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Student A performed gravimetric analysis for sulfate in her unknown using the same procedures we did . Results of her three trials were 68.6%, 66.2% and 67.1% sulfate. Student B analyzed the same unknown his results were 66.7%, 66.6% and 66.5%. The unknown was sodium sulfate. Calculate a percent e rror fo r Student A and for Student B using as the accepted value the theoretical value for sulfate in sodium sulfate based on molar mass . [Y ou may use the Internet, a textbook, the Chemistry C ommunity Learning Center (CCLC) or any other resource for an explanation/description of percent error .] Which student , A or B, was more accurate? Which student was more precise? Explain your answers

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saadhussain514

Answer:

The answer is Sodium Sulfate = Na2SO4  

Explanation:

Molar mass of sulfate = 1 (S) + 4 (O) = 1 (32) + 4 (16) = 32 + 64 = 96  

Molar mass of sodium sulfate = 2 (23) + 96 = 46 + 96 = 142  

% of Sulfate = (96/142)*100 = 67.6%  

Percent mistake in Studen A,  

(I) % mistake = (67.6 - 68.6)/67.6 = 1.48  

(ii) % mistake = (67.6 - 66.2)/67.6 = 2.07  

(iii) % mistake = (67.6 - 67.1)/67.6 = 0.74  

For understudy B  

(I) % mistake = (67.6 - 66.7)/67.6 = 1.33  

(ii) % mistake = (67.6 - 66.6)/67.6 = 1.48  

(iii) % mistake = (67.6 - 66.5)/67.6 = 1.63  

Sutdent An is some how exact.  

Understudy B is exact however not precise.