A 12.51 g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (definite proportions) how many grams of sodium does a 7.49 g sample of sodium bromide contain?
It will contain 2.04 g of sodium.
We can work out the empirical formula of sodium bromide by writing down the ratio by which they combine in grams:
Na : Br
= 22.34 : 77.66
( I got 77.66 by subtracting 22.34 from 100 as it is a % age)
Now we get the ratio in moles by dividing each by the mass of I mole .
(Ar[Na]=23 Ar[Br]=80)
= 22.34/23 : 77.66/80
= 0.97 : 0.97
= 1 : 1
So the empirical formula is NaBr.
The Mr of sodium bromide is therefore 23 + 80 = 103.
This means that 103g of sodium bromide must contain 23g of sodium.
So 1g of sodium bromide must contain 23/103 g of sodium.
So 9.13 g of sodium bromide must contain 23/103 x 9.13 = 2.04 g of sodium.
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