Question

What is the percentage of water in the following compound? Answer using three significant figures. Sodium carbonate decahydrate, Na2CO3 • 10H2O % by mass H2O

Answer 1

The trick for this problem is to understand atomic mass: the fact that different atoms have different masses. What we need to do is add up all the atomic masses of the compound and work out the ratio of mass of water to the mass of sodium carbonate. Atomic masses are often given for each atom in the periodic table, but you can look them up on google too.

You can do this by adding up individual atoms for each molecule, or you can shortcut and lookup the molar mass of the compound (i.e.the task already done for you). 

The molar mass of water is 18.01g/mole so for 10 moles of water we have a mass of 180.1g.


The molar mass of sodium carbonate is 106g/mole (google).

So the total mass of the sodium carbonate decahydrate compound is 180.1+106 = 286.1g, of which water would make up 180.1g, so the percentage of water is is 180.1/286.1 = 0.629, so we can round this to 63%

:)

Answer 2

The percentage by mass of H₂O in Na₂CO₃.10H₂O is 62.970%.

First, we will determine the mass of 1 mole of Na₂CO₃.10H₂O by adding the masses of the elements that form it.

$mNa_2CO_3.10H_2O = 2 mNa + 1mC + 13 mO + 20 mH\\= 2 (22.98 g) + 1(12.01 g)+ 13 (16.00 g) + 20 (1.01 g) = 286.17 g$

Then, we follow the same procedure to calculate the mass of H₂O in  1 mole of  Na₂CO₃.10H₂O.

$mH_2O = 20mH + 10mO = 20(1.01 g) + 10(16.00 g) = 180.20 g$

The percentage by mass of water in Na₂CO₃.10H₂O is:

$\% H_2O = \frac{mH_2O}{mNa_2CO_3.10H_2O} \times 100 \% = \frac{180.20 g}{286.17 g} \times 100 \% = 62.970%$

The percentage by mass of H₂O in Na₂CO₃.10H₂O is 62.970%.

You can learn more about percentage by mass here: https://brainly.com/question/12073721