Part A. At present, automobile batteries are sealed. When lead storage batteries discharge, they produce hydrogen. Suppose the void volume in the battery is 100 mL at 1 atm of pressure and 25°C. What would be the pressure increase if 8×10−2 g H2 were produced by the discharge of the battery? atm
Part B. A certain compound containing only carbon and hydrogen was found to have a vapor density of 2.550 g/L at 100°C and 760 mm Hg. If the empirical formula of this compound is CH, what is the molecular formula of this compound?

Part A. At present, automobile batteries are sealed. When lead storage batteries discharge, they produce hydrogen. Suppose the void volume in the battery is 100 mL at 1 atm of pressure and 25°C. What would be the pressure increase if 8 × 10⁻² g H₂ were produced by the discharge of the battery?

We can find the pressure for this amount of Hydrogen using the ideal gas equation.

[tex]P.V=n.R.T\\P.V=\frac{m}{M} .R.T\\P=\frac{m.R.T}{M.V} =\frac{8 \times 10^{-2}g \times (0.082atm.L/mol.K) \times 298K }{(2g/mol) \times 0.1L} =10atm[/tex]

Part B. A certain compound containing only carbon and hydrogen was found to have a vapor density of 2.550 g/L at 100°C and 760 mmHg. If the empirical formula of this compound is CH, what is the molecular formula of this compound?

First, we have to look for the molar mass of the compound through the following expression:

[tex]P.M=\rho .R.T\\M = \frac{\rho .R.T}{P} =\frac{2.550g/L \times (0.08206atm.l/mol.K) \times 373.15K }{760mmHg} .\frac{760mmHg}{1atm} =78.1g/mol[/tex]

Now, we need the relation between the molar mass of the molecular formula and the molar mass of the empirical formula.

[tex]\frac{78.1g/mol}{13.0g/mol} =6.01 \approx 6[/tex]

Finnally, we multiply the empirical formula by this coefficient to get the molecular formula.

(CH) × 6 = C₆H₆