Question

find the empirical formula and molecular formula for the compound which contains 51.28%, 9.40%, 27.35%, and 11.97% of carbon,hydrogen, oxygen and nitrogen respectively. molar mass of the compound is 234g/mol.
Atomic mass: C= 12 H=1 O=16. N=14​

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Mass of Carbon = 51.28 g

✭ Mass of Hydrogen = 9.40 g

✭ Mass of Oxygen = 27.35 g

✭ Mass of Nitrogen = 11.97 g

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ The empirical formula and the Molecular formula

$\displaystyle\large\underline{\sf\gray{Solution}}$

First we have to find the simplest whole number ratio of all the elements and that has been provided as a table

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

$\boxed{\begin{array}{c|c|c|c}\sf Element&\sf Mass&\sf No \ of \ Moles &\sf Whole \ No \ Ratios\\\frac{\qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\\\sf C&\sf 51.28 &\sf \dfrac{51.28}{12} = 4.27 & \sf\dfrac{4.27}{0.86} = 5\\\\ \\\sf H &\sf 9.4&\sf \dfrac{9.4}{1} = 9.4&\sf \dfrac{9.4}{0.86} = 11 \\\\\\\sf O&\sf 27.35 &\sf \dfrac{27.35}{14} = 1.71&\sf \dfrac{1.71}{0.86} = 2 \\\\\\\sf N&\sf 11.97&\sf \dfrac{11.97}{14} = 0.86 &\sf \dfrac{0.86}{0.86} = 1\end{array}}$

Hence the empirical formula of the element is C₅H₁₁NO₂

Now we have to calculate the molecular mass of the element,

➝ $\displaystyle\sf Molecular \ mass = 5 \times 12 + 11 \times 1 + 14 + 2 \times 16$

➝ $\displaystyle\sf Molecular \ mass = 60 + 11 + 14 + 32$

➝ $\displaystyle\sf \green{Molecular \ mass = 117}$

We know that,

$\displaystyle\sf \dfrac{Molar \ mass}{Molecular \ mass} = n$

• Molar Mass = 244
• Molecular Mass = 117

Substituting the values,

»» $\displaystyle\sf \dfrac{234}{117} = n$

»» $\displaystyle\sf \orange{n = 2}$

And Finally,

➳ $\displaystyle\sf Molecular \ formula = n \times Empirical \ formula$

➳ $\displaystyle\sf Molecular \ formula = 2 \times (C_5H_{11}NO_2)$

➳ $\displaystyle\sf \pink{Molecular formula = C_{10}H_{22}N_2O_4}$

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