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Alex777 [14]
3 years ago
7

What is the percentage of water in the following compound? Answer using three significant figures. Sodium carbonate decahydrate,

Na2CO3 • 10H2O % by mass H2O
Chemistry
2 answers:
Marrrta [24]3 years ago
4 0

Answer:

% of water = 62.94%

Explanation: up

To calculate the amount of water in

Na2CO3.10H2O

Calculating the molecular mass of

Na2CO3

Na2=2*23=46g/mol of Na

C= 1*12=12g/mol of C

O3= 3*16=48g/mol of O

Molecular mass of Na2CO3=46+12+48=106g/mol

Molecular mass of H2O=18g/mol

10.H2O= 10*18= 180g/mol

Na2CO3.10H2O=106+180=286g/mol

%of water = 180/286*100%=62.94%

Karo-lina-s [1.5K]3 years ago
3 0
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%

:)




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