Question

2. A chemist makes a standard by diluting 10.408 mg standard to 100 mL in a volumetric flask. The standard bottle reads: "For quantitative applications, use a value of 0.997 mg standard per mg of material on an as-is basis." The chemist makes a sample by dissolving ten tablets in a 1000 mL volumetric flask. When the analysis was completed, the following absorbances were obtained: Average of all standard injections = 5820178, Average of duplicate sample injections = 5600881. Given the formula below and the above information, what is the percent label claim of the 10 mg tablet under analysis?

mg/tablet = (Abs sample x wgt sample x purity std x volume sample) / (Abs std x volume std x number of tablets)

Answer 1

Here, Abs sample = 5600881; weight sample = 10.408 mg; purity std = 99.7 % = 0.997 mg; volume sample = 1000ml

Abs std = 5820178; volume std = 100 ml; number of tablets = 10

Putting all values in given equation

mg/tablet = (Abs sample x wgt sample x purity std x volume sample) / (Abs std x volume std x number of tablets)

= (5600881 x 10.408 mg x 0.997 mg x 1000 ml ) / ( 5820178 x 100 ml x 10)

= 1.00 mg/tablet

Percent label claim = mg/tablet *100 = 100 mg/ml