# A sample of a volatile liquid is vaporized completely in a 240 cm3 flask at 135°C and 1200 mm Hg...?

A sample of a volatile liquid is vaporized completely in a 240 cm3 flask at 135°C and 1200 mm Hg. The condensed vapor (liquid) weighs 0.837 grams. What is the molecular weight? The liquid contains 64.9% carbon, 13.5% hydrogen, and 21.7% oxygen. What is the molecular formula?

V = 0.240 L

p = 1200 / 760 = 1.58 atm

T = 135 + 273 = 408 K

n = pV / RT = 0.240 x 1.58 / 0.0821 x 408 = 0.0113

0.0113 ( moles ) : 0.837 g = 1 mole : x g

x = 74.1 g / mol This is the molecular weight

We consider 100 g of this compound . There are

64.9 g of C . 64.9 / 12.011 = 5.4

13.5 g of H. 13.5 / 1.00794 = 13.4

21.7 g of O . 21.7 / 15.9994 = 1.36

we divide for the smallest number

5.4 / 1.36 = 4

13.4 / 1.36 = 9.85

1.36 / 1.36 = 1

The empirical formula would be C4H10 O ( mass = 74 g/mol)

The molecular weight of the empirical formula is the same so the molecular formula is C4H10 O

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**Answer:**V = 0.240 L

p = 1200 / 760 = 1.58 atm

T = 135 + 273 = 408 K

n = pV / RT = 0.240 x 1.58 / 0.0821 x 408 = 0.0113

0.0113 ( moles ) : 0.837 g = 1 mole : x g

x = 74.1 g / mol This is the molecular weight

We consider 100 g of this compound . There are

64.9 g of C . 64.9 / 12.011 = 5.4

13.5 g of H. 13.5 / 1.00794 = 13.4

21.7 g of O . 21.7 / 15.9994 = 1.36

we divide for the smallest number

5.4 / 1.36 = 4

13.4 / 1.36 = 9.85

1.36 / 1.36 = 1

The empirical formula would be C4H10 O ( mass = 74 g/mol)

The molecular weight of the empirical formula is the same so the molecular formula is C4H10 O

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