Sapling Learning Map dab macmilan learning A sample containing a mixture of SrCl
ID: 1045180 • Letter: S
Question
Sapling Learning Map dab macmilan learning A sample containing a mixture of SrCl2 6H20 (MW 266.62 g/mol) and CsCI (MW 168.36 g/mol) originally weighs 1.6240 g. Upon heating the sample to 320 °c, the waters of hydration are driven off SrCl2 6H20, leaving the anhydrous SrCl2. After cooling the sample in a desiccator, it has a mass of 1.3750 g. Calculate the weight percent of Sr, Cs, and Cl in the original sample. °C ? 32 SrCl;"6H2O(s) SrCl2(s) + 6H20g) Number wt % Sr Number wt % Cs Number wt % CI =Explanation / Answer
Ans. Mass of water lost = Original sample mass – Dried mass
= 1.6240 g – 1.3750 g = 0.2490 g
Moles of H2O lost = Mass / MW = 0.2490 g / (18.01528 g/ mol) = 0.013822 mol
# 1 mol SrCl2 consists of 6 mol H2O.
So,
Moles of SrCl2 = (1/6) x Moles of H2O = (1/6) x 0.013822 mol = 0.002304 mol
# Mass of Sr = 0.002304 mol x (87.62 g/ mol) = 0.2018 g
Now,
% Sr = (Mass of Sr/ Original sample mass) x 100
= (0.2018 g / 1.6240) x 100
= 12.4287 %
# Mass of SrCl2 = 0.002304 mol x (158.5254 g/ mol) = 0.3652 g
# Mass of CsCl = Dried mass – Mass of SrCl2 = 1.3750 g – 0.3652 g = 1.0098 g
Moles of CsCl = 1.0098 g / (168.35813g/ mol) = 0.005998 mol
Since 1 mol CsCl consists of 1 mol Cs, the moles of Cs is equal to the moles of CsCl.
So,
Mass of Cs = 0.005998 mol x 132.90543 g/ mol = 0.7972 g
Now,
% Cs = (0.7972 g / 1.6240 g) x 100 = 49.0871 %
# 1 mol SrCl2 has 2 mol Cl whereas 1 mol CsCl has 1 mol Cl.
So,
Total moles of Cl in sample = 2 x Moles of SrCl2 + Moles of CsCl
= (2 x 0.002304 mol) + 0.005998 mol
= 0.010605 mol
# Mass of Cl = 0.010605 mol x (35.4527 g/ mol) = 0.3760 mol
Now,
% Cl = (0.3760 g / 1.6240 g) x 100 = 23.1518%
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