Question

A 35.40 gram hydrate of sodium carbonate, Na2CO3•nH2O, is heated to a constant mass. Its final weight is 30.2 g. What is formula for the hydrate?A. Na2CO3∙1H2OB. Na2CO3∙2H2OC. Na2CO3D. Na2CO3∙3H2O

Answer 1

First, we have to calculate the molecular weights of each molecule:

`\begin{gathered} Na_2CO_3\text{ : 23*2+12+16*3= 106 g/mol} \\ H_2O\text{ : 1*2+16= 18 g/mol} \end{gathered}`

Then, we have to calculate the number of grams of water. We can calculate them because the process of evaporation lets us know the water amount that was retired:

`g\text{ H}_2O\text{ = 35.40 g - 30.2 g=5.2 g H}_2O`

Then, we're gonna convert the grams of sodium carbonate alone (30.2 g) and the grams of water to moles:

`\begin{gathered} 30.2\text{ g Na}_2CO_3\text{ * }\frac{1\text{ mol}}{106\text{ g}}=\text{ 0.2849 mol Na}_2CO_3\text{ }\approx0.3\text{ mol Na}_2CO_3 \\ \\ 5.2\text{ g H}_2O\text{ * }\frac{1\text{ mol}}{18\text{ g}}=\text{ 0.288 mol H}_2O\text{ }\approx\text{ 0.3 mol H}_2O \end{gathered}`

It means that the mole relation is 1:1 approx, as it is the same amount for both. Then, the formula is going to be:

`Na_2CO_3\text{ . 1H}_2O`

It means that the answer is A.