Question

solve the attachmentl​

Answer 1

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The smallest value we got is 2.01

The empirical formula becomes CH₂Cl

Molecular weight of Empirical formula = 12 + 2(1) + 35.5 = 49.5

We are already given that molecular weight as 99.

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Relation between Empirical formula and Molecular formula is given by ,

$\star\large{\boxed{ \sf{ \purple{Molecular\:Formula=Empirical\:formula\times\:n}}}}⋆$

n is some integer which is given by ,

$\star\large{\boxed{\purple{ \sf{ n = \dfrac{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}$

We know the values for calculating the value of n . So , by substituting we get ;

$\begin{gathered}\\ : \implies \sf \: n = \frac{99}{49.5} \\ \\ \\ : \implies \sf {\boxed {\underline{ \sf{n = 2}}}}\end{gathered}$

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Applying the first formulae , (i.e , relation between empirical formulae and Molecular formulae) we get ;

$\begin{gathered}\\ \implies \sf Molecular\:Formula=Empirical\:formula \times \: n \\ \\ \\ \implies \sf Molecular\:Formula=CH_2Cl \times 2 \\ \\ \\ \implies \sf \: Molecular\:Formula=C_2H_4Cl_2\end{gathered}$

Hence , The molecular formula of the given compound is C₂H₄Cl₂.

Answer 2

Answer:

Calculate moles of each element - divide

% by atomic mass

C=19.998/12.011 = 1.665

H= 3.331/ 1.008 = 3.305

N= 23.320 / 14.007 = 1.665

O= 53.302 / 15.999 = 3.332

Divide through by smallest

C=1

H=2

N=1

O = 2 ( small rounding is accepted)

Empirical formula = CH2NO2

Formula mass = 12.011 +1.008'2 + 14.007 + 15.999:2 = 60.03 g/formula unit

Formula units in 1 molecule = 120.07 /

60.03 = 2

Molecular formula = ( CH2NO2)2 = C2H4N204 (1,2-dinitroethane.)