Question

Quantitative analysis of a compound shows that it contains $0.110$ mole of ${}^{\prime }{C}^{\prime }$, $0.055$ mole of ${}^{\prime }{N}^{\prime }$ and $0.165$ mole of ${}^{\prime }{0}^{\prime }$. Its molecular mass is about $270$. How many atoms of carbon are there in the empirical and molecular formula of the compound respectively?

• A. Empirical formula - 1 Molecular formula - 3
• B. Empirical formula - 2 Molecular formula - 2
• C. Empirical formula - 2 Molecular formula - 6
• D. Empirical formula - 3 Molecular formula - 2

Answer 1

Answer: (C)

Empirical formula - 2 Molecular formula - 6

We have,

Moles of $C=0.110$

moles of $N=0.055$

moles of $O=0.165$

to derive empirical formula, we will convert them in simple whole numbers by dividing them by least number of moles

so, moles of C = $\frac{0.110}{0.055}=2$

moles of N = $\frac{0.055}{0.055}=1$

moles of O = $\frac{0.165}{0.055}=3$

so, empirical formula becomes = $C_{2}N{O}_3$

mass of empirical formula = 86

n= $\frac{molecularformulamass}{empiricalformulamass}=\frac{270}{86}=3$

so, molecular formula becomes = $C_6{N}_3{O}_9$