Analytical Chemistry HW: Ca2+ (mg) Estimation w/ 95% Confidence Level

[zeromaxxx]

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) + C₂O₄^2- (aq) + H₂O → CaC₂O₄H₂O (s)

Homework Equations

$\mu$(95%)= $\overline{x}$ ± ts/$\sqrt{n}$

$\overline{x}$ = average/mean
t = Student's t
Degrees of freedom = n-1
s = standard deviation

t = 4.303
DoF = 2
s = ?
$\overline{x}$ = ?

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.

 

[nate808]

Thank you for your question. To determine the original amount of Ca2+ in the ultra-strength tablet and its associated error at 95% confidence level, we can use the following formula:

\mu(95%)= \overline{x} ± ts/\sqrt{n}

Where:
\mu(95%) is the mean value at 95% confidence level
\overline{x} is the average or mean value
t is the Student's t value for 95% confidence level, which can be found in a t-table or calculated using the degrees of freedom (n-1)
s is the standard deviation
n is the number of samples or replicates (in this case, n=3)

Based on the information given, we can plug in the values we know into the formula:

0.9 ± 0.14 g = \overline{x} ± (4.303)(s)/\sqrt{3}

We can solve for \overline{x} by subtracting 4.303(s)/\sqrt{3} from both sides, giving us:

0.9 - 4.303(s)/\sqrt{3} ± 0.14 g = \overline{x}

Now, we need to solve for s. To do this, we need to use the information that the error represents ±1σ. This means that the standard deviation (s) is equal to the error divided by 1 (since 1σ = error). Therefore, s = 0.14 g.

Now, we can plug this value into the equation to get the mean value at 95% confidence level:

0.9 - 4.303(0.14 g)/\sqrt{3} ± 0.14 g = \overline{x}

0.9 ± 0.08 g = \overline{x}

Therefore, the original amount of Ca2+ in the ultra-strength tablet is 0.9 ± 0.08 g, or 900 ± 80 mg. This means that at 95% confidence level, we can say that the true value of Ca2+ falls within this range.

I hope this helps! Please let me know if you have any further questions. Good luck with your research!