You and your lab partner are asked to determine the density of an aluminum bar. the mass is known accurately (to four significant figures). you use a simple metric ruler to determine its size and calculate the results in
a. your partner uses a precision micrometer and obtains the results in
b. method a (g/cm3 ) method b (g/cm3 ) 2.2 2.703 2.3 2.701 2.7 2.705 2.4 5.811 the accepted density of aluminum is 2.702g/cm3 .
a. calculate the average density for each method. should all the experimental results be included in your calculations? if not, justify and omissions.
b. calculate the percent error for each methodâs average value.
c. which methodâs average value is more precise? which method is more accurate?

The given data is tabulated below

Measurement Method A  Method B
--------------------  --------------  --------------
           1                  2.2           2.703
           2                 2.3            2.701
           3                 2.7             2.705
           4                 2.4             5.811

A graph of the data reveals that 5.811 from measurement B is an outlier, and should be rejected.

Part a: Calculate average density.
Method A:
Avg. density = (2.2+2.3+2.7+2.4)/4 = 2.4  g/cm³
Method B:
Avg. density = (2.703+2.701+2.705)/3 = 2.703 g/cm³

Average densities are
2.4 g/cm³ by method A
2.703 g/cm³ by method B, with the outlier removed.

Part b: Calculate the percent error.
The true value is 2.702 g/cm³
Method A:
%error = 100*(|2.4 - 2.702|/2.702) = 11.2%
Method B:
%error = 100*(|2.703 - 2.702|/2.702) = 0.04%

The percent error is
11.2 by method A,
0.04% by method B.

Part c: 
Method B is more accurate and more precise because its average value is closest to the true value and its %error is extremely small.

It won't lose or gain electrons if it's a covalent bond it shares the electrons with another atom or atoms. if it is an ionic bond it will lose the 3 electrons on the outer shell and become a positively charged ion.  credit goes to @sunshinebaby of answers.yahoo.com.