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Empirical formula

Is this correct?
Reply 2
No.

Work to more significant figures - you should only ever round at the very last step.

You'll get horrible decimals for the two values, so you just divide both of those decimals by the smaller or the two values, which will give you a 1:something ratio. Turn that ratio into a whole number ratio, i.e. 3:4 or 2:3. If it gives 4:5.001 that'll be close enough.
ImageUploadedByStudent Room1444557370.698875.jpg
Hey I'm doing the same thing right now! Do you know if I'm doing this right?
Original post by <Hannah_Banana>
ImageUploadedByStudent Room1444557370.698875.jpg
Hey I'm doing the same thing right now! Do you know if I'm doing this right?


Looks right to me. If you want to present it in a clearer way for yourself, you could do it as three columns of ratios:

Rb : Ag : I
7.42/85.5 : 37.48/107.9 : 55.10/126.9
0.08678... : 0.3474... : 0.4342...
1 : 4.003... : 5.003...
1 : 4 : 5

Therefore RbAg(4)I(5)

The way you've presented it is completely fine, and if it makes sense to do it how you've done it, please carry on. I find this way a bit more useful as the numbers are side by side and it's a lot more clear to me since empirical formula is made up of the ratios of each element in the compound.
Original post by GetOverHere
Looks right to me. If you want to present it in a clearer way for yourself, you could do it as three columns of ratios:

Rb : Ag : I
7.42/85.5 : 37.48/107.9 : 55.10/126.9
0.08678... : 0.3474... : 0.4342...
1 : 4.003... : 5.003...
1 : 4 : 5

Therefore RbAg(4)I(5)

The way you've presented it is completely fine, and if it makes sense to do it how you've done it, please carry on. I find this way a bit more useful as the numbers are side by side and it's a lot more clear to me since empirical formula is made up of the ratios of each element in the compound.


Thanks that's useful

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