Answer:
[tex]C_6H_{15}O_6[/tex]
Explanation:
Hello,
In this case, one computes the molar mass for the given empirical formula as follows:
[tex]M_{empirical}=12*2+1*5+16*2=61g/mol[/tex]
In such a way, one computes how many times the molecular formula's molecular mass is contained into the empirical formula's molecular mass in order to determine the whole number relating them as shown below:
[tex]\frac{183.2g/mol}{61g/mol}=3[/tex]
Finally, the molecular formula is three times the empirical formula, hence:
[tex]C_6H_{15}O_6[/tex]
Best regards.
