A sample of pure compound contains 1.15 g of sodium, 3.01 × 1022 atoms of carbon and 0.1 mol of oxygen atom. Its empirical formula is

Question

A sample of pure compound contains 1.15 g of sodium, 3.01 × 1022 atoms of carbon and 0.1 mol of oxygen atom. Its empirical formula is (A) Na2CO3 (B) NaCO2 (C) Na2CO (D) Na2CO2

Tardigrade Admin · Accepted Answer

Moles of sodium =(1.15 g/23 g mol-1)=0.05 mol Moles of carbon =(3.01×1022/6.02×1023)=0.05 mol Moles of oxygen = 0.1 Mole ratio, Na: C: O = 1: 1: 2 ∴ E.F. =NaCO2

Q. A sample of pure compound contains 1.15 g of sodium, $3.01 \times 1 0^{22}$ atoms of carbon and 0.1 mol of oxygen atom. Its empirical formula is

$N a_{2} C O_{3}$

$N a C O_{2}$

$N a_{2} CO$

$N a_{2} C O_{2}$

Moles of sodium $= \frac{1.15 g}{23 g m o l ^{- 1}} = 0.05 m o l$
Moles of carbon $= \frac{3.01 \times 1 0 ^{22}}{6.02 \times 1 0 ^{23}} = 0.05 m o l$
Moles of oxygen $= 0.1$
Mole ratio, $N a : C : O = 1 : 1 : 2$
$∴ E . F . = N a C O_{2}$

Q. A sample of pure compound contains 1.15 g of sodium, 3.01×1022 atoms of carbon and 0.1 mol of oxygen atom. Its empirical formula is

Na2​CO3​

NaCO2​

Na2​CO

Na2​CO2​