posted by .

I am trying to calculate my percent error for an experiment conducted. I need to compare my average density calculated to the accepted average density but I am having trouble doing this because of the conversions.

My average density was 0.12 g/mm^3. The accepted value is for iron which is 7.8 X 10^3 kg/m^3.

So to get 0.12 g to kg I multiplied by 1000 and got 120.

Then I subtracted 120 from 7.80 X 10^3.

Is this correct so far??

You got it backwards:

0.12kg is 120g
0.12g = 0.00012kg

.12g * 1kg/1000g = .00012kg

Oops. Forgot to include the volume:

.12g/mm^3 * 1kg/1000g * 1m^3/(1000mm)^3 = 120000kg/m^3

Oh ok so now to find the percent error do I subtract 120,000 from 7800 and then multiply by 100?

well, if the correct value is 10, and you get 12, then the percent error is (12-10)/10 * 100 = 20%

Yours is figured the same way, but the values are so different, I wonder whether there is something wrong here.

Yes my data is probably wrong I'm just not sure where I messed up

## Similar Questions

1. ### Science

What is the mass of 15mL sample of mercury?
2. ### chemistry

In a laboratory experiment, the density of a concentrated sugar solution was determined by measuring the volume of the solution and corresponding mass. Three students each made a set of measurements using a different balance and used …
3. ### Chemistry

The mass of the earth is known to be about 5.98 X10^24kg assume that it is approx. spherical with an average diameter of 1.276X10^7 METERS. from these data calculate the average density of the earth(g/cm^3) could the earth be made …
4. ### chemistry - density and error (check)

Calculate the density of earth and the density error. i got the answer 5.57*10^3k +/- 9.60 *10^7kg/m is that the correct answer?

An experiment was done where a tiny steel ball was measured with a micrometer. I know I already posted a question about this before and was given a respond but after trying to fix my answers I am still lost. If someone could please …

An experiment was done where a tiny steel ball was measured with a micrometer. I know I already posted a question about this before and was given a respond but after trying to fix my answers I am still lost. If someone could please …