A3.87mg sample of an organic compound gave 5.80mg co² and 1.58mg h²o on combustion. deduce the empirical formula of the compound.

C₃H₄.

Explanation:

The no. of moles of CO₂ produced = mass/molar mass = 5.80 mg/44.0 g/mol = 0.1318 mmol.

Which is corresponding to 0.1318 mmol of C.

The no. of moles of H₂O produced = mass/molar mass = 1.58 mg/18.0 g/mol = 0.08778 mmol.

which corresponds to (0.08778 x 2) = 0.1756 mmoles of H.

The ratio of the number of moles of hydrogen to carbon in the composition of the compound will be 0.1756/0.1318 = 1.332 = 1 + 1/3 = 4/3.

PRESSURE: atmospheres or mm Hg; 1 atm = 760 mm Hg

TEMPERATURE: Kelvin, K, which is o

C + 273

STP: Standard Temperature and Pressure: 273 K and 1 atm (or 760 mm Hg)

BOYLE'S LAW (temperature is constant): PV = constant

This is an inverse relationship: if one variable increases the other must

decrease.

CHARLES' LAW (pressure is constant): V = constant x T

This is a direct relationship: if one variable increases so does the other.

GAY-LUSSAC'S LAW (volume is constant): P = constant x T

This is a direct relationship: if one variable increases so does the other.

IDEAL GAS LAW: PV = nRT, where

P = pressure of the gas

Explanation: