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Quote:

$Na_2CO_3 + xH_2O$

2.558g of $Na_2CO_3$ loses 1.61g $H_2O$ upon heating at 125 degrees, leaving 0.948g $Na_2CO_3$ ?

The molar masses given are $Na_CO_3=106.0\,\, H_2O=18.02$

$Na_2CO_3\cdot xH_2O = Na_2CO_3 + xH_2O$

Note, you're starting from a hydrate going to the free molecules. For every molecule of Sodium Carbonate in the original hydrate, you get one mole of sodium carbonate (dry) + x moles of water. Start with the right side:

$0.948 \, g \, Na_2CO_3 \,\div\, \frac {106.0\,g}{mole} =8.94\times 10^{-3}\,moles\,of\,Na_2CO_3$

$2.558\,g\,of\,hydrate - 0.948\,g\,of\,Na_2CO_3 = 1.61\,g\,of\,H_2O$

$1.61\,g\,of\,H_2O \,\div\, \frac {18.02\,g}{mole}=8.93\times 10^{-2}\,moles\,of\,H_2O$

We take the number of moles of Na2CO3 and divide that into the number of moles of H2O:

$\frac {x}{1} = \frac {8.93\times 10^{-2}}{8.94 \times 10^{-3}} \approx 10$

We note that x will always be some small, integer. Thus, we can round off $9.9888143 \approx 10$