You have done an experiment and found that 1.017 g of an unknown liquid\'s vapor

Question

You have done an experiment and found that 1.017 g of an unknown liquid's vapor is contained in a flask that has a volume of 254.6 mL at 98.9 kPa of pressure and 102.2 °C. Another experiment was performed and the analysis showed the compound contained 57.15% C, 4.80% H, and 38.07% O. In the molecular formula, how many carbons are there? You have done an experiment and found that 1.017 g of an unknown liquid's vapor is contained in a flask that has a volume of 254.6 mL at 98.9 kPa of pressure and 102.2 °C. Another experiment was performed and the analysis showed the compound contained 57.15% C, 4.80% H, and 38.07% O. In the molecular formula, how many carbons are there?

Explanation / Answer

m = 1.017 g of vapor

V =254.6 mL = 0.2546 L

P = 98.9 kPa = 98.9/101.3 = 0.97630 atm

T = 102.2°C = 375.2 K

from here, use ideal gas law to calculate moles:

P V= nRT

n = PV/(RT) = 0.97630 *0.2546/ / (0.082*375.2 )

n = 0.00807913 mol of gas

now...

the molar mass -->

MM = mass of species / mol of species = 1.017 /0.00807913 = 125.87 g/mol

now...

calculate empirical formula

assume a basis of 100 g so:

mass of C = 57.15 g

mol of C = mass/MW = 57.15/12 = 4.7625

mass of H = 4.8g

mol of H = mass/MW = 4.8/2 = 2.4

mass of O = 38.07 g

mol of O = mass/MW = 38.07 /16 = 2.379375

ratio is clearly:

C4.7625 H 2.4 O 2.379375 ---> C2HO

so

MW of empirical = 12*2 + 1*1 + 1*16 =41g/mol

for overall formula

ratio = MW speices / MW empirical = 125.87 /41 = 3.07

so

C2HO x3 --> C6H3O3

short answer --> 6 carbons are present

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