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Q. A sample of pure compound contains 1.15 g of sodium, $3.01 \times 10^{22}$ atoms of carbon and 0.1 mol of oxygen atom. Its empirical formula is

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Solution:

Moles of sodium $=\frac{1.15\,g}{23\,g\,mol^{-1}}=0.05\,mol$
Moles of carbon $=\frac{3.01\times10^{22}}{6.02\times10^{23}}=0.05\,mol$
Moles of oxygen $= 0.1$
Mole ratio, $Na : C : O = 1 : 1 : 2$
$\therefore E.F. =NaCO_{2}$