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### Rounding

Posted: Tue Oct 02, 2018 2:26 pm
When figuring out the empirical ratio, what happens if you divide by the smallest number, and it doesn't come to a clear solution like 1: 1.333 : 1 ? In the pre-assessment question number 19, the ratio was 1 : 2.38 : 1 ? How do you handle this?

### Re: Rounding

Posted: Tue Oct 02, 2018 2:37 pm
.38 is extremely close to .4 so I thought of 2.38 as 2.4 then to turn 2.4 into a whole number you multiply it by 5 so 2.4(5) is 12
Ratio is then 5:12:5 since 5(1:2.38:1)~ 5:12:5.

### Re: Rounding

Posted: Tue Oct 02, 2018 2:42 pm
I found it easy to turn the decimal into a fraction and seeing the number from that standpoint. For example, 1.33 is approximately 4/3 and in order to make that a whole number, you just need to multiply everything by 3!

### Re: Rounding

Posted: Tue Oct 02, 2018 3:45 pm
If it's something large like 2.38, or as long as it's not very near the whole number, I think you have to find another number to multiply it by. So if it's like 1.01 or 4.99 then it's fine to treat it as 1 and 5 respectively, but if it's 1.2 or 4.8 then you should definitely find another number to multiply it by!

### Re: Rounding

Posted: Tue Oct 02, 2018 7:20 pm
I like the idea of turning the decimal into a fraction, it helps me know exactly what I need to multiply by to get a whole number!

### Re: Rounding

Posted: Tue Oct 02, 2018 10:35 pm
What happens when you are dividing the given molar mass by the molar mass of the empirical formula in order to find the molecular formula and get a number like 2.18. Do i just round this to 2 and multiply the empirical formula by 2?

### Re: Rounding

Posted: Wed Oct 03, 2018 10:52 pm
Michelle Nwufo 3A wrote:What happens when you are dividing the given molar mass by the molar mass of the empirical formula in order to find the molecular formula and get a number like 2.18. Do i just round this to 2 and multiply the empirical formula by 2?

Is there a certain problem you are referring to so we could see as reference? And if you're referring to the given molar mass (from mass spectrometry) and are dividing by the total molar mass of the empirical formula then it should be very close to 2 and not 2.18 since the Empirical Formula is giving us the relative ratio of atoms.