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marysya [2.9K]
3 years ago
15

Student A performed gravimetric analysis for sulfate in her unknown using the same procedures we did . Results of her three tria

ls 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
Chemistry
1 answer:
Andre45 [30]3 years ago
8 0

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 = \left | \frac{Experimental - Theoretical}{Theoretical} \right |

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 = \left | \frac{67.3 - 67.6}{67.6} \right |

% 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 = \left | \frac{66.6 - 67.6}{67.6} \right |  

% 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.


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