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Calculate Percentage Error?

Hey, I'm having trouble calculation percentage error.

The question is: "The balance has a maximum error of +-0.005 g in each reading. Calculate the percentage error in your mass of sodium hydrogencarbonate"

So far I have -

Mass of NaHCO3 - 2.10g
How many times it's been read - once so 1 x 0.005 = 0.005

Then I used the calculation
Percentage error = maximum error / measurement value x 100

0.005 / 2.10 x 100 = 0.238%? This doesn't seem right to me.
I also tried it the other way
2.10 / 0.005 x 100 = 4200% which is definitely wrong...

What am I doing wrong?!
Sorry you've not had any responses about this. :frown: Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses. :redface:

I'm going to quote in Tank Girl now so she can move your thread to the right place if it's needed. :h: :yy:

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Reply 2
Avatar for Sam1500
Sam1500
Original post by KingAuthor
Hey, I'm having trouble calculation percentage error.

The question is: "The balance has a maximum error of +-0.005 g in each reading. Calculate the percentage error in your mass of sodium hydrogencarbonate"

So far I have -

Mass of NaHCO3 - 2.10g
How many times it's been read - once so 1 x 0.005 = 0.005

Then I used the calculation
Percentage error = maximum error / measurement value x 100

0.005 / 2.10 x 100 = 0.238%? This doesn't seem right to me.
I also tried it the other way
2.10 / 0.005 x 100 = 4200% which is definitely wrong...

What am I doing wrong?!


Hi, did you manage to solve it?

I am having a similar problem.

Each balance reading has an uncertainty of ±5.00 mg.
Calculate the percentage error in the initial mass of ore used.
The initial pass of the ore had a mass of 6.26g.
Markscheme answers.
0.01/6.26 * 100 =0.160% Why isnt it 0.005/6.26*100?
Reply 3
Avatar for Zacken
Zacken
22
Original post by Sam1500
Hi, did you manage to solve it?

I am having a similar problem.

Each balance reading has an uncertainty of ±5.00 mg.
Calculate the percentage error in the initial mass of ore used.
The initial pass of the ore had a mass of 6.26g.
Markscheme answers.
0.01/6.26 * 100 =0.160% Why isnt it 0.005/6.26*100?


Was it measured twice?
Reply 4
Avatar for Sam1500
Sam1500
Original post by Zacken
Was it measured twice?


This is the original questions


A sample of strontium ore is known to contain strontium oxide, strontium carbonate and some inert impurities. To determine the mass of strontium carbonate present, a student weighed a sample of the solid ore and then heated it in a crucible for 5 minutes. The sample was allowed to cool and then reweighed. This heating, cooling and reweighing was carried out three times.

Mass of crucible/ g 9.85
Mass of crucible and ore sample / g 16.11
mass of crucible and sample after first heating 14.66
Mass of crucible and sample after second heating 14.58
Reply 5
Avatar for xell
xell
1
Yeah you use it twice, first time for the mass of the crucible, and then the ore sample. As you're working out the difference between those two values, you've used the mass balance twice.

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