I have to obtain an empirical formula from:

1 answer:

7 0

When finding empirical formulas decimals are fine so you wouldn't have to multiply all of them to turn into whole numbers. What you would do is divide each of those numbers by the smallest of those numbers and take the nearest whole number so if you had:

1.81 mol .6 mol and .3.61 mol you would divide each of those numbers by .6
and you would have 3, 1, and 6

for your problem you would divide them all by 1 and Carbon would be 2 (if your teacher normally has you round down at .5's then it would be 1) hydrogen would be 3 and oxygen would be 1.
C2H3O

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Data:
$V_{initial} = 590\:mL$
$T_{initial} = -55.0^0C$
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → $T_{initial} = 218.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 30.0^0C$
TK = TC + 273
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By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }$
Product of extremes equals product of means:
$218.0* V_{f} = 590*303.0$
$218.0 V_{f} = 178770$
$V_{f} = \frac{178770}{218.0}$
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