A gaseous hydrocarbon collected over water at a temperature of 21° C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. What is the molecular mass of the hydrocarbon?

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A gaseous hydrocarbon collected over water at a temperature of 21° C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. What is the molecular mass of the hydrocarbon?

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Answer 1 · 2015-12-24T19:48:47

$58.9 g/mol$ $"58.9 g/mol"$

Explanation:

Before focusing on anything else, use the vapor pressure of water at $21^{\circ} C$ $21^@"C"$ to determine the actual pressure of the gaseous hydrocarbon.

As you know, gases collected over water will also contain water vapor. Simply put, the volume of gas collected over water will contain molecules of hydrocarbon and molecules of water.

This means that you can use Dalton's Law of partial pressures to determine the exact pressure of the hydrocarbon. At ${21.0}^{\circ} C$ $21.0^@"C"$ , water vapor has a pressure of about $18.59 torr$ $"18.59 torr"$

http://www.endmemo.com/chem/vaporpressurewater.php

This means that you can write

$P_{mixture} = P_{water} + P_{hydrocarbon}$ $P_"mixture" = P_"water" + P_"hydrocarbon"$

$P_{hydrocarbon} = 753 torr - 18.59 torr = 734.41 torr$ $P_"hydrocarbon" = "753 torr" - "18.59 torr" = "734.41 torr"$

Your next step will be to use the ideal gas law equation to determine how many moles of this gaseous hydrocarbon you have in this sample.

$P V = n R T$ $color(blue)(PV = nRT)" "$ , where

$P$ $P$ - the pressure of the gas
$V$ $V$ - the volume it occupies
$n$ $n$ - the number of mole of gas
$R$ $R$ - the universal gas constant, usually given as $0.0821 \frac{atm \cdot L}{mol \cdot K}$ $0.0821("atm" * "L")/("mol" * "K")$
$T$ $T$ - the temperature of the gas, expressed in Kelvin

Plug in your values and solve for $n$ $n$ , but make sure that the units you have for pressure, temperature, and volume match those used in the expression of the universal gas constant

$P V = n R T \Rightarrow n = \frac{P V}{R T}$ $PV = nRT implies n = (PV)/(RT)$

$n = (734.41/760color(red)(cancel(color(black)("atm"))) * 48.1 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821( color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 21.0)color(red)(cancel(color(black)("K"))))$

$n = "0.0019247 moles"$

So, your hydrocarbon sample has a mass of $"0.1133 g"$ and it contains $0.0019247$ moles, which means that its molar mass, which tells you what the mass of one mole of a substance is, will be

$1 color(red)(cancel(color(black)("mole"))) * "0.1133 g"/(0.0019247color(red)(cancel(color(black)("moles")))) = "58.87 g"$

Rounded to three sig figs, the answer will be

$M_M = color(green)("58.9 g/mol")$

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A gaseous hydrocarbon collected over water at a temperature of 21° C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. What is the molecular mass of the hydrocarbon?

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