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Calculate the percentage composition of ...

Calculate the percentage composition of various elements in :
Sodium carbonate, `Na_2CO_3`
Given that the relative atomic masses of O = 16, Na = 23 and C = 12.

Text Solution

Verified by Experts

Relative molecular mass of `Na_2CO_3`
= 23 x 2 + 12 + 16 x 3
= 46 + 12 + 48 = 106 amu
Since 106 g of `Na_2CO_2` contains 46 g of sodium
`therefore` 100 g of `Na_2CO_3` contains = `(46xx100)/106`of sodium
`=4600/106` =43.4 g of sodium
Similarly, 106 g `Na_2CO_3` contains 12 g of carbon
`therefore` 100 g of `Na_2CO_3` contains `=(12xx100)/106` of sodium
`=1200/106`=11.3 g of Carbon
Again, 106 g of `Na_2CO_3` contains 48 g of oxygen .
`therefore` 100 g of `Na_2CO_3` contains =`(48xx100)/106`
=`4800/106` = 45.3 g of Oxygen
In `Na_2CO_3:Na` = 43.4% , C=11.3% and O=45.3%
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