A thermal decomposition experiment

Reference no: EM13831356

Priya, you helped me with a problem yesterday but I am stuck again on the last part. I was pretty much on the right path when I asked you for help, and I thought the last questions would be a piece of cake, but my boyfriend are fighting over how to get the right answer on the last questions. I say the way is, 48/181.8419=.2639655657*100=26.39655657, and that would be 26.40% room for error, but he is saying that you would subtract 9.24-8.95/9.24=3.14... to ultimate get 3.14% room for error, and looking it up, I saw an answer around 93% or something like that....please help! Could you also explain how you cam to your conclusion?

A student performs a thermal decomposition experiment using 35.00 g of lithium iodate, LiIO₃. The weight of the crucible decreases by 8.95 g after heating. The theoretical mass of oxygen in LiIO₃ is 9.24 g. What is the percent error in the student's experiment?A student calculates a 75% percent error. What can be said about the experimental value obtained in the experiment?A student calculates a 0.54% percent error. What can be said about the experimental value obtained in the experiment?

Reference no: EM13831356

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A thermal decomposition experiment : A student performs a thermal decomposition experiment using 35.00 g of lithium iodate, LiIO3. The weight of the crucible decreases by 8.95 g after heating. The theoretical mass of oxygen in LiIO3 is 9.24 g. What is the percent error in the student's..

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