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Chemistry 9 Online
OpenStudy (anonymous):

C 70.55 % H 10.66 % O 18.80 % Enter the subscripts of the empirical formula. The next step is to determine the molecular formula from mass spectrum. The laboratory sent these results: formula weight was found between 168.24 and 173.24 Enter the subscripts of the molecular formula. If one of the subscripts is 1, you must enter it here -- although this is not normal

OpenStudy (abmon98):

Make sure the sum of percentages add up to 100%. Note % as mass (g) C H O 70.55g 10.66g 18.80g Divide Each element by their molar mass 70.55/12 10.66/1 18.80/16 5.879166 moles of C 10.66 moles of H 1.175 moles of O Divide by the smallest figure which is 1.175 5.879/1.175 10.66/1.175 1.175/1.175 5.00-C 9-H 1-O C5H9O-This is the smallest whole ratio Empirical Formula

OpenStudy (abmon98):

I think you should take the average value by adding 168.24 and 173.24 and divide by 2 we get 170.74 gram 170.74=85x, x is the factor multiplied by the empirical mass to obtain the molecular mass. x=2

OpenStudy (anonymous):

the second one have to be in this form please C H O

OpenStudy (abmon98):

Molecular formula- C10H18O2

OpenStudy (anonymous):

thank you so much please what is your real name and you are so kind

OpenStudy (anonymous):

please can you help with other question?

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