What is the percent composition of water in #Na_2S • 9H_2O?#

Question

Jeremiah Adams · Accepted Answer

#%H_2O# #~=# #70%# #%H_2O="Mass of water"/"Mass of hydrate"=(9xx18.01*g*mol^-1)/(240.18*g*mol^-1)xx100%=??%#

Cameron Andrews · Answer

undefined To find the percent composition of water in Na₂S • 9H₂O, first calculate the molar mass of water and the total molar mass of the compound. Then divide the molar mass of water by the total molar mass of the compound and multiply by 100 to get the percent composition.

Molar mass of water (H₂O) = 2(1.008 g/mol) + 1(15.999 g/mol) = 18.015 g/mol

Total molar mass of Na₂S • 9H₂O:
Na: 2(22.99 g/mol) = 45.98 g/mol
S: 1(32.07 g/mol) = 32.07 g/mol
H₂O: 9(18.015 g/mol) = 162.135 g/mol

Total molar mass = 45.98 g/mol + 32.07 g/mol + 162.135 g/mol = 240.185 g/mol

Percent composition of water:
(162.135 g/mol ÷ 240.185 g/mol) × 100 = 67.5%

Therefore, the percent composition of water in Na₂S • 9H₂O is 67.5%.

Wyatt Adkins · Answer

undefined To find the percent composition of water in Na2S • 9H2O, first calculate the molar mass of water (H2O) and the molar mass of the entire compound (Na2S • 9H2O). Then, divide the molar mass of water by the molar mass of the compound and multiply by 100 to obtain the percent composition.