A compound contains 43.4% sodium, 11.3% carbon and 45.3% oxygen.
Determine its empirical formula. (Atomic mass of $N a = 23 C = 12, O = 16$ ).

N a_{2} C O_{3}

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N a C O_{3}

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N a_{2} C O_{4}

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N a_{4} C O_{3}

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Solution

The correct option is A $N a_{2} C O_{3}$
Dividing their percentages by their atomic weights.
$S o d i u m = \frac{43.4}{23} = 1.88$
$C a r b o n = \frac{11.3}{12} = 0.941$
$O x y g e n = \frac{45.3}{16} = 2.83$
Dividing all these numbers with simple numbers
$S o d i u m = \frac{1.88}{0.94} = 2$
$C a r b o n = \frac{0.94}{0.94} = 1$
$O x y g e n = \frac{2.83}{0.94} = 3$
Their ratio $2 : 1 : 3$
Empirical formula = $N a_{2} C O_{3}$

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Similar questions

Q. A substance on analysis gave the following percentage composition:

N a = 43.4 %, C = 11.3 %

and

O = 45.3 %

. Calculate the empirical weight (Atomic weight in amu of

N a = 23, C = 12, O = 16

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A compound contains 43.4% sodium, 11.3% carbon and 45.3% oxygen.Determine its empirical formula. (Atomic mass of Na=23C=12, O=16).

The correct option is A Na2CO3Dividing their percentages by their atomic weights.Sodium=43.423=1.88Carbon=11.312=0.941Oxygen=45.316=2.83Dividing all these numbers with simple numbersSodium=1.880.94=2Carbon=0.940.94=1Oxygen=2.830.94=3Their ratio 2:1:3 Empirical formula = Na2CO3

A compound contains 43.4% sodium, 11.3% carbon and 45.3% oxygen.
Determine its empirical formula. (Atomic mass of $N a = 23 C = 12, O = 16$ ).