How would you find the molecular formula for a compound with molar mass 180 amu, that is composed of 40% carbon 6,67% hydrogen 53.3 oxygen?

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Answer 1 · 2016-01-03T15:19:12

$C_6H_12O_6$

We assume 100 g of compound.

Elemental composition is divided thru by the atomic mass of each element:

$C: (40*cancelg)/(12.011*cancelg*mol^-1)$ $=$ $3.33*mol$ .

$H: (6.67*cancelg)/(1.00794*cancelg*mol^-1)$ $=$ $6,62*mol$ .

$O: (53.3*cancelg)/(15.99*cancelg*mol^-1)$ $=$ $3.33*mol$ .

So the empirical formula is $CH_2O$ , after we divide thru by the lowest quotient.

Now we know that the molecular formula is always a whole number mulitple of the empirical formula:

i.e. $(12.011+2xx1.0074+15.99)_n*g*mol^-1$ $=$ $180*g*mol^-1.$

Thus $n$ $=$ $6$ , and the molecular formula is $C_6H_12O_6$ .