Rounding molar ratios

Fiona Grant 1I
Posts: 30
Joined: Fri Apr 06, 2018 11:03 am

Rounding molar ratios

When finding the ratio of moles in order to determine an empirical formula, what are the rules for rounding? For example, if i were to divide by the smallest number while calculating the ratios and one of my results is 15.80, how do I know if I can round this up to the nearest whole number, or if I need to multiply all answers by some number to get actual whole numbers?

Natalie Noble 1G
Posts: 30
Joined: Thu Feb 01, 2018 3:02 am
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Re: Rounding molar ratios

General rounding rules are if you get 1.99 or .99 of anything you can just round up to the nearest whole number (1.99 becomes 2). However if it turns out to be a recognizable decimal such as .33 .66 .25 .5 etc you multiply it by whatever you need to get a whole number and you multiply the rest of your answers by the same number (1.5 * 2 becomes 3, multiple your other answers by 2 as well). I am not sure what you would do if the answer you get is 15.8 but I believe for the most part you would not get that because in most questions they give us the numbers are recognizable decimals. But again I am unsure about 15.8

AnthonyDis1A
Posts: 30
Joined: Fri Apr 06, 2018 11:05 am

Re: Rounding molar ratios

I think that if you get a weird decimal by the time you start dividing by the smallest mole amount, it's likely that the calculations before that are incorrect. There were a few problems where I tried rounding up .7 and .8, and got really strange numbers for my answers. My work before that point was usually wrong.

YeseniaGomez_1L
Posts: 32
Joined: Fri Apr 06, 2018 11:02 am
Been upvoted: 1 time

Re: Rounding molar ratios

When I practice these problems, I usually round to the nearest whole number when I get something that .90 or above, so far this has worked good for me and the person who replied before says rounding .70 and .8 may give you weird answers. I would suggest that if you are unsure the you should multiply until you are sure you can round to a whole number.