The molar mass of a given gas can be obtained through the equation Pre-lab L09 M

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The molar mass of a given gas can be obtained through the equation

Pre-lab L09 Molar Mass of a Gas Please review the background knowledge posted on I College. Below is a brief recap. Answer the following questions. The molar mass of a given gat can be obtained through the equation, M_gas = m/n m is mass (in gram) and n is the number of mole From the ideal gas law, PV = nRT, P is the pressure of the gas, in atm V is the volume of the gas, in Liter(L) T is the temperature of &e; gas, in Kelvin(K) R is a constant, 0.0821 L-atm K^-1 mole^-1 Rearrange the equation to express n, the number of mole as, | n = PV/RT Replacing the n in the above molar mass equation, the molar mass can be expressed as, M_gas = m/n Thus, the molar mass of a gas can be determined by measuring the temperature, pressure, volume and mass of a sample gas. A compound has the empirical formula, CHCl. A 256-mL flask, at l00 degree C and 0.967atm, contains 0.800-g of the gaseous compound. Calculate the molar mats of this compound, and determine its molecular formula. Show your work. b. Put the values you determined in previous step into the equation, M_gas = mRt/PV to determine the molar mass of the gas. The empirical formal of the gas is CHCl. The molecular formula of the gas is (CHCl). Calculate the formula mass of the empirical formula. Use the ratio of the molar mass of the gas and the formula mass to determine the factor "n". Write out the molecular formula of the gas. 2. An unknown diatomic gas a density of 3.164_gL at standard temperature (273K) and standard pressure (1.00 atm). At standard temperature and pressure, one mole of any gas occupies 22.4 L of volume. What is the identity of this gas? Show your work to explain why.

Explanation / Answer

From ideal gas equation Mass molar= mass * RT/PV , R is gas constant

1. No of moles n = PV/RT = 0.987 * 0.256 / 0.0821* 373 = 0.00825

Molar mass = mass in gms * RT / PV = 0.800 / 0.00825 = 96.97 g/mol.

Empirical Formula of CHCL = 12 + 1+ 35.35 = 48.5 empirical Formula shows ratio of elements in a compound but not actual no of atoms in molecule.

96.97/ 48.5 = 2

2 * empirical Formula = molecular Formula = C2H2Cl2

1) a. Volume of gas in litres = 0.256 L

100 degrees + 273 = 373 K

0.967 atm

0.800 g

R = 0.0821 L-atm-K ^-1 mole ^-1

b. Molar mass = Mass * RT / PV = mass / n = 96.97 g/mol

c. Since n = 2 Molecular Formula =C2H2Cl2. Molecular Formula is no and kinds of atoms in a molecule.

2. Volume of gas = 22.4 litres.

Density = 3.164 g/L

Mass = density * volume = 3.164* 22.4 = 70.873 g since at standard temperature and pressure the total mad comes to be 71 and it's diatomic 71/2 is 35.5 which is mass and or mol. Wt of chlorine. Thus the unknown gas must be Cl2.

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