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

Let us first calculate the accepted value of sulfate using the molar mass of sodium sulfate.

The molar mass of sodium sulfate is 2 Na + S + 4 O = 2 (22.99) + 32.07 + 4 (16)

Molar mass of Sodium sulfate = 142.05

Molar mass of sulfate = S + 4 (O) = 32.07 + 4 (16) = 96.07

% of sulfate in sodium sulfate = (Molar mass of SO4 / Molar mass of Na2SO4) x 100

% of sulfate = (96.07/142.05) x 100

% of sulfate = 67.6%

Theoretical value for % Sulfate is 67.6%

Percent error is calculated as  

Percent Error =

Let us calculate mean for student A.

Mean for student A = (68.6 + 66.2 + 67.1) / 3 = 201.9/3 = 67.3 %

% error for student A =

% error for student A = 0.44%

Mean for student B = (66.7 +66.6 + 66.5)/3 = 199.8 /3 = 66.6 %

% error for student B =  

% error for student B = 1.48%

Accuracy is the closeness of experimental values to the true value. This is defined in terms of absolute and percent errors.

Since percent error for student A is lower, we can say that student A was more accurate.

The precision of the data depends on the closeness of experimental values to each other. We can see that experimental values for student B were close to each other.

Therefore we can say that student B was more precise.

c.

explanation:

thats how its supposed to be