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3 years ago

If a molecule has an empirical formula of C2H2O and a molecular mass of 84.0 g/mol, what is the molecular formula?

Chemistry

2 answers:

Ksivusya [100]3 years ago

7 0

Answer:

$C_2 H_2 O \times 2=C_4 H_4 O_2$ is the molecular formula

Explanation:

Empirical formula is the simplest form of a Molecular formula.

For example

Molecular formula of glucose is $C_6 H_{12} O_6$

Its empirical formula is $C_1 H_2 O_1$

The subscript numbers of the molecular formula is divided by a common number n=6 here.

Another example:

Molecular formula of Octane is $C_8 H_{18}$ and Its empirical formula will be $C_4 H_9$

Subscripts divided by n=2.

To find n we make use of the formula

$\frac {(molecular mass )}{(empirical formula mass)}=n$

$= \frac {84.0g}{(C_2H_2O mass)}\\\\=\frac {84.0g}{(24+2+16)g}=\frac {84.0g}{42g}=2$

So, we see n=2 here

Empirical formula × n = molecular formula

$C_2 H_2 O \times 2=C_4 H_4 O_2$ is the molecular formula

(Answer)

Please note:

Molar mass is the mass of 1 mole of the substance and its unit is g/mol .

n is the number of moles by which the empirical formula is multiplied to get the molecular formula of the compound.

lord [1]3 years ago

6 0

Answer:

=C₄H₄O₂

Explanation:

Given the empirical formula of a molecule, the he the quotient of the molecular mas and and the empirical mass=constant.

84.0 g/mol/mass of(C₂H₂O)=constant

=84/(12×2+1×2×16)

=84/42

Therefore, the molecular formula is (C₂H₂O)₂=C₄H₄O₂

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$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

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$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

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