Which among the following contain 43.4% of sodium?

sodium bicarbonate

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sodium nitrate

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sodium carbonate

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sodium chloride

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Solution

The correct option is C
sodium carbonate

Explanation for the correct option:
Formula used to find the % of an element in a compound:
$% o f a n e l e m e n t = \frac{W e i g h t o f t h e e l e m e n t i n t h e s u b s \tan c e}{M o l e c u l a r w e i g h t o f t h e s u b s \tan c e} \times 100$
Molecular weight of sodium carbonate (Na₂CO₃)= 106 u
Weight of sodium in sodium carbonate (Na₂CO₃)= 46 u
$% o f s o d i u m i n s o d i u m c a r b o n a t e = \frac{46}{106} \times 100 = 43.4 %$
Explanation for the incorrect option:
(i) Molecular weight of sodium bicarbonate (NaHCO₃)= 84 u
Weight of sodium in sodium bicarbonate (NaHCO₃)= 23 u
$% o f s o d i u m i n s o d i u m b i c a r b o n a t e = \frac{23}{84} \times 100 = 27.38 %$
(ii) Molecular weight of sodium nitrate (NaNO₃)= 85 u
Weight of sodium in sodium nitrate (NaNO₃)= 23 u
$% o f s o d i u m i n s o d i u m n i t r a t e = \frac{23}{85} \times 100 = 27.05 %$
(iv) Molecular weight of sodium chloride (NaCl)= 58.5 u
Weight of sodium in sodium chloride (NaCl)= 23 u
$% o f s o d i u m i n s o d i u m c h l o r i d e = \frac{23}{58.5} \times 100 = 39.31 %$
Hence option (c) is correct.

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Similar questions

A substance contains 43.4% sodium, 11.3% carbon and 43.3% oxygen. Calculate its empirical formula.

N a = 23 C = 12, O = 16

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