Question

calculate the following % errors. show your numbers plugged into the equation along with your answer.

1. joshua uses his thermometer and finds the boiling point of ethyl alcohol to be 75.0°c. he looks in a reference book and finds that the actual boiling point of ethyl alcohol is 80.0°c. what is his percent error?
2. the density of water at 4°c is known to be 1.00 g/ml. kayla experimentally found the density of water to be 1.075 g/ml. what is her percent error?
3. an object has a mass of 35.0 grams. on anthonys balance, it weighs 34.85 grams. what is the percent error of his balance?
4. marina thinks the cost to rent scooters is $100.00. the actual price was $150.00. find the % error.
5. the handbook of chemistry and physics lists the density of a certain liq be 0.7988 g/ml. taylor experimentally finds this liquid to have a density of 0.7925 g/ml. the teacher allows up to +/- 0.500% error to make an \a\ on the lab. did she make an \a\? prove your answer.

Explanation:

Step1: Recall the formula for percent error

The formula for percent error is $\text{Percent Error}=\frac{\vert\text{Experimental Value}-\text{True Value}\vert}{\text{True Value}}\times100\%$

Step2: Solve problem 1

For Joshua:
Experimental Value = $75.0^{\circ}C$, True Value = $80.0^{\circ}C$
$\text{Percent Error}=\frac{\vert75.0 - 80.0\vert}{80.0}\times100\%=\frac{5.0}{80.0}\times100\% = 6.25\%$

Step3: Solve problem 2

For Kayla:
Experimental Value = $1.075\ g/mL$, True Value = $1.00\ g/mL$
$\text{Percent Error}=\frac{\vert1.075 - 1.00\vert}{1.00}\times100\%=\frac{0.075}{1.00}\times100\% = 7.5\%$

Step4: Solve problem 3

For Anthony:
Experimental Value = $34.85\ g$, True Value = $35.0\ g$
$\text{Percent Error}=\frac{\vert34.85 - 35.0\vert}{35.0}\times100\%=\frac{0.15}{35.0}\times100\%\approx0.43\%$

Step5: Solve problem 4

For Marina:
Experimental Value = $\$100.00$, True Value = $\$150.00$
$\text{Percent Error}=\frac{\vert100 - 150\vert}{150}\times100\%=\frac{50}{150}\times100\%\approx33.33\%$

Step6: Solve problem 5

For Taylor:
Experimental Value = $0.7925\ g/mL$, True Value = $0.7988\ g/mL$
$\text{Percent Error}=\frac{\vert0.7925 - 0.7988\vert}{0.7988}\times100\%=\frac{0.0063}{0.7988}\times100\%\approx0.79\%$
Since $0.79\%>0.500\%$, she did not make an "A".

Answer:

1. $6.25\%$
2. $7.5\%$
3. $\approx0.43\%$
4. $\approx33.33\%$
5. No, because her percent error is approximately $0.79\%$ which is greater than $0.500\%$.