Question

a sample of an unknown compound is vaporized at 110. °c. the gas produced has a volume of 1190. ml at a pressure of 1.00 atm, and it weighs 1.21 g. assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert units

Convert volume \(V = 1190\space mL\) to \(V=1.19\space L\), temperature \(T=(110 + 273.15)\space K=383.15\space K\), pressure \(P = 1.00\space atm\), and \(R = 0.0821\space L\cdot atm/(mol\cdot K)\)

Step2: Use ideal gas law \(PV=nRT\) to find the number of moles \(n\)

$$n=\frac{PV}{RT}$$

Substitute \(P = 1.00\space atm\), \(V = 1.19\space L\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 383.15\space K\)

$$n=\frac{1.00\times1.19}{0.0821\times383.15}$$

$$n=\frac{1.19}{31.456615}$$

$$n\approx0.0378\space mol$$

Step3: Calculate molar mass \(M\)

Molar mass \(M=\frac{m}{n}\), where \(m = 1.21\space g\) and \(n\approx0.0378\space mol\)

$$M=\frac{1.21}{0.0378}$$

$$M\approx32.0\space g/mol$$

Answer:

\(32.0\space g/mol\)