Error and Percent Error

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NOTES:

actual = 500.0 ft

Trials: Error Percent error

503.5 ft - 500.0 ft = 3.5 / 500.0 ft * 100% = 0.7%

502.8 ft - 500.0 ft = 2.8 ft / 500.0 ft * 100% = 0.52%

497.4 ft - 500.0 ft = -2.6 ft /

error= measured - actual

value - value

percent error = error / actual value * 100 %

percent error = measured value -- actual value / actual value * 100%

1. Joshua uses his thermometer and finds the boiling point of ethyl alcohol to be 75o C. He looks in a reference book and finds that the actual boiling point of ethyl alcohol is 80oC. What is his percent error?

75 C - 80 C = 5 C

5 C / 80 = 0.0625

0.0625 * 100% = 6.25 %

Percent error = 6.25%

6. An object has a mass of 35.0 grams. On Anthony’s balance, it weighs 34.85 grams. What is the percent error of his balance?

34.85 - 35.00 = 0.15

0.15 / 35 = 0.0042

0.0042 * 100% = 0.42

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PERCENT ERROR PRACTICE PROBLEMS

Name: Emilia Lehnhoff

Date: 23/02/2021

Per: A

Percent Error Practice Problems

Show all work, including a formula and correct units after EVERY number on EVERY problem, or you will not receive credit for that problem

A student estimated a mass to be 250g, but upon carefully measuring it, found the actual mass to be 240g. What is the percent error?

Measured Value: 250g

Actual Value: 240g

250g - 240g = 10g

10g/250g = 0.0416

0.0416g * 100% = 4.16%

Percent Error: 4.16%

2. A student measured the temperature of boiling water and got an experimental reading of 97.5oC. What is the percent error?

Measured Value: 97.5oC.

Actual Value: 100oC

97.5oC - 100.0oC = 2.5oC

2.5oC/100oC = 0.025oC

0.025oC * 100% = 2.5%

Percent Value: 2.5%

3. An experimental measurement was taken of 10.4 mL and the actual measurement was 9.7 mL. What is the percent error?

Measured Value: 10.4mL

Actual Value: 9.7mL

10.4mL - 9.7 mL =0.7mL

0.7mL/9.7mL = 0.0721mL

0.0721mL * 100% = 7.21%

Percent Error: 7.21%

4. Experimental = 62.5 g and accepted value = 65.2 g. Solve for percent error.

Measured Value: 65.5g

Actual Value: 65.2g

65.5g - 65.2g = 0.3g

0.3g/65.2 = 0.0046g

0.0046g * 100% = 0.46%

Percent Error: 0.46%

5. Joe measured the volume of an object 5 times, and got results of: 34.5 mL, 34.9mL, 34.2 mL, 33.4 mL, and 35.9mL. What is his average experimental result?

34.5 mL + 34.9mL + 34.2 mL + 33.4 mL + 35.9mL = 172.9 mL

172.9 mL/5= 34.58mL

Average Value: 34.58mL

6. The actual volume of the object was 34.1 mL. What is the percent error of his average result?

Measured Value: 34.58 mL

Actual Value: 34.1 mL

34.58mL - 34.1 mL = 0.48mL

0.48mL / 34.1mL = 0.014 mL

0.014mL * 100% = 1.4%

Percent Error: 1.4%

7. A student estimated a mass to be 325g, but upon carefully measuring it, found the actual mass to be 342g. What is the percent error?

Measured Value: 325g

Actual Value: 342g

325g - 342g = 17g

17g/342g = 0.0497

0.0497 * 100% = 4.97%

Percent Error: 4.97%

8. Joshua uses his thermometer and finds he boiling of ethyl alcohol to be 75 C He looks in a reference book and find that the actual bling point ethyl alcohol is 80 C. What is the percent error?

Measured Value: 75 c

Actual Value: 80 C

75 C - 80 C= 5 C

5 C / 80 C = 0.0625 C

0.0625 C * 100% = 6.25

Percent Error: 6.25%

9. An object has a mass of 35.0 grams. On Anthony’s balance, it weighs 34.85 grams. What is the percent error of his balance?

Measured Value: 34.85

Actual Value: 35.0 g

34.85g - 35.0 g = 0.15g

0.15g / 35.0g = 0.0042 g

0.0042 * 100% = 0.42%

Percent error: 0.42%

10. The concentration determined for an unknown sample of hydrochloric acid by a student is 0.1355 M. According to the instructor’s information, the true molarity (M) of this solution is 0.1364 M. What is the percent error in this experiment?

Measured Value: 0.1355 M

Actual Value: 0.1364 M

0.1355 M - 0.1364 M = 0.0009 M

0.0009 M / 0.1364 M = 0.0065 M

0.0065 M* 100% = 0.65%

Percent error: 0.65%

11. The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted value of this melting point is 173.2 oC. What is the percent error in this reading?

Measured Value: 174.5C

Actual Value: 173.2C

174.5C -173.2 C = 1.3 C

1.3 C / 173.2 = 0.0075

0.0075 C * 100% = 0.75

Percent error: 0.75%

12. An object with an actual mass of exactly 0.54 g is given to 2 students. One student obtains a mass of 0.59 g for the object, while another says the mass is 0.49 g. Which of the students, if either has the greater percent error?

Measured Value: 0.59g

Actual Value: 0.54g

0.59g - 0.54g = 0.05g

0.05g /0.54g = 0.092g

0.092g * 100% = 9.2%

Percent error: 9.2%

Measured Value: 0.49g

Actual Value: 0.54g

0.49g - 0.54 g = 0.05g

0.05g/0.54g = 0.092g

0.092g * 100% = 9.2%

Percent error: 9.2%

Both of them have the same percent error.