers

Answer 1
C4H6Br4
Because 373.69 is the mass we want, but the empirical mass so far is 186.85

If you notice, 186.85 x 2 is 373.69
So now we times every number in
C2H3Br2 by 2
So c4h6Br4
Answer 2

Empirical is the simplest whole number ratio of elements in a compound. Molecular formula is n times the empirical formula, where n is integers 1,2,... The molecular formula is  C[tex]_4[/tex]H[tex]_6[/tex]Br[tex]_4[/tex]. The correct option is option E.

What is empirical formula?

Empirical formula gives the proportion of the element that is present in a compound. It does not give information about the actual number or the arrangements of the atoms.

n=empirical formula mass÷ molecular formula mass

n=1,2,3....

empirical formula mass = 373.69 g/mol

molecular formula mass=  186.85 g/mol

substituting all the given values in the above formula, we get

n=  186.85 g/mol÷  373.69 g/mol

   = 2

Multiply C[tex]_2[/tex]H[tex]_3[/tex]Br[tex]_2[/tex] by 2, which comes out to be C[tex]_4[/tex]H[tex]_6[/tex]Br[tex]_4[/tex].

Therefore, the molecular formula is C[tex]_4[/tex]H[tex]_6[/tex]Br[tex]_4[/tex]. The correct option is option E.

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Answers

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Answers

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Answers

Answer:

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Answers

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Answers

1. The mass percentage of each element in aluminum oxide Al₂O are:

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ii. The mass percent of O in Al₂O₃ is 47.1%

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Therefore,

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