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zeromaxxx
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Homework Statement
CaCO3 acts as an antacid for temporary relief of heartburn/acid-reflux. Free calcium ions in solution can be analyzed gravimetrically after acid digestion and precipitation with oxalate in basic solution to form an insoluble calcium oxalate monohydrate salt that can be accurately weighed, CaC2O4∙H2O (s). What is the original amount of Ca2+ (mg) and error at 95% confidence level derived from an ultra-strength tablet if (1.3240 ± 0.0150)g of CaC2O4∙H2O (s) was weighed, where error represents ±1σ based on triplicate analysis (n=3)?
Ca2+ (aq) + C2O42- (aq) + H2O → CaC2O4H2O (s)
Homework Equations
[itex]\mu[/itex](95%)= [itex]\overline{x}[/itex] ± ts/[itex]\sqrt{n}[/itex]
[itex]\overline{x}[/itex] = average/mean
t = Student's t
Degrees of freedom = n-1
s = standard deviation
t = 4.303
DoF = 2
s = ?
[itex]\overline{x}[/itex] = ?
The Attempt at a Solution
I've calculated the amount of Ca2+ based on the amount of CaC2O4∙H2O using mole to mole ratios which turns out to be 0.9 ± 0.14 g or 900 ± 140 mg (please correct if I'm wrong). What I don't know how to approach is getting the 95% confidence level with the associated error. Any advise on how to solve this would be appreciated.